factoring - anna kuczynska...factoring factoring is the reverse process of multiplication. factoring...
TRANSCRIPT
214
Factoring
Factoring is the reverse process of multiplication. Factoring polynomials in algebra has similar role as factoring numbers in arithmetic. Any number can be expressed as a product of prime numbers. For example, 6 = 2 β 3. Similarly, any polynomial can be expressed as a product of prime polynomials, which are polynomials that cannot be factored any further. For example, π₯π₯2 + 5π₯π₯ + 6 = (π₯π₯ + 2)(π₯π₯ + 3). Just as factoring numbers helps in simplifying or adding fractions, factoring polynomials is very useful in
simplifying or adding algebraic fractions. In addition, it helps identify zeros of polynomials, which in turn allows for solving higher degree polynomial equations.
In this chapter, we will examine the most commonly used factoring strategies with particular attention to special factoring. Then, we will apply these strategies in solving polynomial equations.
F.1 Greatest Common Factor and Factoring by Grouping
Prime Factors
When working with integers, we are often interested in their factors, particularly prime factors. Likewise, we might be interested in factors of polynomials.
Definition 1.1 To factor a polynomial means to write the polynomial as a product of βsimplerβ polynomials. For example,
5π₯π₯ + 10 = 5(π₯π₯ + 2), or π₯π₯2 β 9 = (π₯π₯ + 3)(π₯π₯ β 3).
In the above definition, βsimplerβ means polynomials of lower degrees or polynomials with coefficients that do not contain common factors other than 1 or β1. If possible, we would like to see the polynomial factors, other than monomials, having integral coefficients and a positive leading term.
When is a polynomial factorization complete?
In the case of natural numbers, the complete factorization means a factorization into prime numbers, which are numbers divisible only by their own selves and 1. We would expect that similar situation is possible for polynomials. So, which polynomials should we consider as prime?
Observe that a polynomial such as β4π₯π₯ + 12 can be written as a product in many different ways, for instance
β(4π₯π₯ + 12), 2(β2π₯π₯ + 6), 4(βπ₯π₯ + 3), β4(π₯π₯ β 3), β12 οΏ½13π₯π₯ + 1οΏ½, etc.
Since the terms of 4π₯π₯ + 12 and β2π₯π₯ + 6 still contain common factors different than 1 or β1, these polynomials are not considered to be factored completely, which means that they should not be called prime. The next two factorizations, 4(βπ₯π₯ + 3) and β4(π₯π₯ β 3) are both complete, so both polynomials βπ₯π₯ + 3 and π₯π₯ β 3 should be considered as prime. But what about the last factorization, β12 οΏ½1
3π₯π₯ + 1οΏ½? Since the remaining binomial 1
3π₯π₯ + 1
does not have integral coefficients, such a factorization is not always desirable.
215
Here are some examples of prime polynomials:
any monomials such as β2π₯π₯2, ππππ2, or 13π₯π₯π₯π₯;
any linear polynomials with integral coefficients that have no common factors other than 1 or β1, such as π₯π₯ β 1 or 2π₯π₯ + 5;
some quadratic polynomials with integral coefficients that cannot be factored into any lower degree polynomials with integral coefficients, such as π₯π₯2 + 1 or π₯π₯2 + π₯π₯ + 1.
For the purposes of this course, we will assume the following definition of a prime polynomial.
Definition 1.2 A polynomial with integral coefficients is called prime if one of the following conditions is true
- it is a monomial, or - the only common factors of its terms are ππ or βππ and it cannot be factored into any
lower degree polynomials with integral coefficients.
Definition 1.3 A factorization of a polynomial with integral coefficients is complete if all of its factors are prime.
Here is an example of a polynomial factored completely:
β6π₯π₯3 β 10π₯π₯2 + 4π₯π₯ = β2π₯π₯(3π₯π₯ β 1)(π₯π₯ + 2)
In the next few sections, we will study several factoring strategies that will be helpful in finding complete factorizations of various polynomials.
Greatest Common Factor
The first strategy of factoring is to factor out the greatest common factor (GCF).
Definition 1.4 The greatest common factor (GCF) of two or more terms is the largest expression that is a factor of all these terms.
In the above definition, the βlargest expressionβ refers to the expression with the most factors, disregarding their signs.
To find the greatest common factor, we take the product of the least powers of each type of common factor out of all the terms. For example, suppose we wish to find the GCF of the terms
6π₯π₯2π₯π₯3, β18π₯π₯5π₯π₯, and 24π₯π₯4π₯π₯2.
First, we look for the GCF of 6, 18, and 24, which is 6. Then, we take the lowest power out of π₯π₯2, π₯π₯5, and π₯π₯4, which is π₯π₯2. Finally, we take the lowest power out of π₯π₯3, π₯π₯, and π₯π₯2, which is π₯π₯. Therefore,
GCF(6π₯π₯2π₯π₯3 , β 18π₯π₯5π₯π₯, 24π₯π₯4π₯π₯2) = 6π₯π₯2π₯π₯
This GCF can be used to factor the polynomial 6π₯π₯2π₯π₯3 β 18π₯π₯5π₯π₯ + 24π₯π₯4π₯π₯2 by first seeing it as
6π₯π₯2π₯π₯ β π₯π₯2 β 6π₯π₯2π₯π₯ β 3π₯π₯3 + 6π₯π₯2π₯π₯ β 4π₯π₯2π₯π₯,
216
and then, using the reverse distributing property, βpullingβ the 6π₯π₯2π₯π₯ out of the bracket to obtain
6π₯π₯2π₯π₯(π₯π₯2 β 3π₯π₯3 + 4π₯π₯2π₯π₯).
Note 1: Notice that since 1 and β1 are factors of any expression, the GCF is defined up to the sign. Usually, we choose the positive GCF, but sometimes it may be convenient to choose the negative GCF. For example, we can claim that
GCF(β2π₯π₯,β4π₯π₯) = 2 or GCF(β2π₯π₯,β4π₯π₯) = β2,
depending on what expression we wish to leave after factoring the GCF out:
β2π₯π₯ β 4π₯π₯ = 2βπππππππππππππππππΊπΊπΊπΊπΊπΊ
(βπ₯π₯ β 2π₯π₯)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½πππππππππππππππππππππππππππππππππππ‘π‘π‘π‘
or β2π₯π₯ β 4π₯π₯ = β2οΏ½πππππππππππππππππΊπΊπΊπΊπΊπΊ
(π₯π₯ + 2π₯π₯)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½πππππππππππππππππππππππππππππππππππ‘π‘π‘π‘
Note 2: If the GCF of the terms of a polynomial is equal to 1, we often say that these terms do not have any common factors. What we actually mean is that the terms do not have a common factor other than 1, as factoring 1 out does not help in breaking the original polynomial into a product of simpler polynomials. See Definition 1.1.
Finding the Greatest Common Factor
Find the Greatest Common Factor for the given expressions.
a. 6π₯π₯4(π₯π₯ + 1)3, 3π₯π₯3(π₯π₯ + 1), 9π₯π₯(π₯π₯ + 1)2 b. 4ππ(π₯π₯ β π₯π₯), 8ππ(π₯π₯ β π₯π₯) c. ππππ2, ππ2ππ, ππ, ππ d. 3π₯π₯β1π₯π₯β3, π₯π₯β2π₯π₯β2π§π§
a. Since GCF(6, 3, 9) = 3, the lowest power out of π₯π₯4, π₯π₯3, and π₯π₯ is π₯π₯, and the lowest power out of (π₯π₯ + 1)3, (π₯π₯ + 1), and (π₯π₯ + 1)2 is (π₯π₯ + 1), then
GCF(6π₯π₯4(π₯π₯ + 1)3, 3π₯π₯3(π₯π₯ + 1), 9π₯π₯(π₯π₯ + 1)2) = ππππ(ππ + ππ)
b. Since π₯π₯ β π₯π₯ is opposite to π₯π₯ β π₯π₯, then π₯π₯ β π₯π₯ can be written as β(π₯π₯ β π₯π₯). So 4, ππ, and (π₯π₯ β π₯π₯) is common for both expressions. Thus,
GCFοΏ½4ππ(π₯π₯ β π₯π₯), 8ππ(π₯π₯ β π₯π₯)οΏ½ = πππ π (ππ β ππ)
Note: The Greatest Common Factor is unique up to the sign. Notice that in the above example, we could write π₯π₯ β π₯π₯ as β(π₯π₯ β π₯π₯) and choose the GCF to be 4ππ(π₯π₯ β π₯π₯).
c. The terms ππππ2, ππ2ππ, ππ, and ππ have no common factor other than 1, so
GCF(ππππ2, ππ2ππ, ππ, ππ) = ππ
Solution
217
d. The lowest power out of π₯π₯β1 and π₯π₯β2 is π₯π₯β2, and the lowest power out of π₯π₯β3 and π₯π₯β2 is π₯π₯β3. Therefore,
GCF(3π₯π₯β1π₯π₯β3, π₯π₯β2π₯π₯β2π§π§) = ππβππππβππ
Factoring out the Greatest Common Factor Factor each expression by taking the greatest common factor out. Simplify the factors, if possible.
a. 54π₯π₯2π₯π₯2 + 60π₯π₯π₯π₯3 b. ππππ β ππ2ππ(ππ β 1)
c. βπ₯π₯(π₯π₯ β 5) + π₯π₯2(5 β π₯π₯) β (π₯π₯ β 5)2 d. π₯π₯β1 + 2π₯π₯β2 β π₯π₯β3 a. To find the greatest common factor of 54 and 60, we can use the method of dividing
by any common factor, as presented below.
2 54, 603 27, 30 9, 10
So, GCF(54, 60) = 2 β 3 = 6.
Since GCF(54π₯π₯2π₯π₯2, 60π₯π₯π₯π₯3) = 6π₯π₯π₯π₯2, we factor the 6π₯π₯π₯π₯2 out by dividing each term of the polynomial 54π₯π₯2π₯π₯2 + 60π₯π₯π₯π₯3 by 6π₯π₯π₯π₯2, as below.
54π₯π₯2π₯π₯2 + 60π₯π₯π₯π₯3
= ππππππππ(ππππ+ ππππππ) Note: Since factoring is the reverse process of multiplication, it can be checked by finding the product of the factors. If the product gives us the original polynomial, the factorization is correct.
b. First, notice that the polynomial has two terms, ππππ and βππ2ππ(ππ β 1). The greatest common factor for these two terms is ππππ, so we have
ππππ β ππ2ππ(ππ β 1) = πππποΏ½ππ β ππ(ππ β 1)οΏ½
= ππππ(1 β ππ2 + ππ) = ππππ(βππ2 + ππ + 1)
= βπππποΏ½ππππ β ππ β πποΏ½
Solution
no more common factors
for 9 and 10
all common factors are listed in this column
54π₯π₯2π₯π₯2
6π₯π₯π₯π₯2= 9π₯π₯
60π₯π₯π₯π₯3
6π₯π₯π₯π₯2= 10π₯π₯
remember to leave 1 for the first term
simplify and arrange in decreasing powers
take the βββ out
218
Note: Both factorizations, ππππ(βππ2 + ππ + 1) and βππππ(ππ2 β ππ β 1) are correct. However, we customarily leave the polynomial in the bracket with a positive leading coefficient.
c. Observe that if we write the middle term π₯π₯2(5 β π₯π₯) as βπ₯π₯2(π₯π₯ β 5) by factoring the
negative out of the (5 β π₯π₯), then (5 β π₯π₯) is the common factor of all the terms of the equivalent polynomial
βπ₯π₯(π₯π₯ β 5) β π₯π₯2(π₯π₯ β 5) β (π₯π₯ β 5)2.
Then notice that if we take β(π₯π₯ β 5) as the GCF, then the leading term of the remaining polynomial will be positive. So, we factor
βπ₯π₯(π₯π₯ β 5) + π₯π₯2(5β π₯π₯) β (π₯π₯ β 5)2
= βπ₯π₯(π₯π₯ β 5) β π₯π₯2(π₯π₯ β 5) β (π₯π₯ β 5)2
= β(π₯π₯ β 5)οΏ½π₯π₯ + π₯π₯2 + (π₯π₯ β 5)οΏ½
= β(ππ β ππ)οΏ½ππππ + ππππ β πποΏ½
d. The GCF(π₯π₯β1, 2π₯π₯β2, βπ₯π₯β3) = π₯π₯β3, as β3 is the lowest exponent of the common factor π₯π₯. So, we factor out π₯π₯β3 as below.
π₯π₯β1 + 2π₯π₯β2 β π₯π₯β3
= ππβππ οΏ½ππππ + ππππ β πποΏ½
To check if the factorization is correct, we multiply
π₯π₯β3 (π₯π₯2 + 2π₯π₯ β 1)
= π₯π₯β3π₯π₯2 + 2π₯π₯β3π₯π₯ β 1π₯π₯β3
= π₯π₯β1 + 2π₯π₯β2 β π₯π₯β3
Since the product gives us the original polynomial, the factorization is correct.
Factoring by Grouping
Consider the polynomial π₯π₯2 + π₯π₯ + π₯π₯π₯π₯ + π₯π₯. It consists of four terms that do not have any common factors. Yet, it can still be factored if we group the first two and the last two terms. The first group of two terms contains the common factor of π₯π₯ and the second group of two terms contains the common factor of π₯π₯. Observe what happens when we factor each group.
π₯π₯2 + π₯π₯οΏ½οΏ½οΏ½
+ π₯π₯π₯π₯ + π₯π₯οΏ½οΏ½οΏ½
= π₯π₯(π₯π₯ + 1) + π₯π₯(π₯π₯ + 1)
= (π₯π₯ + 1)(π₯π₯ + π₯π₯)
When referring to a common factor, we
have in mind a common factor other
than 1.
the exponent 2 is found by subtracting β3 from β1
the exponent 1 is found by subtracting β3 from β2
add exponents
now (π₯π₯ + 1) is the
common factor of the entire polynomial
simplify and arrange in decreasing powers
219
This method is called factoring by grouping, in particular, two-by-two grouping.
Warning: After factoring each group, make sure to write the β+β or βββ between the terms. Failing to write these signs leads to the false impression that the polynomial is already factored. For example, if in the second line of the above calculations we would fail to write the middle β+β, the expression would look like a product π₯π₯(π₯π₯ + 1) π₯π₯(π₯π₯ + 1), which is not the case. Also, since the expression π₯π₯(π₯π₯ + 1) + π₯π₯(π₯π₯ + 1) is a sum, not a product, we should not stop at this step. We need to factor out the common bracket (π₯π₯ + 1) to leave it as a product.
A two-by-two grouping leads to a factorization only if the binomials, after factoring out the common factors in each group, are the same. Sometimes a rearrangement of terms is necessary to achieve this goal. For example, the attempt to factor π₯π₯3 β 15 + 5π₯π₯2 β 3π₯π₯ by grouping the first and the last two terms,
π₯π₯3 β 15οΏ½οΏ½οΏ½οΏ½οΏ½
+ 5π₯π₯2 β 3π₯π₯οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
= (π₯π₯3 β 15) + π₯π₯(5π₯π₯ β 3)
does not lead us to a common binomial that could be factored out.
However, rearranging terms allows us to factor the original polynomial in the following ways:
π₯π₯3 β 15 + 5π₯π₯2 β 3π₯π₯ or π₯π₯3 β 15 + 5π₯π₯2 β 3π₯π₯
= π₯π₯3 + 5π₯π₯2οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
+ β3π₯π₯ β 15οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
= π₯π₯3 β 3π₯π₯οΏ½οΏ½οΏ½οΏ½οΏ½
+ 5π₯π₯2 β 15οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
= π₯π₯2(π₯π₯ + 5) β 3(π₯π₯ + 5) = π₯π₯(π₯π₯2 β 3) + 5(π₯π₯2 β 3)
= (π₯π₯ + 5)(π₯π₯2 β 3) = (π₯π₯2 β 3)(π₯π₯ + 5) Factoring by grouping applies to polynomials with more than three terms. However, not all such polynomials can be factored by grouping. For example, if we attempt to factor π₯π₯3 +π₯π₯2 + 2π₯π₯ β 2 by grouping, we obtain
π₯π₯3 + π₯π₯2οΏ½οΏ½οΏ½οΏ½οΏ½
+ 2π₯π₯ β 2οΏ½οΏ½οΏ½
= π₯π₯2(π₯π₯ + 1) + 2(π₯π₯ β 1).
Unfortunately, the expressions π₯π₯ + 1 and π₯π₯ β 1 are not the same, so there is no common factor to factor out. One can also check that no other rearrangments of terms allows us for factoring out a common binomial. So, this polynomial cannot be factored by grouping.
Factoring by Grouping
Factor each polynomial by grouping, if possible. Remember to check for the GCF first.
220
a. 2π₯π₯3 β 6π₯π₯2 + π₯π₯ β 3 b. 5π₯π₯ β 5π₯π₯ β πππ₯π₯ + πππ₯π₯ c. 2π₯π₯2π₯π₯ β 8 β 2π₯π₯2 + 8π₯π₯ d. π₯π₯2 β π₯π₯ + π₯π₯ + 1 a. Since there is no common factor for all four terms, we will attempt the two-by-two
grouping method. 2π₯π₯3 β 6π₯π₯2οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
+ π₯π₯ β 3οΏ½οΏ½οΏ½
= 2π₯π₯2(π₯π₯ β 3) + 1(π₯π₯ β 3)
= (ππ β ππ)οΏ½ππππππ + πποΏ½
b. As before, there is no common factor for all four terms. The two-by-two grouping method works only if the remaining binomials after factoring each group are exactly the same. We can achieve this goal by factoring βππ , rather than ππ, out of the last two terms. So,
5π₯π₯ β 5π₯π₯οΏ½οΏ½οΏ½οΏ½οΏ½
βπππ₯π₯ + πππ₯π₯οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
= 5(π₯π₯ β π₯π₯) β ππ(π₯π₯ β π₯π₯)
= (ππ β ππ)οΏ½ππππππ + πποΏ½
c. Notice that 2 is the GCF of all terms, so we factor it out first.
2π₯π₯2π₯π₯ β 8 β 2π₯π₯2 + 8π₯π₯
= 2(π₯π₯2π₯π₯ β 4 β π₯π₯2 + 4π₯π₯)
Then, observe that grouping the first and last two terms of the remaining polynomial does not help, as the two groups do not have any common factors. However, exchanging for example the second with the fourth term will help, as shown below.
= 2(π₯π₯2π₯π₯ + 4π₯π₯οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
βπ₯π₯2 β 4οΏ½οΏ½οΏ½οΏ½οΏ½)
= 2[π₯π₯(π₯π₯2 + 4) β (π₯π₯2 + 4)]
= πποΏ½ππππ + πποΏ½(ππ β ππ)
d. The polynomial π₯π₯2 β π₯π₯ + π₯π₯ + 1 does not have any common factors for all four terms. Also, only the first two terms have a common factor. Unfortunately, when attempting to factor using the two-by-two grouping method, we obtain
π₯π₯2 β π₯π₯ + π₯π₯ + 1
= π₯π₯(π₯π₯ β 1) + (π₯π₯ + 1),
which cannot be factored, as the expressions π₯π₯ β 1 and π₯π₯ + 1 are different.
One can also check that no other arrangement of terms allows for factoring of this polynomial by grouping. So, this polynomial cannot be factored by grouping.
Solution
write the 1 for the second term
reverse signs when βpullingβ a βββ out
reverse signs when βpullingβ a βββ out
the square bracket is
essential here because of the factor of 2
now, there is no need for the square bracket as multiplication is associative
221
Factoring in Solving Formulas
Solve ππππ = 3ππ + 5 for ππ.
First, we move the terms containing the variable ππ to one side of the equation,
ππππ = 3ππ + 5 ππππ β 3ππ = 5,
and then factor ππ out ππ(ππ β 3) = 5.
So, after dividing by ππ β 3, we obtain ππ = ππππβππ
.
F.1 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list: common factor, distributive, grouping, prime, product.
1. To factor a polynomial means to write it as a ____________ of simpler polynomials.
2. The greatest ______________ _____________ of two or more terms is the product of the least powers of each type of common factor out of all the terms.
3. To factor out the GCF, we reverse the ______________ property of multiplication.
4. A polynomial with four terms having no common factors can be still factored by _____________ its terms.
5. A ___________ polynomial, other than a monomial, cannot be factored into two polynomials, both different that 1 or β1.
Concept Check True or false.
6. The polynomial 6π₯π₯ + 8π₯π₯ is prime.
7. The factorization 12π₯π₯ β 3
4π₯π₯ = 1
4(2π₯π₯ β 3π₯π₯) is essential to be complete.
8. The GCF of the terms of the polynomial 3(π₯π₯ β 2) + π₯π₯(2 β π₯π₯) is (π₯π₯ β 2)(2 β π₯π₯). Concept Check Find the GCF with a positive coefficient for the given expressions.
9. 8π₯π₯π₯π₯, 10π₯π₯π§π§, β14π₯π₯π₯π₯ 10. 21ππ3ππ6, β35ππ7ππ5, 28ππ5ππ8
11. 4π₯π₯(π₯π₯ β 1), 3π₯π₯2(π₯π₯ β 1) 12. βπ₯π₯(π₯π₯ β 3)2, π₯π₯2(π₯π₯ β 3)(π₯π₯ + 2)
13. 9(ππ β 5), 12(5β ππ) 14. (π₯π₯ β 2π₯π₯)(π₯π₯ β 1), (2π₯π₯ β π₯π₯)(π₯π₯ + 1)
Solution
222
15. β3π₯π₯β2π₯π₯β3, 6π₯π₯β3π₯π₯β5 16. π₯π₯β2(π₯π₯ + 2)β2, βπ₯π₯β4(π₯π₯ + 2)β1 Factor out the greatest common factor. Leave the remaining polynomial with a positive leading coeficient. Simplify the factors, if possible.
17. 9π₯π₯2 β 81π₯π₯ 18. 8ππ3 + 24ππ 19. 6ππ3 β 3ππ2 β 9ππ4
20. 6ππ3 β 36ππ4 + 18ππ2 21. β10ππ2π π 2 + 15ππ4π π 2 22. 5π₯π₯2π₯π₯3 β 10π₯π₯3π₯π₯2
23. ππ(π₯π₯ β 2) + ππ(π₯π₯ β 2) 24. ππ(π₯π₯2 β 3) β 2(π₯π₯2 β 3)
25. (π₯π₯ β 2)(π₯π₯ + 3) + (π₯π₯ β 2)(π₯π₯ + 5) 26. (ππ β 2)(ππ + 3) + (ππ β 2)(ππ β 3)
27. π₯π₯(π₯π₯ β 1) + 5(1 β π₯π₯) 28. (4π₯π₯ β π₯π₯) β 4π₯π₯(π₯π₯ β 4π₯π₯)
29. 4(3 β π₯π₯)2 β (3 β π₯π₯)3 + 3(3 β π₯π₯) 30. 2(ππ β 3) + 4(ππ β 3)2 β (ππ β 3)3 Factor out the least power of each variable.
31. 3π₯π₯β3 + π₯π₯β2 32. ππβ2 + 2ππβ4 33. π₯π₯β4 β 2π₯π₯β3 + 7π₯π₯β2
34. 3ππβ5 + ππβ3 β 2ππβ2 35. 3π₯π₯β3π₯π₯ β π₯π₯β2π₯π₯2 36. β5π₯π₯β2π₯π₯β3 + 2π₯π₯β1π₯π₯β2 Factor by grouping, if possible.
37. 20 + 5π₯π₯ + 12π₯π₯ + 3π₯π₯π₯π₯ 38. 2ππ3 + ππ2 β 14ππ β 7 39. ππππ β ππππ + ππππ β ππππ
40. 2π₯π₯π₯π₯ β π₯π₯2π₯π₯ + 6 β 3π₯π₯ 41. 3π₯π₯2 + 4π₯π₯π₯π₯ β 6π₯π₯π₯π₯ β 8π₯π₯2 42. π₯π₯3 β π₯π₯π₯π₯ + π₯π₯2 β π₯π₯2π₯π₯
43. 3ππ2 + 9ππππ β ππππ β 3ππ2 44. 3π₯π₯2 β π₯π₯2π₯π₯ β π₯π₯π§π§2 + 3π§π§2 45. 2π₯π₯3 β π₯π₯2 + 4π₯π₯ β 2
46. π₯π₯2π₯π₯2 + ππππ β πππ₯π₯2 β πππ₯π₯2 47. π₯π₯π₯π₯ + ππππ + πππ₯π₯ + πππ₯π₯ 48. π₯π₯2π₯π₯ β π₯π₯π₯π₯ + π₯π₯ + π₯π₯
49. π₯π₯π₯π₯ β 6π₯π₯ + 3π₯π₯ β 18 50. π₯π₯πππ₯π₯ β 3π₯π₯ππ + π₯π₯ β 5 51. πππππ₯π₯ππ + 2ππππ + π₯π₯ππ + 2
Factor completely. Remember to check for the GCF first.
52. 5π₯π₯ β 5πππ₯π₯ + 5ππππππ β 5ππππ 53. 6πππ π β 14π π + 6ππ β 14
54. π₯π₯4(π₯π₯ β 1) + π₯π₯3(π₯π₯ β 1) β π₯π₯2 + π₯π₯ 55. π₯π₯3(π₯π₯ β 2)2 + 2π₯π₯2(π₯π₯ β 2) β (π₯π₯ + 2)(π₯π₯ β 2) Discussion Point
56. One of possible factorizations of the polynomial 4π₯π₯2π₯π₯5 β 8π₯π₯π₯π₯3 is 2π₯π₯π₯π₯3(2π₯π₯π₯π₯2 β 4). Is this a complete factorization?
Use factoring the GCF strategy to solve each formula for the indicated variable.
57. π΄π΄ = π·π· + π·π·ππ, for π·π· 58. ππ = 12ππππ + 1
2ππππ, for ππ
59. 2ππ + ππ = ππππ, for ππ 60. π€π€ππ = 3ππ β π₯π₯, for ππ
223
Analytic Skills Write the area of each shaded region in factored form. 61. 62.
63. 64.
4π₯π₯
π₯π₯ π₯π₯
ππ π π ππ
ππ
224
F.2 Factoring Trinomials
In this section, we discuss factoring trinomials. We start with factoring quadratic trinomials of the form π₯π₯2 + πππ₯π₯ + ππ, then quadratic trinomials of the form πππ₯π₯2 + πππ₯π₯ + ππ, where ππ β 1, and finally trinomials reducible to quadratic by means of substitution.
Factorization of Quadratic Trinomials ππππ + ππππ + ππ
Factorization of a quadratic trinomial π₯π₯2 + πππ₯π₯ + ππ is the reverse process of the FOIL method of multiplying two linear binomials. Observe that
(π₯π₯ + ππ)(π₯π₯ + ππ) = π₯π₯2 + πππ₯π₯ + πππ₯π₯ + ππππ = π₯π₯2 + (ππ + ππ)π₯π₯ + ππππ
So, to reverse this multiplication, we look for two numbers ππ and ππ, such that the product ππππ equals to the free term ππ and the sum ππ + ππ equals to the middle coefficient ππ of the trinomial.
π₯π₯2 + ππβ(ππ+ππ)
π₯π₯ + ππβππππ
= (π₯π₯ + ππ)(π₯π₯ + ππ)
For example, to factor π₯π₯2 + 5π₯π₯ + 6, we think of two integers that multiply to 6 and add to 5. Such integers are 2 and 3, so π₯π₯2 + 5π₯π₯ + 6 = (π₯π₯ + 2)(π₯π₯ + 3). Since multiplication is commutative, the order of these factors is not important.
This could also be illustrated geometrically, using algebra tiles.
The area of a square with the side length π₯π₯ is equal to π₯π₯2. The area of a rectangle with the dimensions π₯π₯ by 1 is equal to π₯π₯, and the area of a unit square is equal to 1. So, the trinomial π₯π₯2 + 5π₯π₯ + 6 can be represented as
To factor this trinomial, we would like to rearrange these tiles to fulfill a rectangle.
The area of such rectangle can be represented as the product of its length, (π₯π₯ + 3), and width, (π₯π₯ + 2) which becomes the factorization of the original trinomial. In the trinomial examined above, the signs of the middle and the last terms are both positive. To analyse how different signs of these terms influence the signs used in the factors, observe the next three examples.
π₯π₯2 π₯π₯
1
π₯π₯2 π₯π₯ π₯π₯ π₯π₯ π₯π₯ π₯π₯
1 1 1 1 1 1
π₯π₯2 π₯π₯ π₯π₯ π₯π₯
π₯π₯ π₯π₯ 1 1
1 1 1 1
π₯π₯ + 3 οΏ½β―β―β―β―β―β―β―β―β―β―β―β―β―β―β―οΏ½
οΏ½β―β―β―β―β―β―β―β―β―β―β―β―οΏ½
π₯π₯ +
2
225
To factor π₯π₯2 β 5π₯π₯ + 6, we look for two integers that multiply to 6 and add to β5. Such integers are β2 and β3, so π₯π₯2 β 5π₯π₯ + 6 = (π₯π₯ β 2)(π₯π₯ β 3). To factor π₯π₯2 + π₯π₯ β 6, we look for two integers that multiply to β6 and add to 1. Such integers are β2 and 3, so π₯π₯2 + π₯π₯ β 6 = (π₯π₯ β 2)(π₯π₯ + 3). To factor π₯π₯2 β π₯π₯ β 6, we look for two integers that multiply to β6 and add to β1. Such integers are 2 and β3, so π₯π₯2 β π₯π₯ β 6 = (π₯π₯ + 2)(π₯π₯ β 3). Observation: The positive constant ππ in a trinomial π₯π₯2 + πππ₯π₯ + ππ tells us that the integers ππ and ππ in the factorization (π₯π₯ + ππ)(π₯π₯ + ππ) are both of the same sign and their sum is the middle coefficient ππ. In addition, if ππ is positive, both ππ and ππ are positive, and if ππ is negative, both ππ and ππ are negative.
The negative constant ππ in a trinomial π₯π₯2 + πππ₯π₯ + ππ tells us that the integers ππ and ππ in the factorization (π₯π₯ + ππ)(π₯π₯ + ππ) are of different signs and a difference of their absolute values is the middle coefficient ππ. In addition, the integer whose absolute value is larger takes the sign of the middle coefficient ππ. These observations are summarized in the following Table of Signs.
Assume that |ππ| β₯ |ππ|. sum ππ product ππ ππ ππ comments
+ + + + ππ is the sum of ππ and ππ β + β β ππ is the sum of ππ and ππ + β + β ππ is the difference |ππ| β |ππ| β β β + ππ is the difference |ππ| β |ππ|
Factoring Trinomials with the Leading Coefficient Equal to 1
Factor each trinomial, if possible.
a. π₯π₯2 β 10π₯π₯ + 24 b. π₯π₯2 + 9π₯π₯ β 36 c. π₯π₯2 β 39π₯π₯π₯π₯ β 40π₯π₯2 d. π₯π₯2 + 7π₯π₯ + 9 a. To factor the trinomial π₯π₯2 β 10π₯π₯ + 24, we look for two integers with a product of 24
and a sum of β10. The two integers are fairly easy to guess, β4 and β6. However, if one wishes to follow a more methodical way of finding these numbers, one can list the possible two-number factorizations of 24 and observe the sums of these numbers.
product = ππππ (pairs of factors of 24)
sum = βππππ (sum of factors)
ππ β ππππ 25 ππ β ππππ 14 ππ β ππ 11 ππ β ππ 10
Solution
π©π©π©π©π©π©π©π©π©π©!
For simplicity, the table doesnβt include signs of the
integers. The signs are determined according to
the Table of Signs.
226
Since the product is positive and the sum is negative, both integers must be negative. So, we take β4 and β6.
Thus, π₯π₯2 β 10π₯π₯ + 24 = (ππ β ππ)(ππ β ππ). The reader is encouraged to check this factorization by multiplying the obtained binomials.
b. To factor the trinomial π₯π₯2 + 9π₯π₯ β 36, we look for two integers with a product of β36
and a sum of 9. So, let us list the possible factorizations of 36 into two numbers and observe the differences of these numbers.
product = βππππ (pairs of factors of 36)
sum = ππ (difference of factors)
ππ β ππππ 35 ππ β ππππ 16 ππ β ππππ 9 ππ β ππ 5 ππ β ππ 0
Since the product is negative and the sum is positive, the integers are of different signs
and the one with the larger absolute value assumes the sign of the sum, which is positive. So, we take 12 and β3.
Thus, π₯π₯2 + 9π₯π₯ β 36 = (ππ + ππππ)(ππ β ππ). Again, the reader is encouraged to check this factorization by mltiplying the obtained binomials.
c. To factor the trinomial π₯π₯2 β 39π₯π₯π₯π₯ β 40π₯π₯2, we look for two binomials of the form
(π₯π₯+ ?π₯π₯)(π₯π₯+ ?π₯π₯) where the question marks are two integers with a product of β40 and a sum of 39. Since the two integers are of different signs and the absolute values of these integers differ by 39, the two integers must be β40 and 1.
Therefore, π₯π₯2 β 39π₯π₯π₯π₯ β 40π₯π₯2 = (ππ β ππππππ)(ππ + ππ). Suggestion: Create a table of pairs of factors only if guessing the two integers with the given product and sum becomes too difficult. d. When attempting to factor the trinomial π₯π₯2 + 7π₯π₯ + 9, we look for a pair of integers
that would multiply to 9 and add to 7. There are only two possible factorizations of 9: 9 β 1 and 3 β 3. However, neither of the sums, 9 + 1 or 3 + 3, are equal to 7. So, there is no possible way of factoring π₯π₯2 + 7π₯π₯ + 9 into two linear binomials with integral coefficients. Therefore, if we admit only integral coefficients, this polynomial is not factorable.
Factorization of Quadratic Trinomials ππππππ + ππππ + ππ with ππ β 0
Before discussing factoring quadratic trinomials with a leading coefficient different than 1, let us observe the multiplication process of two linear binomials with integral coefficients.
(πππ₯π₯ + ππ)(π©π©π₯π₯ + ππ) = πππππ₯π₯2 + πππππ₯π₯ + πππππ₯π₯ + ππππ = ππβπππ©π©
π₯π₯2 + ππβ(ππππ+π©π©ππ)
π₯π₯ + ππβππππ
This row contains the solution, so there is no need to list any of the
subsequent rows.
227
To reverse this process, notice that this time, we are looking for four integers ππ, ππ, ππ, and ππ that satisfy the conditions
ππππ = ππ, ππππ = ππ, ππππ + ππππ = ππ,
where ππ, ππ, ππ are the coefficients of the quadratic trinomial that needs to be factored. This produces a lot more possibilities to consider than in the guessing method used in the case of the leading coefficient equal to 1. However, if at least one of the outside coefficients, ππ or ππ, are prime, the guessing method still works reasonably well.
For example, consider 2π₯π₯2 + π₯π₯ β 6. Since the coefficient ππ = 2 = ππππ is a prime number, there is only one factorization of ππ, which is 1 β 2. So, we can assume that ππ = 2 and ππ =1. Therefore,
2π₯π₯2 + π₯π₯ β 6 = (2π₯π₯ Β± |ππ|)(π₯π₯ β |ππ|)
Since the constant term ππ = β6 = ππππ is negative, the binomial factors have different signs in the middle. Also, since ππππ is negative, we search for such ππ and ππ that the inside and outside products differ by the middle term ππ = π₯π₯, up to its sign. The only factorizations of 6 are 1 β 6 and 2 β 3. So we try
2π₯π₯2 + π₯π₯ β 6 = (2π₯π₯ Β± 1)(π₯π₯ β 6)
2π₯π₯2 + π₯π₯ β 6 = (2π₯π₯ Β± 6)(π₯π₯ β 1)
2π₯π₯2 + π₯π₯ β 6 = (2π₯π₯ Β± 2)(π₯π₯ β 3)
2π₯π₯2 + π₯π₯ β 6 = (2π₯π₯ Β± 3)(π₯π₯ β 2)
Then, since the difference between the inner and outer products should be positive, the larger product must be positive and the smaller product must be negative. So, we distribute the signs as below.
2π₯π₯2 + π₯π₯ β 6 = (2π₯π₯ β 3)(π₯π₯ + 2)
In the end, it is a good idea to multiply the product to check if it results in the original polynomial. We leave this task to the reader. What if the outside coefficients of the quadratic trinomial are both composite? Checking all possible distributions of coefficients ππ, ππ, ππ, and ππ might be too cumbersome. Luckily, there is another method of factoring, called decomposition.
π₯π₯
12π₯π₯ differs by 11π₯π₯ β too much
6π₯π₯ 2π₯π₯
differs by 4π₯π₯ β still too much
2π₯π₯ 6π₯π₯
differs by 4π₯π₯ β still too much
3π₯π₯ 4π₯π₯
differs by π₯π₯ β perfect!
β3π₯π₯ 4π₯π₯
Observe that these two trials can be disregarded at once as 2 is not a common factor
of all the terms of the trinomial, while it is a
common factor of the terms of one of the binomials.
228
The decomposition method is based on the reverse FOIL process. Suppose the polynomial 6π₯π₯2 + 19π₯π₯ + 15 factors into (πππ₯π₯ + ππ)(πππ₯π₯ + ππ). Observe that the FOIL multiplication of these two binomials results in the four term polynomial,
πππππ₯π₯2 + πππππ₯π₯ + πππππ₯π₯ + ππππ,
which after combining the two middle terms gives us the original trinomial. So, reversing these steps would lead us to the factored form of 6π₯π₯2 + 19π₯π₯ + 15.
To reverse the FOIL process, we would like to:
β’ Express the middle term, 19π₯π₯, as a sum of two terms, πππππ₯π₯ and πππππ₯π₯, such that the product of their coefficients, ππππππππ, is equal to the product of the outside coefficients ππππ = 6 β 15 = 90.
β’ Then, factor the four-term polynomial by grouping.
Thus, we are looking for two integers with the product of 90 and the sum of 19. One can check that 9 and 10 satisfy these conditions. Therefore,
6π₯π₯2 + 19π₯π₯ + 15
= 6π₯π₯2 + 9π₯π₯ + 10π₯π₯ + 15
= 3π₯π₯(2π₯π₯ + 3) + 5(2π₯π₯ + 3)
= (2π₯π₯ + 3)(3π₯π₯ + 5)
Factoring Trinomials with the Leading Coefficient Different than 1
Factor completely each trinomial.
a. 6π₯π₯3 + 14π₯π₯2 + 4π₯π₯ b. β6π₯π₯2 β 10 + 19π₯π₯ c. 18ππ2 β 19ππππ β 12ππ2 d. 2(π₯π₯ + 3)2 + 5(π₯π₯ + 3) β 12 a. First, we factor out the GCF, which is 2π₯π₯. This gives us
6π₯π₯3 + 14π₯π₯2 + 4π₯π₯ = 2π₯π₯(3π₯π₯2 + 7π₯π₯ + 2)
The outside coefficients of the remaining trinomial are prime, so we can apply the guessing method to factor it further. The first terms of the possible binomial factors must be 3π₯π₯ and π₯π₯ while the last terms must be 2 and 1. Since both signs in the trinomial are positive, the signs used in the binomial factors must be both positive as well. So, we are ready to give it a try:
2π₯π₯(3π₯π₯ + 2 )(π₯π₯ + 1 ) or 2π₯π₯(3π₯π₯ + 1 )(π₯π₯ + 2 )
The first distribution of coefficients does not work as it would give us 2π₯π₯ + 3π₯π₯ = 5π₯π₯ for the middle term. However, the second distribution works as π₯π₯ + 6π₯π₯ = 7π₯π₯, which matches the middle term of the trinomial. So,
6π₯π₯3 + 14π₯π₯2 + 4π₯π₯ = ππππ(ππππ + ππ)(ππ + ππ)
Solution
This product is often referred to as the
master product or the ππππ-product.
2π₯π₯ 3π₯π₯
π₯π₯ 6π₯π₯
229
b. Notice that the trinomial is not arranged in decreasing order of powers of π₯π₯. So, first, we rearrange the last two terms to achieve the decreasing order. Also, we factor out the β1, so that the leading term of the remaining trinomial is positive.
β6π₯π₯2 β 10 + 19π₯π₯ = β6π₯π₯2 + 19π₯π₯ β 10 = β(6π₯π₯2 β 19π₯π₯ + 10)
Then, since the outside coefficients are composite, we will use the decomposition method of factoring. The ππππ-product equals to 60 and the middle coefficient equals to β19. So, we are looking for two integers that multiply to 60 and add to β19. The integers that satisfy these conditions are β15 and β4. Hence, we factor
β(6π₯π₯2 β 19π₯π₯ + 10)
= β(6π₯π₯2 β 15π₯π₯ β 4π₯π₯ + 10)
= β[3π₯π₯(2π₯π₯ β 5) β 2(2π₯π₯ β 5)]
= β(ππππ β ππ)(ππππ β ππ)
c. There is no common factor to take out of the polynomial 18ππ2 β 19ππππ β 12ππ2. So, we will attempt to factor it into two binomials of the type (ππππ Β± ππππ)(ππππ β ππππ), using the decomposition method. The ππππ-product equals β12 β 18 = β2 β 2 β 2 β 3 β 3 β 3 and the middle coefficient equals β19. To find the two integers that multiply to the ππππ-product and add to β19, it is convenient to group the factors of the product
2 β 2 β 2 β 3 β 3 β 3
in such a way that the products of each group differ by 19. It turns out that grouping all the 2βs and all the 3βs satisfy this condition, as 8 and 27 differ by 19. Thus, the desired integers are β27 and 8, as the sum of them must be β19. So, we factor
18ππ2 β 19ππππ β 12ππ2
= 18ππ2 β 27ππππ + 8ππππ β 12ππ2
= 9ππ(2ππ β 3ππ) + 4ππ(2ππ β 3ππ)
= (ππππ β ππππ)(ππππ + ππππ)
d. To factor 2(π₯π₯ + 3)2 + 5(π₯π₯ + 3) β 12, first, we notice that treating the group (π₯π₯ + 3) as another variable, say ππ, simplify the problem to factoring the quadratic trinomial
2ππ2 + 5ππ β 12
This can be done by the guessing method. Since
2ππ2 + 5ππ β 12 = (2ππ β 3)(ππ + 4),
then
2(π₯π₯ + 3)2 + 5(π₯π₯ + 3) β 12 = [2(π₯π₯ + 3) β 3][(π₯π₯ + 3) + 4]
= (2π₯π₯ + 6 β 3)(π₯π₯ + 3 + 4)
= (ππππ + ππ)(ππ + ππ)
remember to reverse the sign!
the square bracket is
essential because of the negative sign outside
β3ππ 8ππ
IN FORM
230
Note 1: Polynomials that can be written in the form ππ( )ππ + ππ( ) + ππ, where ππ β 0 and ( ) represents any nonconstant polynomial expression, are referred to as quadratic in form. To factor such polynomials, it is convenient to replace the expression in the bracket by a single variable, different than the original one. This was illustrated in Example 2d by substituting ππ for (π₯π₯ + 3). However, when using this substitution method, we must remember to leave the final answer in terms of the original variable. So, after factoring, we replace ππ back with (π₯π₯ + 3), and then simplify each factor.
Note 2: Some students may feel comfortable factoring polynomials quadratic in form directly, without using substitution.
Application of Factoring in Geometry Problems If the area of a trapezoid is 2π₯π₯2 + 5π₯π₯ + 2 square meters and the lengths of the two parallel sides are π₯π₯ and π₯π₯ + 1 meters, then what polynomial represents the height of the trapezoid?
Using the formula for the area of a trapezoid, we write the equation 12β(ππ + ππ) = 2π₯π₯2 + 5π₯π₯ + 2
Since ππ + ππ = π₯π₯ + (π₯π₯ + 1) = 2π₯π₯ + 1, then we have 12β(2π₯π₯ + 1) = 2π₯π₯2 + 5π₯π₯ + 2,
which after factoring the right-hand side gives us 12β(2π₯π₯ + 1) = (2π₯π₯ + 1)(π₯π₯ + 2).
To find β, it is enough to divide the above equation by the common factor (2π₯π₯ + 1) and then multiply it by 2. So,
β = 2(π₯π₯ + 2) = ππππ + ππ.
F.2 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list:
decomposition, guessing, multiplication, prime, quadratic, sum, variable. 1. Any factorization can be checked by using _________________.
Solution
π₯π₯
π₯π₯ + 1
β
231
2. To factor a quadratic trinomial with a leading coefficient equal to 1, we usually use the ______________ method.
3. To factor π₯π₯2 + πππ₯π₯ + ππ using the guessing method, write the trinomial as (π₯π₯+ ? )(π₯π₯+ ? ), where the question marks are two factors of ππ whose _______ is ππ.
4. To factor a quadratic trinomial with a leading coefficient different than 1, we usually use the ______________ method. If one of the outside coefficients is a __________ number, we can still use the guessing method.
5. To factor polynomials that are _____________ in form, it is convenient to substitute a single variable (different than the original one) for the expression that appears in the first and the second power. However, the final factorization must be expressed back in the original __________.
Concept Check
6. If πππ₯π₯2 + πππ₯π₯ + ππ has no monomial factor, can either of the possible binomial factors have a monomial factor?
7. Is (2π₯π₯ + 5)(2π₯π₯ β 4) a complete factorization of the polynomial 4π₯π₯2 + 2π₯π₯ β 20?
8. When factoring the polynomial β2π₯π₯2 β 7π₯π₯ + 15, students obtained the following answers: (β2π₯π₯ + 3)(π₯π₯ + 5), (2π₯π₯ β 3)(βπ₯π₯ β 5), or β(2π₯π₯ β 3)(π₯π₯ + 5)
Which of the above factorizations are correct?
9. Is the polynomial π₯π₯2 β π₯π₯ + 2 factorable or is it prime?
Concept Check Fill in the missing factor.
10. π₯π₯2 β 4π₯π₯ + 3 = ( )(π₯π₯ β 1) 11. π₯π₯2 + 3π₯π₯ β 10 = ( )(π₯π₯ β 2)
12. π₯π₯2 β π₯π₯π₯π₯ β 20π₯π₯2 = (π₯π₯ + 4π₯π₯)( ) 13. π₯π₯2 + 12π₯π₯π₯π₯ + 35π₯π₯2 = (π₯π₯ + 5π₯π₯)( )
Factor, if possible.
14. π₯π₯2 + 7π₯π₯ + 12 15. π₯π₯2 β 12π₯π₯ + 35 16. π₯π₯2 + 2π₯π₯ β 48
17. ππ2 β ππ β 42 18. π₯π₯2 + 2π₯π₯ + 3 19. ππ2 β 12ππ β 27
20. ππ2 β 15ππ + 56 21. π₯π₯2 + 3π₯π₯ β 28 22. 18 β 7ππ β ππ2
23. 20 + 8ππ β ππ2 24. π₯π₯2 β 5π₯π₯π₯π₯ + 6π₯π₯2 25. ππ2 + 9ππππ + 20ππ2
Factor completely.
26. βπ₯π₯2 + 4π₯π₯ + 21 27. βπ₯π₯2 + 14π₯π₯ + 32 28. ππ4 β 13ππ3 β 30ππ2
29. π₯π₯3 β 15π₯π₯2 + 54π₯π₯ 30. β2π₯π₯2 + 28π₯π₯ β 80 31. β3π₯π₯2 β 33π₯π₯ β 72
32. π₯π₯4π₯π₯ + 7π₯π₯2π₯π₯ β 60π₯π₯ 33. 24ππππ2 + 6ππ2ππ2 β 3ππ3ππ2 34. 40 β 35π‘π‘15 β 5π‘π‘30
35. π₯π₯4π₯π₯2 + 11π₯π₯2π₯π₯ + 30 36. 64ππ β 12ππ5 β ππ9 37. 24 β 5π₯π₯ππ β π₯π₯2ππ
232
Discussion Point
38. A polynomial π₯π₯2 + ππ π₯π₯ + 75 with an unknown coefficient ππ by the middle term can be factored into two binomials with integral coefficients. What are the possible values of ππ?
Concept Check Fill in the missing factor.
39. 2π₯π₯2 + 7π₯π₯ + 3 = ( )(π₯π₯ + 3) 40. 3π₯π₯2 β 10π₯π₯ + 8 = ( )(π₯π₯ β 2)
41. 4π₯π₯2 + 8π₯π₯ β 5 = (2π₯π₯ β 1)( ) 42. 6π₯π₯2 β π₯π₯ β 15 = (2π₯π₯ + 3)( )
Factor completely.
43. 2π₯π₯2 β 5π₯π₯ β 3 44. 6π₯π₯2 β π₯π₯ β 2 45. 4ππ2 + 17ππ + 4
46. 6π‘π‘2 β 13π‘π‘ + 6 47. 10π₯π₯2 + 23π₯π₯ β 5 48. 42ππ2 + 5ππ β 25
49. 3ππ2 β 27ππ + 24 50. β12π₯π₯2 β 2π₯π₯ + 30 51. 6π₯π₯2 + 41π₯π₯π₯π₯ β 7π₯π₯2
52. 18π₯π₯2 + 27π₯π₯π₯π₯ + 10π₯π₯2 53. 8 β 13ππ + 6ππ2 54. 15 β 14ππ β 8ππ2
55. 30π₯π₯4 + 3π₯π₯3 β 9π₯π₯2 56. 10π₯π₯3 β 6π₯π₯2 + 4π₯π₯4 57. 2π₯π₯6 + 7π₯π₯π₯π₯3 + 6π₯π₯2
58. 9π₯π₯2π₯π₯2 β 4 + 5π₯π₯π₯π₯ 59. 16π₯π₯2π₯π₯3 + 3π₯π₯ β 16π₯π₯π₯π₯2 60. 4ππ4 β 28ππ2ππ + 49ππ2
61. 4(π₯π₯ β 1)2 β 12(π₯π₯ β 1) + 9 62. 2(ππ + 2)2 + 11(ππ + 2) + 15 63. 4π₯π₯2ππ β 4π₯π₯ππ β 3
Discussion Point
64. A polynomial 2π₯π₯2 + ππ π₯π₯ β 15 with an unknown coefficient ππ by the middle term can be factored into two binomials with integral coefficients. What are the possible values of ?
Analytic Skills
65. If the volume of a box is π₯π₯3 + π₯π₯2 β 2π₯π₯ cubic feet and the height of this box is (π₯π₯ β 1) feet, then what polynomial represents the area of the bottom of the box?
66. A ceremonial red carpet is rectangular in shape and covers 2π₯π₯2 + 11π₯π₯ + 12 square feet. If the width of the carpet is (π₯π₯ + 4) feet, express the length, in feet.
π₯π₯ β 1
π₯π₯ + 4
233
F.3 Special Factoring and a General Strategy of Factoring
Recall that in Section P2, we considered formulas that provide a shortcut for finding special products, such as a product of two conjugate binomials,
(ππ + ππ)(ππ β ππ) = ππππ β ππππ,
or the perfect square of a binomial,
(ππ Β± ππ)2 = ππππ Β± ππππππ + ππππ.
Since factoring reverses the multiplication process, these formulas can be used as shortcuts in factoring binomials of the form ππππ β ππππ (difference of squares), and trinomials of the form ππππ Β± ππππππ + ππππ (perfect square). In this section, we will also introduce a formula for factoring binomials of the form ππππ Β± ππππ (sum or difference of cubes). These special product factoring techniques are very useful in simplifying expressions or solving equations, as they allow for more efficient algebraic manipulations.
At the end of this section, we give a summary of all the factoring strategies shown in this chapter.
Difference of Squares Out of the special factoring formulas, the easiest one to use is the difference of squares,
ππππ β ππππ = (ππ + ππ)(ππ β ππ) Figure 3.1 shows a geometric interpretation of this formula. The area of the yellow square, ππ2, diminished by the area of the blue square, ππ2, can be rearranged to a rectangle with the length of (ππ + ππ) and the width of (ππ β ππ). To factor a difference of squares ππππ β ππππ, first, identify ππ and ππ, which are the expressions being squared, and then, form two factors, the sum (ππ + ππ), and the difference (ππ β ππ), as illustrated in the example below.
Factoring Differences of Squares
Factor each polynomial completely.
a. 25π₯π₯2 β 1 b. 3.6π₯π₯4 β 0.9π₯π₯6 c. π₯π₯4 β 81 d. 16 β (ππ β 2)2 a. First, we rewrite each term of 25π₯π₯2 β 1 as a perfect square of an expression.
ππ ππ
25π₯π₯2 β 1 = (5π₯π₯)2 β 12
Then, treating 5π₯π₯ as the ππ and 1 as the ππ in the difference of squares formula ππ2 β ππ2 = (ππ + ππ)(ππ β ππ), we factor:
Solution
Figure 3.1
ππ
ππ ππ β ππ ππ
ππβππ
ππ ππ + ππ
ππ2 ππ2
234
ππ2 β ππ2 = (ππ + ππ)(ππ β ππ)
25π₯π₯2 β 1 = (5π₯π₯)2 β 12 = (ππππ + ππ)(ππππ β ππ)
b. First, we factor out 0.9 to leave the coefficients in a perfect square form. So,
3.6π₯π₯4 β 0.9π₯π₯6 = 0.9(4π₯π₯4 β π₯π₯6)
Then, after writing the terms of 4π₯π₯4 β π₯π₯6 as perfect squares of expressions that correspond to ππ and ππ in the difference of squares formula ππ2 β ππ2 = (ππ + ππ)(ππ β ππ), we factor
ππ ππ
0.9(4π₯π₯4 β π₯π₯6) = 0.9[(2π₯π₯2)2 β (π₯π₯3)2] = ππ.πποΏ½ππππππ + πππποΏ½οΏ½ππππππ β πππποΏ½
c. Similarly as in the previous two examples, π₯π₯4 β 81 can be factored by following the difference of squares pattern. So,
π₯π₯4 β 81 = (π₯π₯2)2 β (9)2 = (π₯π₯2 + 9)(π₯π₯2 β 9)
However, this factorization is not complete yet. Notice that π₯π₯2 β 9 is also a difference of squares, so the original polynomial can be factored further. Thus,
π₯π₯4 β 81 = (π₯π₯2 + 9)(π₯π₯2 β 9) = οΏ½ππππ + πποΏ½(ππ + ππ)(ππ β ππ) Attention: The sum of squares, π₯π₯2 + 9, cannot be factored using real coefficients.
Generally, except for a common factor, a quadratic binomial of the form ππππ + ππππ is not factorable over the real numbers.
d. Following the difference of squares formula, we have
16 β (ππ β 2)2 = 42 β (ππ β 2)2
= [4 + (ππ β 2)][4 β (ππ β 2)]
= (4 + ππ β 2)(4 β ππ + 2) work out the inner brackets
= (ππ + ππ)(ππ β ππ) combine like terms
Perfect Squares
Another frequently used special factoring formula is the perfect square of a sum or a difference.
ππππ + ππππππ + ππππ = (ππ + ππ)ππ or
ππππ β ππππππ + ππππ = (ππ β ππ)ππ
Figure 3.2 shows the geometric interpretation of the perfect square of a sum. We encourage the reader to come up with a similar interpretation of the perfect square of a difference.
Remember to use brackets after the
negative sign!
Figure 3.2
ππ + ππ
ππ + ππ ππ2 ππππ
ππ2 ππππ
235
To factor a perfect square trinomial ππππ Β± ππππππ+ ππππ, we find ππ and ππ, which are the expressions being squared. Then, depending on the middle sign, we use ππ and ππ to form the perfect square of the sum (ππ + ππ)ππ, or the perfect square of the difference (ππ β ππ)ππ.
Identifying Perfect Square Trinomials Decide whether the given polynomial is a perfect square.
a. 9π₯π₯2 + 6π₯π₯ + 4 b. 9π₯π₯2 + 4π₯π₯2 β 12π₯π₯π₯π₯ c. 25ππ4 + 40ππ2 β 16 d. 49π₯π₯6 + 84π₯π₯π₯π₯3 + 36π₯π₯2 a. Observe that the outside terms of the trinomial 9π₯π₯2 + 6π₯π₯ + 4 are perfect squares, as
9π₯π₯2 = (3π₯π₯)2 and 4 = 22. So, the trinomial would be a perfect square if the middle terms would equal 2 β 3π₯π₯ β 2 = 12π₯π₯. Since this is not the case, our trinomial is not a perfect square.
Attention: Except for a common factor, trinomials of the type ππππ Β± ππππ + ππππ are not factorable over the real numbers!
b. First, we arrange the trinomial in decreasing order of the powers of π₯π₯. So, we obtain
9π₯π₯2 β 12π₯π₯π₯π₯ + 4π₯π₯2. Then, since 9π₯π₯2 = (3π₯π₯)2, 4π₯π₯2 = (2π₯π₯)2, and the middle term (except for the sign) equals 2 β 3π₯π₯ β 2 = 12π₯π₯, we claim that the trinomial is a perfect square. Since the middle term is negative, this is the perfect square of a difference. So, the trinomial 9π₯π₯2 β 12π₯π₯π₯π₯ + 4π₯π₯2 can be seen as
ππ2 β 2 ππ ππ + ππ2 = ( ππ β ππ )2
(3π₯π₯)2 β 2 β 3π₯π₯ β 2π₯π₯ + (2π₯π₯)2 = (3π₯π₯ β 2π₯π₯)2 c. Even though the coefficients of the trinomial 25ππ4 + 40ππ2 β 16 and the distribution
of powers seem to follow the pattern of a perfect square, the last term is negative, which makes it not a perfect square.
d. Since 49π₯π₯6 = (7π₯π₯3)2, 36π₯π₯2 = (6π₯π₯)2, and the middle term equals 2 β 7π₯π₯3 β 6π₯π₯ =
84π₯π₯π₯π₯3, we claim that the trinomial is a perfect square. Since the middle term is positive, this is the perfect square of a sum. So, the trinomial 9π₯π₯2 β 12π₯π₯π₯π₯ + 4π₯π₯2 can be seen as
ππ2 + 2 ππ ππ + ππ2 = ( ππ β ππ )2
(7π₯π₯3)2 + 2 β 7π₯π₯3 β 6π₯π₯ + (6π₯π₯)2 = (7π₯π₯3 β 6π₯π₯)2
Factoring Perfect Square Trinomials Factor each polynomial completely.
a. 25π₯π₯2 + 10π₯π₯ + 1 b. ππ2 β 12ππππ + 36ππ2 c. ππ2 β 8ππ + 16 β 49ππ2 d. β4π₯π₯2 β 144π₯π₯8 + 48π₯π₯5
Solution
236
a. The outside terms of the trinomial 25π₯π₯2 + 10π₯π₯ + 1 are perfect squares of 5ππ and 1, and the middle term equals 2 β 5π₯π₯ β 1 = 10π₯π₯, so we can follow the perfect square formula. Therefore,
25π₯π₯2 + 10π₯π₯ + 1 = (ππππ + ππ)ππ b. The outside terms of the trinomial ππ2 β 12ππππ + 36ππ2 are perfect squares of ππ and 6ππ,
and the middle term (disregarding the sign) equals 2 β ππ β 6ππ = 12ππππ, so we can follow the perfect square formula. Therefore,
ππ2 β 12ππππ + 36ππ2 = (ππ β ππππ)ππ c. Observe that the first three terms of the polynomial ππ2 β 8ππ + 16 β 49ππ2 form a
perfect square of ππβ 6 and the last term is a perfect square of 7ππ. So, we can write
ππ2 β 8ππ + 16 β 49ππ2 = (ππ β 6)2 β (7ππ)2
Notice that this way we have formed a difference of squares. So we can factor it by following the difference of squares formula
(ππ β 6)2 β (7ππ)2 = (ππβ ππβ πππ©π©)(ππβ ππ + πππ©π©)
d. As in any factoring problem, first we check the polynomial β4π₯π₯2 β 144π₯π₯8 + 48π₯π₯5 for a common factor, which is 4π₯π₯2. To leave the leading term of this polynomial positive, we factor out β4π₯π₯2. So, we obtain
β4π₯π₯2 β 144π₯π₯8 + 48π₯π₯5
= β4π₯π₯2 (1 + 36π₯π₯6 β 12π₯π₯3)
= β4π₯π₯2 (36π₯π₯6 β 12π₯π₯3 + 1)
= βππππππ οΏ½ππππππ β πποΏ½ππ
Sum or Difference of Cubes
The last special factoring formula to discuss in this section is the sum or difference of cubes.
ππππ + ππππ = (ππ + ππ)οΏ½ππππ β ππππ + πππποΏ½ or
ππππ β ππππ = (ππ β ππ)οΏ½ππππ + ππππ + πππποΏ½
The reader is encouraged to confirm these formulas by multiplying the factors in the right-hand side of each equation. In addition, we offer a geometric visualization of one of these formulas, the difference of cubes, as shown in Figure 3.3.
Solution
fold to the perfect square form
arrange the polynomial in decreasing powers
This is not in factored form yet!
Figure 3.3
ππ3 β ππ3 = ππ2(ππ β ππ)+ππππ(ππ β ππ) + ππ2(ππ β ππ)
ππ
ππ β ππ ππ
ππ
ππ
ππ β ππ
ππ β ππ
ππ
ππ
ππ3
= (ππ β ππ)(ππ2+ππππ + ππ2)
ππ3
237
Hints for memorization of the sum or difference of cubes formulas: β’ The binomial factor is a copy of the sum or difference of the terms that were originally
cubed. β’ The trinomial factor follows the pattern of a perfect square, except that the middle term
is single, not doubled. β’ The signs in the factored form follow the pattern Same-Opposite-Positive (SOP).
Factoring Sums or Differences of Cubes Factor each polynomial completely.
a. 8π₯π₯3 + 1 b. 27π₯π₯7π₯π₯ β 125π₯π₯π₯π₯4 c. 2ππ6 β 128 d. (ππ β 2)3 + ππ3
a. First, we rewrite each term of 8π₯π₯3 + 1 as a perfect cube of an expression.
ππ ππ
8π₯π₯3 + 1 = (2π₯π₯)2 + 12
Then, treating 2π₯π₯ as the ππ and 1 as the ππ in the sum of cubes formula ππ3 + ππ3 =(ππ + ππ)(ππ2 β ππππ + ππ2), we factor:
ππππ + ππππ = (ππ + ππ) οΏ½ ππππ β ππ ππ + πππποΏ½
8π₯π₯3 + 1 = (2π₯π₯)2 + 12 = (2π₯π₯ + 1)((2π₯π₯)2 β 2π₯π₯ β 1 + 12)
= (ππππ + ππ)οΏ½ππππππ β ππππ + πποΏ½
Notice that the trinomial 4π₯π₯2 β 2π₯π₯ + 1 in not factorable anymore. b. Since the two terms of the polynomial 27π₯π₯7π₯π₯ β 125π₯π₯π₯π₯4 contain the common factor
π₯π₯π₯π₯, we factor it out and obtain
27π₯π₯7π₯π₯ β 125π₯π₯π₯π₯4 = π₯π₯π₯π₯(27π₯π₯6 β 125π₯π₯3)
Observe that the remaining polynomial is a difference of cubes, (3π₯π₯2)3 β (5π₯π₯)3. So, we factor,
27π₯π₯7π₯π₯ β 125π₯π₯π₯π₯4 = π₯π₯π₯π₯[(3π₯π₯2)3 β (5π₯π₯)3]
( ππ + ππ ) οΏ½ ππππ β ππ ππ + πππποΏ½
= π₯π₯π₯π₯(3π₯π₯2 β 5π₯π₯)[(3π₯π₯2)2 + 3π₯π₯2 β 5π₯π₯ + (5π₯π₯)2]
= πππποΏ½ππππππ β πππποΏ½οΏ½ππππππ + ππππππππππ+ πππππππποΏ½ c. After factoring out the common factor 2, we obtain
2ππ6 β 128 = 2(ππ6 β 64)
Solution
Difference of squares or difference of cubes?
Quadratic trinomials of the form ππππ Β± ππππ + ππππ
are not factorable!
238
Notice that ππ6 β 64 can be seen either as a difference of squares, (ππ3)2 β 82, or as a difference of cubes, (ππ2)3 β 43. It turns out that applying the difference of squares formula first leads us to a complete factorization while starting with the difference of cubes does not work so well here. See the two approaches below.
(ππ3)2 β 82 (ππ2)3 β 43
= (ππ3 + 8)(ππ3 β 8) = (ππ2 β 4)(ππ4 + 4ππ2 + 16) = (ππ + 2)(ππ2 β 2ππ + 4)(ππ β 2)(ππ2 + 2ππ + 4) = (ππ + 2)(ππ β 2)(ππ4 + 4ππ2 + 16)
Therefore, the original polynomial should be factored as follows:
2ππ6 β 128 = 2(ππ6 β 64) = 2[(ππ3)2 β 82] = 2(ππ3 + 8)(ππ3 β 8)
= ππ(π©π© + ππ)οΏ½π©π©ππ β πππ©π© + πποΏ½(π©π© β ππ)οΏ½π©π©ππ + πππ©π© + πποΏ½
d. To factor (ππ β 2)3 + ππ3, we follow the sum of cubes formula (ππ + ππ)(ππ2 β ππππ + ππ2) by assuming ππ = ππ β 2 and ππ = ππ. So, we have
(ππ β 2)3 + ππ3 = (ππ β 2 + ππ) [(ππ β 2)2 β (ππ β 2)ππ + ππ2]
= (ππ β 2 + ππ) [ππ2 β 2ππππ + 4 β ππππ + 2ππ + ππ2]
= (ππ β ππ + ππ) οΏ½ππππ β ππππππ + ππ + ππππ + πππποΏ½
General Strategy of Factoring
Recall that a polynomial with integral coefficients is factored completely if all of its factors are prime over the integers.
How to Factorize Polynomials Completely?
1. Factor out all common factors. Leave the remaining polynomial with a positive leading term and integral coefficients, if possible.
2. Check the number of terms. If the polynomial has
- more than three terms, try to factor by grouping; a four term polynomial may require 2-2, 3-1, or 1-3 types of grouping.
- three terms, factor by guessing, decomposition, or follow the perfect square formula, if applicable.
- two terms, follow the difference of squares, or sum or difference of cubes formula, if applicable. Remember that sum of squares, ππππ + ππππ, is not factorable over the real numbers, except for possibly a common factor.
There is no easy way of factoring this trinomial!
4 prime factors, so the factorization is complete
239
3. Keep in mind the special factoring formulas:
Difference of Squares ππππ β ππππ = (ππ + ππ)(ππ β ππ) Perfect Square of a Sum ππππ + ππππππ + ππππ = (ππ + ππ)ππ Perfect Square of a Difference ππππ β ππππππ + ππππ = (ππ β ππ)ππ Sum of Cubes ππππ + ππππ = (ππ + ππ)οΏ½ππππ β ππππ + πππποΏ½ Difference of Cubes ππππ β ππππ = (ππ β ππ)οΏ½ππππ + ππππ + πππποΏ½ 4. Keep factoring each of the obtained factors until all of them are prime over the
integers.
Multiple-step Factorization Factor each polynomial completely.
a. 80π₯π₯5 β 5π₯π₯ b. 4ππ2 β 4ππ + 1 β ππ2 c. (5ππ + 8)2 β 6(5ππ + 8) + 9 d. (ππ β 2ππ)3 + (ππ + 2ππ)3
a. First, we factor out the GCF of 80π₯π₯5 and β5π₯π₯, which equals to 5π₯π₯. So, we obtain
80π₯π₯5 β 5π₯π₯ = 5π₯π₯(16π₯π₯4 β 1)
Then, we notice that 16π₯π₯4 β 1 can be seen as the difference of squares (4π₯π₯2)2 β 12. So, we factor further
80π₯π₯5 β 5π₯π₯ = 5π₯π₯(4π₯π₯2 + 1)(4π₯π₯2 β 1)
The first binomial factor, 4π₯π₯2 + 1, cannot be factored any further using integral coefficients as it is the sum of squares, (2π₯π₯)2 + 12. However, the second binomial factor, 4π₯π₯2 β 1, is still factorable as a difference of squares, (2π₯π₯)2 β 12. Therefore,
80π₯π₯5 β 5π₯π₯ = πππποΏ½ππππππ + πποΏ½(ππππ + ππ)(ππππ β ππ)
This is a complete factorization as all the factors are prime over the integers. b. The polynomial 4ππ2 β 4ππ + 1 β ππ2 consists of four terms, so we might be able to
factor it by grouping. Observe that the 2-2 type of grouping has no chance to succeed, as the first two terms involve only the variable ππ while the second two terms involve only the variable ππ. This means that after factoring out the common factor in each group, the remaining binomials would not be the same. So, the 2-2 grouping would not lead us to a factorization. However, the 3-1 type of grouping should help. This is because the first three terms form the perfect square, (2ππ β 1)2, and there is a subtraction before the last term ππ2, which is also a perfect square. So, in the end, we can follow the difference of squares formula to complete the factoring process.
4ππ2 β 4ππ + 1οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β ππ2οΏ½ = (2ππ β 1)2 β ππ2 = (ππππ β ππ β ππ)(ππππ β ππ + ππ)
Solution
3-1 type of grouping
repeated difference of squares
240
c. To factor (5ππ + 8)2 β 6(5ππ + 8) + 9, it is convenient to substitute a new variable, say ππ, for the expression 5ππ + 8. Then,
(5ππ + 8)2 β 6(5ππ + 8) + 9 = ππ2 β 6ππ + 9
= (ππ β 3)2
= (5ππ + 8 β 3)2
= (5ππ + 5)2
Notice that 5ππ + 5 can still be factored by taking the 5 out. So, for a complete factorization, we factor further
(5ππ + 5)2 = οΏ½5(ππ + 1)οΏ½2 = ππππ(ππ + ππ)ππ
d. To factor (ππ β 2ππ)3 + (ππ + 2ππ)3, we follow the sum of cubes formula (ππ + ππ)(ππ2 β
ππππ + ππ2) by assuming ππ = ππ β 2ππ and ππ = ππ + 2ππ. So, we have
(ππ β 2ππ)3 + (ππ + 2ππ)3
= (ππ β 2ππ + ππ + 2ππ) [(ππ β 2ππ)2 β (ππ β 2ππ)(ππ + 2ππ) + (ππ + 2ππ)2]
= 2ππ [ππ2 β 4ππππ + 4ππ2 β (ππ2 β 4ππ2) + ππ2 + 4ππππ + 4ππ2]
= 2ππ (2ππ2 + 8ππ2 β ππ2 + 4ππ2) = πππποΏ½ππππ + πππππππποΏ½
F.3 Exercises
Vocabulary Check Complete each blank with one of the suggested words, or the most appropriate term or
phrase from the given list: difference of cubes, difference of squares, perfect square, sum of cubes, sum of squares.
1. If a binomial is a ________________________________ its factorization has the form (ππ + ππ)(ππ β ππ).
2. Trinomials of the form ππ2 Β± 2ππππ + ππ2 are ____________________________ trinomials.
3. The product (ππ + ππ)(ππ2 β ππππ + ππ2) is the factorization of the __________________________.
4. The product (ππ β ππ)(ππ2 + ππππ + ππ2) is the factorization of the __________________________.
5. A ________________________ is not factorable.
6. Quadratic trinomials of the form ππ2 Β± ππππ + ππ2 _________________ππππππ /ππππππ πππππ‘π‘
factorable.
go back to the original variable
perfect square! factoring by substitution
multiple special formulas and simplifying
Remember to represent the new variable by a
different letter than the original variable!
241
Concept Check Determine whether each polynomial is a perfect square, a difference of squares, a sum or difference of cubes, or neither.
7. 0.25π₯π₯2 β 0.16π₯π₯2 8. π₯π₯2 β 14π₯π₯ + 49
9. 9π₯π₯4 + 4π₯π₯2 + 1 10. 4π₯π₯2 β (π₯π₯ + 4)2
11. 125π₯π₯3 β 64 12. π₯π₯12 + 0.008π₯π₯3
13. βπ₯π₯4 + 16π₯π₯4 14. 64 + 48π₯π₯3 + 9π₯π₯6
15. 25π₯π₯6 β 10π₯π₯3π₯π₯2 + π₯π₯4 16. β4π₯π₯6 β π₯π₯6
17. β8π₯π₯3 + 27π₯π₯6 18. 81π₯π₯2 β 16π₯π₯
Concept Check
19. The binomial 4π₯π₯2 + 64 is an example of a sum of two squares that can be factored. Under what conditions can the sum of two squares be factored?
20. Insert the correct signs into the blanks. a. 8 + ππ3 = (2 ___ ππ)(4 ___ 2ππ ___ ππ2) b. ππ3 β 1 = (ππ ___ 1)(ππ2 ___ ππ ___ 1) Factor each polynomial completely, if possible.
21. π₯π₯2 β π₯π₯2 22. π₯π₯2 + 2π₯π₯π₯π₯ + π₯π₯2 23. π₯π₯3 β π₯π₯3
24. 16π₯π₯2 β 100 25. 4π§π§2 β 4π§π§ + 1 26. π₯π₯3 + 27
27. 4π§π§2 + 25 28. π₯π₯2 + 18π₯π₯ + 81 29. 125β π₯π₯3
30. 144π₯π₯2 β 64π₯π₯2 31. ππ2 + 20ππππ + 100ππ2 32. 27ππ3ππ6 + 1
33. 9ππ4 β 25ππ6 34. 25 β 40π₯π₯ + 16π₯π₯2 35. ππ6 β 64ππ3
36. 16π₯π₯2π§π§2 β 100π₯π₯2 37. 4 + 49ππ2 + 28ππ 38. π₯π₯12 + 0.008π₯π₯3
39. ππ4 β 9ππ2 40. 9ππ2 β 12ππππ β 4ππ2 41. 18β ππ3
42. 0.04π₯π₯2 β 0.09π₯π₯2 43. π₯π₯4 + 8π₯π₯2 + 1 44. β 127
+ π‘π‘3
45. 16π₯π₯6 β 121π₯π₯2π₯π₯4 46. 9 + 60ππππ + 100ππ2ππ2 47. βππ3ππ3 β 125ππ6
48. 36ππ2ππ β 1 49. 9ππ8 β 48ππ4ππ + 64ππ2 50. 9π₯π₯3 + 8
51. (π₯π₯ + 1)2 β 49 52. 14π’π’2 β π’π’π’π’ + π’π’2 53. 2π‘π‘4 β 128π‘π‘
54. 81 β (ππ + 3)2 55. π₯π₯2ππ + 6π₯π₯ππ + 9 56. 8 β (ππ + 2)3
57. 16π§π§4 β 1 58. 5ππ3 + 20ππ2 + 20ππ 59. (π₯π₯ + 5)3 β π₯π₯3
60. ππ4 β 81ππ4 61. 0.25π§π§2 β 0.7π§π§ + 0.49 62. (π₯π₯ β 1)3 + (π₯π₯ + 1)3
63. (π₯π₯ β 2π₯π₯)2 β (π₯π₯ + π₯π₯)2 64. 0.81ππ8 + 9ππ4 + 25 65. (π₯π₯ + 2)3 β (π₯π₯ β 2)3
242
Factor each polynomial completely.
66. 3π₯π₯3 β 12π₯π₯2π₯π₯ 67. 2π₯π₯2 + 50ππ2 β 20πππ₯π₯ 68. π₯π₯3 β π₯π₯π₯π₯2 + π₯π₯2π₯π₯ β π₯π₯3
69. π₯π₯2 β 9ππ2 + 12π₯π₯ + 36 70. 64π’π’6 β 1 71. 7ππ3 + ππ6 β 8
72. β7ππ2 + 2ππ3 + 4ππ β 14 73. ππ8 β ππ8 74. π₯π₯9 β π₯π₯
75. (π₯π₯2 β 2)2 β 4(π₯π₯2 β 2) β 21 76. 8(ππ β 3)2 β 64(ππ β 3) + 128 77. ππ2 β ππ2 β 6ππ β 9
78. 25(2ππ β ππ)2 β 9 79. 3π₯π₯2π₯π₯2π§π§ + 25π₯π₯π₯π₯π§π§2 + 28π§π§3 80. π₯π₯8ππ β π₯π₯2
81. π₯π₯6 β 2π₯π₯5 + π₯π₯4 β π₯π₯2 + 2π₯π₯ β 1 82. 4π₯π₯2π₯π₯4 β 9π₯π₯4 β 4π₯π₯2π§π§4 + 9π§π§4 83. ππ2π€π€+1 + 2πππ€π€+1 + ππ
243
F.4 Solving Polynomial Equations and Applications of Factoring
Many application problems involve solving polynomial equations. In Chapter L, we studied methods for solving linear, or first-degree, equations. Solving higher degree polynomial equations requires other methods, which often involve factoring. In this chapter, we study solving polynomial equations using the zero-product property, graphical connections between roots of an equation and zeros of the corresponding function, and some application problems involving polynomial equations or formulas that can be solved by factoring.
Zero-Product Property
Recall that to solve a linear equation, for example 2π₯π₯ + 1 = 0, it is enough to isolate the variable on one side of the equation by applying reverse operations. Unfortunately, this method usually does not work when solving higher degree polynomial equations. For example, we would not be able to solve the equation π₯π₯2 β π₯π₯ = 0 through the reverse operation process, because the variable π₯π₯ appears in different powers.
So β¦ how else can we solve it?
In this particular example, it is possible to guess the solutions. They are π₯π₯ = 0 and π₯π₯ = 1.
But how can we solve it algebraically?
It turns out that factoring the left-hand side of the equation π₯π₯2 β π₯π₯ = 0 helps. Indeed, π₯π₯(π₯π₯ β 1) = 0 tells us that the product of π₯π₯ and π₯π₯ β 1 is 0. Since the product of two quantities is 0, at least one of them must be 0. So, either π₯π₯ = 0 or π₯π₯ β 1 = 0, which solves to π₯π₯ = 1.
The equation discussed above is an example of a second degree polynomial equation, more commonly known as a quadratic equation.
Definition 4.1 A quadratic equation is a second degree polynomial equation in one variable that can be written in the form,
ππππππ + ππππ + ππ = ππ,
where ππ, ππ, and ππ are real numbers and ππ β 0. This form is called standard form.
One of the methods of solving such equations involves factoring and the zero-product property that is stated below.
For any real numbers ππ and ππ,
ππππ = ππ if and only if ππ = ππ or ππ = ππ
This means that any product containing a factor of 0 is equal to 0, and conversely, if a product is equal to 0, then at least one of its factors is equal to 0.
The implication βif ππ = ππ or ππ = ππ, then ππππ = ππβ is true by the multiplicative property of zero.
To prove the implication βif ππππ = ππ, then ππ = ππ or ππ = ππβ, let us assume first that ππ β 0. (As, if ππ = 0, then the implication is already proven.)
Zero-Product Theorem
Proof
244
Since ππ β 0, then 1ππ exists. Therefore, both sides of ππππ = 0 can be multiplied by 1
ππ and we
obtain
1ππβ ππππ =
1ππβ 0
ππ = 0,
which concludes the proof.
Attention: The zero-product property works only for a product equal to 0. For example, the fact that ππππ = ππ does not mean that either ππ or ππ equals to 1.
Using the Zero-Product Property to Solve Polynomial Equations
Solve each equation.
a. (π₯π₯ β 3)(2π₯π₯ + 5) = 0 b. 2π₯π₯(π₯π₯ β 5)2 = 0 a. Since the product of π₯π₯ β 3 and 2π₯π₯ + 5 is equal to zero, then by the zero-product
property at least one of these expressions must equal to zero. So,
π₯π₯ β 3 = 0 or 2π₯π₯ + 5 = 0
which results in π₯π₯ = 3 or 2π₯π₯ = β5 π₯π₯ = β5
2
Thus, οΏ½β ππππ
,πποΏ½ is the solution set of the given equation.
b. Since the product 2π₯π₯(π₯π₯ β 5)2 is zero, then either π₯π₯ = 0 or π₯π₯ β 5 = 0, which solves to π₯π₯ = 5. Thus, the solution set is equal to {ππ,ππ}.
Note 1: The factor of 2 does not produce any solution, as 2 is never equal to 0.
Note 2: The perfect square (π₯π₯ β 5)2 equals to 0 if and only if the base π₯π₯ β 5 equals to 0.
Solving Polynomial Equations by Factoring
To solve polynomial equations of second or higher degree by factoring, we
β’ arrange the polynomial in decreasing order of powers on one side of the equation, β’ keep the other side of the equation equal to 0, β’ factor the polynomial completely, β’ use the zero-product property to form linear equations for each factor, β’ solve the linear equations to find the roots (solutions) to the original equation.
Solution
245
Solving Quadratic Equations by Factoring Solve each equation by factoring.
a. π₯π₯2 + 9 = 6π₯π₯ b. 15π₯π₯2 β 12π₯π₯ = 0
c. (π₯π₯ + 2)(π₯π₯ β 1) = 4(3 β π₯π₯) β 8 d. (π₯π₯ β 3)2 = 36π₯π₯2
a. To solve π₯π₯2 + 9 = 6π₯π₯ by factoring we need one side of this equation equal to 0. So, first, we move the 6π₯π₯ term to the left side of the equation,
π₯π₯2 + 9 β 6π₯π₯ = 0,
and arrange the terms in decreasing order of powers of π₯π₯,
π₯π₯2 β 6π₯π₯ + 9 = 0.
Then, by observing that the resulting trinomial forms a perfect square of π₯π₯ β 3, we factor
(π₯π₯ β 3)2 = 0, which is equivalent to
π₯π₯ β 3 = 0, and finally
π₯π₯ = 3.
So, the solution is π₯π₯ = ππ.
b. After factoring the left side of the equation 15π₯π₯2 β 12π₯π₯ = 0,
3π₯π₯(5π₯π₯ β 4) = 0,
we use the zero-product property. Since 3 is never zero, the solutions come from the equations
π₯π₯ = ππ or 5π₯π₯ β 4 = 0.
Solving the second equation for π₯π₯, we obtain
5π₯π₯ = 4, and finally
π₯π₯ = ππππ.
So, the solution set consists of 0 and ππππ.
c. To solve (π₯π₯ + 2)(π₯π₯ β 1) = 4(3 β π₯π₯) β 8 by factoring, first, we work out the brackets and arrange the polynomial in decreasing order of exponents on the left side of the equation. So, we obtain
π₯π₯2 + π₯π₯ β 2 = 12β 4π₯π₯ β 8
π₯π₯2 + 5π₯π₯ β 6 = 0
(π₯π₯ + 6)(π₯π₯ β 1) = 0
Solution
246
Now, we can read the solutions from each bracket, that is, π₯π₯ = βππ and π₯π₯ = ππ.
Observation: In the process of solving a linear equation of the form πππ₯π₯ + ππ = 0, first we subtract ππ and then we divide by ππ. So the solution, sometimes
referred to as the root, is π₯π₯ = βππππ. This allows us to read the solution
directly from the equation. For example, the solution to π₯π₯ β 1 = 0 is π₯π₯ = 1 and the solution to 2π₯π₯ β 1 = 0 is π₯π₯ = 1
2.
d. To solve (π₯π₯ β 3)2 = 36π₯π₯2, we bring all the terms to one side and factor the obtained difference of squares, following the formula ππ2 β ππ2 = (ππ + ππ)(ππ β ππ). So, we have
(π₯π₯ β 3)2 β 36π₯π₯2 = 0
(π₯π₯ β 3 + 6π₯π₯)(π₯π₯ β 3 β 6π₯π₯) = 0
(7π₯π₯ β 3)(β5π₯π₯ β 3) = 0
Then, by the zero-product property,
7π₯π₯ β 3 = 0 or β5π₯π₯ β 3, which results in
π₯π₯ = ππππ or π₯π₯ = βππ
ππ.
Solving Polynomial Equations by Factoring
Solve each equation by factoring.
a. 2π₯π₯3 β 2π₯π₯2 = 12π₯π₯ b. π₯π₯4 + 36 = 13π₯π₯2
a. First, we bring all the terms to one side of the equation and then factor the resulting polynomial.
2π₯π₯3 β 2π₯π₯2 = 12π₯π₯
2π₯π₯3 β 2π₯π₯2 β 12π₯π₯ = 0
2π₯π₯(π₯π₯2 β π₯π₯ β 6) = 0
2π₯π₯(π₯π₯ β 3)(π₯π₯ + 2) = 0
By the zero-product property, the factors π₯π₯, (π₯π₯ β 3) and (π₯π₯ + 2), give us the corresponding solutions, 0, 3, and β 2. So, the solution set of the given equation is {ππ,ππ,βππ}.
b. Similarly as in the previous examples, we solve π₯π₯4 + 36 = 13π₯π₯2 by factoring and
using the zero-product property. Since
π₯π₯4 β 13π₯π₯2 + 36 = 0
Solution
247
(π₯π₯2 β 4)(π₯π₯2 β 9) = 0
(π₯π₯ + 2)(π₯π₯ β 2)(π₯π₯ + 3)(π₯π₯ β 3) = 0,
then, the solution set of the original equation is {βππ,ππ,βππ,ππ}
Observation: ππ-th degree polynomial equations may have up to ππ roots (solutions).
Factoring in Applied Problems
Factoring is a useful strategy when solving applied problems. For example, factoring is often used in solving formulas for a variable, in finding roots of a polynomial function, and generally, in any problem involving polynomial equations that can be solved by factoring.
Solving Formulas with the Use of Factoring
Solve each formula for the specified variable.
a. π΄π΄ = 2πππ€π€ + 2π€π€π€π€ + 2π€π€ππ, for ππ b. π π = 2ππ+3ππ
, for ππ a. To solve π΄π΄ = 2πππ€π€ + 2π€π€π€π€ + 2π€π€ππ for ππ, we want to keep both terms containing ππ on
the same side of the equation and bring the remaining terms to the other side. Here is an equivalent equation,
π΄π΄ β 2π€π€π€π€ = 2πππ€π€ + 2π€π€ππ,
which, for convenience, could be written starting with β-terms:
2πππ€π€ + 2π€π€ππ = π΄π΄ β 2π€π€π€π€
Now, factoring ππ out causes that ππ appears in only one place, which is what we need to isolate it. So,
(2π€π€ + 2π€π€)ππ = π΄π΄ β 2π€π€π€π€
ππ =π¨π¨ β ππππππππππ + ππππ
Notice: In the above formula, there is nothing that can be simplified. Trying to reduce 2 or 2π€π€ or π€π€ would be an error, as there is no essential common factor that can be carried out of the numerator.
b. When solving π π = 2ππ+3ππ
for ππ, our goal is to, firstly, keep the variable ππ in the numerator and secondly, to keep it in a single place. So, we have
π π =2ππ + 3ππ
π π ππ = 2ππ + 3
Solution
/ Γ· (2π€π€ + 2π€π€)
/ β π‘π‘
/ β2π‘π‘
248
π π ππ β 2ππ = 3
ππ(π π β 2) = 3
ππ =ππ
ππ β ππ.
Finding Roots of a Polynomial Function
A small rocket is launched from the ground vertically upwards with an initial velocity of 128 feet per second. Its height in feet after π‘π‘ seconds is a function defined by
β(π‘π‘) = β16π‘π‘2 + 128π‘π‘.
After how many seconds will the rocket hit the ground? The rocket hits the ground when its height is 0. So, we need to find the time π‘π‘ for which β(π‘π‘) = 0. Therefore, we solve the equation
β16π‘π‘2 + 128π‘π‘ = 0 for π‘π‘. From the factored form
β16π‘π‘(π‘π‘ β 8) = 0
we conclude that the rocket is on the ground at times 0 and 8 seconds. So the rocket hits the ground after 8 seconds from its launch.
Solving an Application Problem with the Use of Factoring
The height of a triangle is 1 meter less than twice the length of the base. The area is 14 m2. What are the measures of the base and the height? Let ππ and β represent the base and the height of the triangle, correspondingly. The first sentence states that β is 1 less than 2 times ππ. So, we record
β = 2ππ β 1.
Using the formula for area of a triangle, π΄π΄ = 12ππβ, and the fact that π΄π΄ = 14, we obtain
14 =12ππ(2ππ β 1).
Since this is a one-variable quadratic equation, we will attempt to solve it by factoring, after bringing all the terms to one side of the equation. So, we have
0 =12ππ(2ππ β 1) β 14
0 = ππ(2ππ β 1) β 28
0 = 2ππ2 β ππ β 28
Solution
Solution
/ β 2
to clear the fraction, multiply each term by 2 before working out the bracket
factor π‘π‘
/ Γ· (π π β 2)
249
0 = (2ππ + 7)(ππ β 4),
which by the zero-product property gives us ππ = β72 or ππ = 4. Since ππ represents the length
of the base, it must be positive. So, the base is 4 meters long and the height is β = 2ππ β1 = 2 β 4 β 1 = ππ meters long.
F.4 Exercises
Vocabulary Check Complete each blank with the most appropriate term from the given list: factored, linear,
n, zero, zero-product.
1. The ________________________ property states that if ππππ = 0 then either ππ = 0 or ππ = 0.
2. When a quadratic equation is solved by factoring, the zero-product property is used to form two ______________ equations.
3. The zero-product property can be applied only when one side of the equation is equal to __________ and the other side is in a ________________ form.
4. An ππ-th degree polynomial equation may have up to _____ solutions. Concept Check True or false.
5. If π₯π₯π₯π₯ = 0 then π₯π₯ = 0 or π₯π₯ = 0.
6. If ππππ = 1 then ππ = 1 or ππ = 1.
7. If π₯π₯ + π₯π₯ = 0 then π₯π₯ = 0 or π₯π₯ = 0.
8. If ππ2 = 0 then ππ = 0.
9. If π₯π₯2 = 1 then π₯π₯ = 1. Concept Check
10. Which of the following equations is not in proper form for using the zero-product property.
a. π₯π₯(π₯π₯ β 1) + 3(π₯π₯ β 1) = 0 b. (π₯π₯ + 3)(π₯π₯ β 1) = 0
c. π₯π₯(π₯π₯ β 1) = 3(π₯π₯ β 1) d. (π₯π₯ + 3)(π₯π₯ β 1) = β3 Solve each equation.
11. 3(π₯π₯ β 1)(π₯π₯ + 4) = 0 12. 2(π₯π₯ + 5)(π₯π₯ β 7) = 0
13. (3π₯π₯ + 1)(5π₯π₯ + 4) = 0 14. (2π₯π₯ β 3)(4π₯π₯ β 1) = 0
15. π₯π₯2 + 9π₯π₯ + 18 = 0 16. π₯π₯2 β 18π₯π₯ + 80 = 0
250
17. 2π₯π₯2 = 7 β 5π₯π₯ 18. 3ππ2 = 14ππ β 8
19. π₯π₯2 + 6π₯π₯ = 0 20. 6π₯π₯2 β 3π₯π₯ = 0
21. (4 β ππ)2 = 0 22. (2ππ + 5)2 = 0
23. 0 = 4ππ2 β 20ππ + 25 24. 0 = 16π₯π₯2 + 8π₯π₯ + 1
25. ππ2 β 32 = β4ππ 26. 19ππ + 36 = 6ππ2
27. π₯π₯2 + 3 = 10π₯π₯ β 2π₯π₯2 28. 3π₯π₯2 + 9π₯π₯ + 30 = 2π₯π₯2 β 2π₯π₯
29. (3π₯π₯ + 4)(3π₯π₯ β 4) = β10π₯π₯ 30. (5π₯π₯ + 1)(π₯π₯ + 3) = β2(5π₯π₯ + 1)
31. 4(π₯π₯ β 3)2 β 36 = 0 32. 3(ππ + 5)2 β 27 = 0
33. (π₯π₯ β 3)(π₯π₯ + 5) = β7 34. (π₯π₯ + 8)(π₯π₯ β 2) = β21
35. (2π₯π₯ β 1)(π₯π₯ β 3) = π₯π₯2 β π₯π₯ β 2 36. 4π₯π₯2 + π₯π₯ β 10 = (π₯π₯ β 2)(π₯π₯ + 1)
37. 4(2π₯π₯ + 3)2 β (2π₯π₯ + 3) β 3 = 0 38. 5(3π₯π₯ β 1)2 + 3 = β16(3π₯π₯ β 1)
39. π₯π₯3 + 2π₯π₯2 β 15π₯π₯ = 0 40. 6π₯π₯3 β 13π₯π₯2 β 5π₯π₯ = 0
41. 25π₯π₯3 = 64π₯π₯ 42. 9π₯π₯3 = 49π₯π₯
43. π₯π₯4 β 26π₯π₯2 + 25 = 0 44. ππ4 β 50ππ2 + 49 = 0
45. π₯π₯3 β 6π₯π₯2 = β8π₯π₯ 46. π₯π₯3 β 2π₯π₯2 = 3π₯π₯
47. ππ3 + ππ2 β 9ππ β 9 = 0 48. 2π₯π₯3 β π₯π₯2 β 2π₯π₯ + 1 = 0
49. 5π₯π₯3 + 2π₯π₯2 β 20π₯π₯ β 8 = 0 50. 2π₯π₯3 + 3π₯π₯2 β 18π₯π₯ β 27 = 0 Discussion Point
51. A student tried to solve the equation π₯π₯3 = 9π₯π₯ by first dividing each side by π₯π₯, obtaining π₯π₯2 = 9. She then solved the resulting equation by the zero-product property and obtained the solution set {β3,3}. Is this a complete solution? Explain your reasoning.
Analytic Skills
52. Given that ππ(π₯π₯) = π₯π₯2 + 14π₯π₯ + 50, find all values of π₯π₯ such that ππ(π₯π₯) = 5.
53. Given that ππ(π₯π₯) = 2π₯π₯2 β 15π₯π₯, find all values of π₯π₯ such that ππ(π₯π₯) = β7.
54. Given that ππ(π₯π₯) = 2π₯π₯2 + 3π₯π₯ and ππ(π₯π₯) = β6π₯π₯ + 5, find all values of π₯π₯ such that ππ(π₯π₯) = ππ(π₯π₯).
55. Given that ππ(π₯π₯) = 2π₯π₯2 + 11π₯π₯ β 16 and β(π₯π₯) = 5 + 2π₯π₯ β π₯π₯2, find all values of π₯π₯ such that ππ(π₯π₯) = β(π₯π₯).
Solve each equation for the specified variable.
56. π·π·πππ‘π‘ = π΄π΄ β π·π·, for π·π· 57. 3ππ + 2ππ = 5 β ππππ, for ππ
58. 5ππ + ππππ = ππ β 2ππ, for ππ 59. πΈπΈ = π π +ππππ
, for ππ
251
60. π§π§ = π₯π₯+2ππ
ππ, for ππ 61. ππ = β2ππ+4
ππ, for ππ
Analytic Skills
Use factoring the GCF out to solve each formula for the indicated variable.
62. An object is thrown downwards, with an initial speed of 16 ft/s, from the top of a building 480 ft high. If the distance travelled by the object, in ft, is given by the function ππ(π‘π‘) = π’π’π‘π‘ + 16π‘π‘2, where π’π’ is the initial speed in ft/s, and π‘π‘ is the time in seconds, then how many seconds later will the object hit the ground?
63. A sandbag is dropped from a hot-air balloon 900 ft above the ground. The height, β, of the sandbag above the ground, in feet, after π‘π‘ seconds is given by the function β(π‘π‘) = 900 β 16π‘π‘2. When will the sandbag hit the ground?
64. The sum of a number and its square is 72. Find the number.
65. The sum of a number and its square is 210. Find the number.
66. The length of a rectangle is 2 meters more than twice the width. The area of the rectangle is 84 m2. Find the length and width of the rectangle.
67. An envelope is 4 cm longer than it is wide. The area is 96 cm2. Find its length and width.
68. The height of a triangle is 8 cm more than the length of the base. The area of the triangle is 64 cm2. Find the base and height of the triangle.
69. A triangular sail is 9 m taller than it is wide. The area is 56 m2. Find the height and the base of the sail.
70. A gardener decides to build a stone pathway of uniform width around her flower bed. The flower bed measures 10 ft by 12 ft. If she wants the area of the bed and the pathway together to be 224 ft2, how wide should she make the pathway?
71. A rectangular flower bed is to be 3 m longer than it is wide. The flower bed will have an area of 108 m2. What will its dimensions be?
72. A picture frame measures 12 cm by 20 cm, and 84 cm2 of picture shows. Find the width of the frame.
73. A picture frame measures 14 cm by 20 cm, and 160 cm2 of picture shows. Find the width of the frame.
74. If each of the sides of a square is lengthened by 6 cm, the area becomes 144 cm2. Find the length of a side of the original square.
75. If each of the sides of a square is lengthened by 4 m, the area becomes 49 m2. Find the length of a side of the original square.
ππ
ππ + 8
2π€π€ + 2
π€π€