polynomial and its types · polynomial and its types by now you are aware of the polynomial...
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Polynomial and Its Types
By now you are aware of the polynomial equation in one variable and
their degrees. In this article, we will look at the various types of
polynomials to establish a foundation for further studies into them.
Introduction to Polynomial Equation
If p(x) is a polynomial equation in x, then the highest power of x in
p(x) is called the degree of the polynomial p(x). So, p(x) =
● 4x + 2 is a polynomial equation in the variable x of degree 1
● 2y2 – 3y + 4 is a polynomial in the variable y of degree 2
● 5x3 – 4x2 + x – 2 is a polynomial in the variable x of degree 3
● 7u6 – 3u4 + 4u2 – 6 is a polynomial in the variable u of degree
6
Further, it is important to note that the following expressions are NOT
polynomials:
● 1 / (x – 1)
● √x + 2
● 1 / (x2 + 2x + 3)
Types of Polynomials
Let’s look at the different types of polynomials that you will come
across while studying them.
Linear Polynomials
Any polynomial with a variable of degree one is a Linear Polynomial.
Some examples of the linear polynomial equation are as follows:
● 2x – 3
● y + √2
● x √3 + 5
● x + 5/11
● 2/3y – 5
Any polynomial where the degree of the variable is greater than 1 is
not a linear polynomial.
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Quadratic Polynomials
Any polynomial with a variable of degree two is a Quadratic
Polynomial. The name ‘quadratic’ is derived from the word ‘quadrate’
which means square. Some examples of the quadratic polynomial
equation are as follows:
● 2x2 + 3x – 5
● y2 – 1
● 2 – x2 + x√3
● u/3 – 2u2 + 5
● v2√5 + 2/3v – 6
● 4z2 + 1/7
To generalize, most quadratic polynomials in x are expressed as, ax2 +
bx + c … where a, b, and c are real numbers where a ≠ 0.
Cubic Polynomials
Any polynomial with a variable of degree three is a Cubic Polynomial.
Some examples of the cubic polynomial equation are as follows:
● x3
● 2 – x3
● x3√2
● x3 – x2 + 3
● 3x3 – 2x2 + x – 1
To generalize, most quadratic polynomials in x are expressed as, ax3 +
bx2 + cx + d … where a, b, c, and d are real numbers. Also, a ≠ 0.
Some more concepts
To begin with, let’s look at the following polynomial p(x),
p(x) = x2 – 3x – 4
Next, let’s put x = 2 in p(x). So, we get p(2) = (2)2 – 3(2) – 4 = 4 – 6 –
4 = –6. Note that the value ‘– 6’ is obtained by replacing x with 2 in
the polynomial x2 – 3x – 4. Hence, it is called ‘the value of x2 – 3x – 4
at x = 2’. Similarly, p(0) is the value of x2 – 3x – 4 at x = 0.
Therefore, we can say, If p(x) is a polynomial in x, and if k is any real
number, then the value obtained by replacing x by k in p(x), is called
the value of p(x) at x = k and is denoted by p(k).
Zero of a Polynomial
So, what is the value of p(x) = x2 – 3x – 4 at x = – 1? p(– 1) = (– 1)2 –
{3(– 1)} – 4 = 1 + 3 – 4 = 0. Also, the value of p(x) = x2 – 3x – 4 at x
= 4 is, p(4) = (4)2 – 3(4) – 4 = 16 – 12 – 4 = 0. In this case, since p(-1)
and p(4) is equal to zero, ‘-1’ and ‘4’ are called zeroes of the quadratic
polynomial x2 – 3x – 4.
Therefore, we can say, A real number k is said to be a zero of a
polynomial p(x) if p(k) = 0. In the previous years, you have already
studied how to find zeroes of a polynomial equation. To elaborate on
it a little more, if ‘k’ is a zero of p(x) = 2x + 3, then
p(k) = 0
Or 2k + 3 = 0
i.e. k = – 3/2
Let’s generalize this. If ‘k’ is a zero of p(x) = ax + b, then p(k) = ak +
b = 0. Or, k = – b/a. In other words, the zero of the linear polynomial
(ax + b) is: – (Constant Term) / (Coefficient of x)
Hence, we can conclude that the zero of a linear polynomial equation
is related to its coefficients. You will study more about if this rule is
applicable to all types of polynomials discussed above.
Solved Examples for You
Question: What are the three types of polynomials and how are they
differentiated?
Solution: The three types of polynomials are:
1. Linear
2. Quadratic
3. Cubic
The linear polynomials have a variable of degree one, quadratic
polynomials have a variable with degree two and cubic polynomials
have a variable with degree three.
Value of Polynomial and Division Algorithm
Arithmetic operations like addition, subtraction, multiplication and
division play a huge and most basic rule in Mathematics. Maths is
made by these operations. All other operations go easy with the
polynomials except the division operation, which gets complex when
dealt with polynomials. But this section will explain to you the
division of polynomials and the division algorithm related to it, from
basics.
So, what’s the basic formula we are learning from the day we solved
our first division problem? This is:
Dividend = Quotient × Divisor + Remainder
Example: Divide the polynomial 2x2+3x+1 by polynomial x+2.
Solution: Divisor= x+2
Dividend=2x2 + 3x + 1
Note: Put the dividend under the division sign and divisor outside the
sign.
Steps for Division of Polynomials
● Step 1: Firstly, Arrange the divisor as well as dividend
individually in decreasing order of their degree of terms.
● Step 2: In case of division we seek to find the quotient. To find
the very first term of the quotient, divide the first term of the
dividend by the highest degree term in the divisor. In the
current case,
2x2/x = 2x.
● Step 3: Write 2x in place of the quotient.
● Step 4: Multiply the divisor by the quotient obtained. Put the
product underneath the dividend.
● Step 5: Subtract the product obtained as happens in case of a
division operation.
● Step 6: Write the result obtained after drawing another bar to
separate it from prior operations performed.
● Step 7: Bring down the remaining terms of the dividend.
● Step 8: Again divide the dividend by the highest degree term of
the remaining divisor. Follow the same prior procedure until
either the remainder becomes zero or its degree is less than the
degree of the divisor.
● Step 9: At this stage, the quotient obtained is our answer.
Quotient Obtained = 2x + 1
Note: Division Algorithms for Polynomials is same as the Long
Division Algorithm In Polynomials
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Division Algorithm For Polynomials
Division algorithm for polynomials states that, suppose f(x) and g(x)
are the two polynomials, where g(x)≠0, we can write:
f(x) = q(x) g(x) + r(x)
which is same as the Dividend = Divisor * Quotient + Remainder and
where r(x) is the remainder polynomial and is equal to 0 and degree
r(x) < degree g(x).
Verification of Division Algorithm
Take the above example and verify it.
Divisor = x+2
Dividend = 2x2 + 3x + 1
Quotient = 2x – 1
Remainder = 0
Applying the Algorithm:
2x2 + 3x + 1 = (x + 2) (2x + 1) + 0
2x2 + 3x + 1 = 2x2 + 3x + 1
Hence verified.
Finding Factors of Polynomials with Division Algorithm
Long division algorithm is used to find out factors of polynomials of
degree greater than equal to two. We’ll be describing the steps to find
out the factors along with an example.
Example: Find roots of cubic polynomial P(x)=3x3 – 5x2 – 11x – 3
Solution
● Step 1: Use the factor theorem to find a factor of the
polynomial.
● Step 2: First divide the whole equation by the coefficient of the
highest degree term of the dividend.
P(x)=3x3 – 5x2 – 11x – 3
On dividing the whole equation by 3,
P(x) =x3 – (5/3)x2 – (11/3)x – 1
● Step 3: Find out factors of the constant term so obtained. In the
present case, factors of the constant term are 1 and -1.
● Step 4: Put the value of x in P(x) = 3x3 – 5x2 – 11x – 3 equal
to 1 and find the remainder. Again put the value of remainder
equal to -1 in and find the remainder using remainder theorem.
Find the value of x for which remainder is zero for the cubic
polynomial.
P (1) = 3(1)3 – 5(1)2 – 11(1) – 3 = -16
P(-1) = 3( -1 )3 – 5( -1 )2 – 11( -1 ) – 3 =0
● Step 5: Remainder is zero for x = -1. So, (x + 1) is a root of the
polynomial.
● Step 6: By Division Algorithm, find out the quotient. It comes
out: 3x2 – 8x – 3
● Step 7: Now, Quotient = 3x2 – 8x – 3
Dividend = (Divisor) * (Quotient) + Remainder
In present case,
3x3 – 5x2 – 11x – 3 = (x + 1) (3x2 – 8x – 3) + 0
By factorizing the quadratic polynomial we shall be able to find out
remaining factors of the cubic polynomial.
● Step 8: Break middle term in terms of a pair of numbers such
that its product is equal to -9 and summation equal to -3.
● Step 9: On factorizing, possible pair of number satisfying both
conditions is (-9, 1). Breaking the middle term,
f(x) = 3x2 – 8x – 3
= 3x2 – 9x + x – 3
● Step 10: Form pairs of terms and factor out GCD of the two
pairs separately. Then again factor out GCD of the remaining
two products.
● Step 11:
f(x) = 3x2 – 8x – 3 = 3x2 – 9x + x – 3
= 3x(x – 3) + 1(x – 3) = (x – 3)(3x + 1)
Now,
3x3 – 5x2 – 11x – 3 = (x + 1) (3x2 – 8x – 3) + 0
= (x + 1) ( x – 3)(3x + 1)
Factors of cubic polynomial are -1, 3 and -1/3.
Degree of Polynomial
Well, before starting, I would like to tell you that this ‘degree’ has
nothing to do with your thermometer’s degree or to your course
completion certification. The term ‘degree’ has come to the important
part of Mathematics, i.e., Polynomials and is adding an essential
meaning to it. So, let’s hit directly to understand the Degree of
Polynomials in details.
Degree of Polynomials
Source: Wikihow
Polynomial in One Variable
The degree of polynomials in one variable is the highest power of the
variable in the algebraic expression. For example, in the following
equation: x2+2x+4. The degree of the equation is 2 .i.e. the highest
power of variable in the equation.
Multivariable polynomial
For a multivariable polynomial, it the highest sum of powers of
different variables in any of the terms in the expression. Take
following example, x5+3x4y+2xy3+4y2-2y+1. It is a multivariable
polynomial in x and y, and the degree of the polynomial is 5 – as you
can see the degree in the terms x5 is 5, x4y it is also 5 (4+1) and so the
highest degree among these individual terms is 5.
A polynomial of two variable x and y, like axrys is the algebraic sum
of several terms of the prior mentioned form, where r and s are
possible integers. Here, the degree of the polynomial is r+s where r
and s are whole numbers.
Note: Exponents of variables of a polynomial .i.e. degree of
polynomials should be whole numbers.
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How to find the Degree of a Polynomial?
There are 4 simple steps are present to find the degree of a
polynomial:-
Example: 6x5+8x3+3x5+3x2+4+2x+4
● Step 1: Combine all the like terms that are the terms of the
variable terms.
(6x5+3x5)+8x3+3x2+2x+(4+4)
● Step 2: Ignore all the coefficients
x5+x3+x2+x+x0
● Step 3: Arrange the variable in descending order of their
powers
x5+x3+x2+x+x0
● Step 4: The largest power of the variable is the degree of the
polynomial
deg(x5+x3+x2+x+x0) = 5
Classification Based on the Degree of the Equation
Based on the degree, the equation can be linear, quadratic, cubic, and
bi-quadratic, and the list goes on.
Name of the Equation Degree of the Equation
Linear Equation 1
Quadratic Equation 2
Cubic Equation 3
Bi-Quadratic equation 4
Importance of Degree of polynomial
Case of Homogeneous Polynomial
The degree of terms is a major deciding factor whether an equation is
homogeneous or not. A polynomial of more that one variable is said to
be homogeneous if the degree of each term is the same. For example,
2x7+5x5y2-3x4y3+4x2y5 is a homogeneous polynomial of degree 7 in x
and y.
Relation of Degree of Polynomials with Zeroes of Equation
Theorem 1: A polynomial f(x) of the nth degree cannot vanish for
more than n values of x unless all its coefficients are zero.
Name of the Equation Degree of the Equation Possible Real Solutions
Linear Equation 1 1
Quadratic Equation 2 2
Cubic Equation 3 3
Bi-Quadratic equation 4 4
The above table shows possible real zeros /solutions; actual real
solutions can be less than the degree of the equation.
Note: A constant polynomial is that whose value remains the same. It
contains no variables. The power of the constant polynomial is Zero.
Well, you can write any constant with a variable having an
exponential power of zero. If the constant term = 4, then the
polynomial form is given by f(x)= 4x0
Before going to start other sections of Polynomials, try to solve the
below-given question.
A Question for You
Question: Find the degree of polynomial x3+4x5+5x4+2x2+x+5.
Solution: x3+4x5+5x4+2x2+x+5
=4x5+5x4+x3+2x2+x+5
=x5+x4+x3+x2+5
Degree of equation is the highest power of x in the given equation .i.e.
5.
Factorization of Polynomials
Do you know what factorization of polynomials means? Now that you
have an idea of what polynomials are, let’s learn how to factorize a
polynomial. This is an important step towards solving an equation in
mathematics. Let’s find out.
Suggested Videos
Standard form of Polynomial H
Factorisation of Polynomials by Common Factor Method
Cyclic Expressions, Cyclic Polynomials H
Before starting, let’s try to solve an example. If you are given 12
chocolates, choose the correct option in which you can divide them,
such that you are left with no chocolates:
● Group of 3 people
● Group of 4 people
● Group of 6 people
Well, the answer is, any one of the options will do. Why so? Well, you
know that 3, 4 and 6 are factors of 12 from your knowledge of number
system, which you have learned long back. So, we can divide the 12
chocolates into these groups.
Factors
What are the factors? When it comes to integers, if a number has an
integral value and that particular value get divided completely by
another number(s) without leaving any remainder other than zero, then
the dividend is said to be a factor(s) of that integer.
OR
When an integer could be written as a product of two or more integers
each then such numbers will be its factor. Why not consider some
examples to understand more. What can be the factors for 12?
12 = 6 × 2
12 = 4 × 3
12 = 12 × 1
So, you can see that a number can be factorized in more than one way.
But it’s not applicable to all the numbers. For a number like 7, 3, 5, 11
(which are prime numbers), these numbers have only one factor other
than itself.
7 = 7 × 1
So, it can be factorized in only one possible way.
Factorization of Polynomials
The same case of numbers also exists with Polynomials. A polynomial
can be written as a product of two or more polynomials of degree less
than or equal to that of it. Each polynomial involved in the product
will be a factor of it.
Graph of a Polynomial (Source: Wikipedia)
The process involved in breaking a polynomial into the product of its
factors is known as the factorization of polynomials. As I’ve told in
the previous sections that Monomial, Binomials are just the other
names for Polynomials. Its always good to start with the smaller and
easiest one. So, let’s hit the Monomials.
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Monomials
Monomials can be factorized in the same way as integers, just by
writing the monomial as the product of its constituent prime factors. In
the case of monomials, these prime factors can be integers as well as
other monomials which cannot be factorized further. Factorize:
● a3 = (a) × (a) × (a)
● 3abc = (3) × (a) × (b) × (c)
We will factorize monomials with binomials, trinomial and
polynomial. But before doing that, we will brush up concepts on GCD
or HCF.
GCD or HCF
For a given set of numbers, the Greatest Common Number that will
divide each of the numbers will be the GCD of that particular set of
numbers. It is also known as HCF of numbers .i.e. Highest Common
Factor.
Steps To Calculate the GCD of Integers
● Step 1: Break the number into the product of its prime factors.
● Step 2: Identify the common factors for the given set of
numbers.
● Step 3: The product of common factors will be gcd of the
number set.
● Step 4: If no common factor is found choose 1 as a common
factor.
Example: Find GCD of 15 and 24.
15 = 3 x 5
24 = 2 x 2 x 2 x 3
GCD of 15 and 24 is 3. This same procedure applies to polynomials.
Steps To Calculate the GCD of Polynomials
● Step 1: For a given set of polynomials, break the polynomial
into its factors such that each factor polynomial cannot be
factorized further.
● Step 2: Identify common terms or polynomials for a given set
of polynomials.
Example: Find GCD of 15ab and 3bc.
15ab = 3 x (5) x (a) x (b)
23bc = 3 x (b) x (c)
Here, GCD of 15ab and 3bc is 3b.
Factoring Binomials
Follow the 4 easy steps to factorize Binomials:
● Step 1: Break each term into its prime factors.
● Step 2: Find common factors or GCD for all individual terms.
● Step 3: Factor out the common factor or GCD.
● Step 4: The remaining terms in individual terms will form
another polynomial. Put them in separate bracket multiplied by
the GCD factored out initially.
● Step 5: The resulting product of terms will be the factors of the
initial polynomial (binomial).
Example: Factorize 15ab + 3bc
= 3 × (5) × (a) × (b) + 3 × (b) × (c) = 3b (5a + c)
Factoring Quadratic Polynomials
Case 1: When the coefficient of x2 is unity.
The general form of an equation is x2+bx+c. Every quadratic equation
can be expressed as: x2+bx+c= (x+d)(x+e). Here, b is the sum of d and
e & c is the product of d and e. Example: (x+2)(x+3) = x2+ 2x + 3x +
6 = x2+ 5x + 6.
Here, 5 = 2 + 3 = d + e = b in general form and 6 = 2 × 3 = d × e = c in
general form. To factorize quadratic polynomial, we shall be looking
for numbers which on multiplication will get equal to c and on
summation equal to b.
Example: Factorize x2+8x+12.
Solution: Below the steps are given for your understanding.
● Step 1: Factorize 12 as 12 = 2 × 6 or = 4 × 3. We have to find a
pair, such that its product is equal to 12 and summation is equal
to 8. Only one such pair is possible i.e. 2 and 6.
2 + 6 = 8
2 × 6 = 12
● Step 2: Break middle term in terms of the summation of a pair
of numbers such that its product is equal to c i.e. 12 in above
case. We will write 8=6+2
x2+ (6+2)x+ 12
= x2+ 6x +2x + 12
● Step 3: Form pairs of terms and factor out GCD of the two
pairs separately.
= x2+ 6x +2x + 12 = (x2+ 6x) +(2x + 12) = x(x+6)+2(x+6)
● Step 4: Again factor out GCD of remaining sum of products.
Follow factorization procedure of binomials as explained
earlier. Factor out (x+6) from sum of product,
=x(x+6)+2(x+6) = (x+6)(x+2)
Example: Do the factorization of polynomials: x2-5x-6
Soluiton: Factors of -6 are 2 × -3; – 3 × 2; 1 × -6; -1 × 6 and the pair
summing up to -5 is (-6,1) as required. Hence,
x2-5x-6 = x2-6x+x-6 = x(x-6)+(x-6) = (x-6)(x+1)
Case 2:When the coefficient of x2 a is an integer other than 1 or -1
General form is given by ax2+bx+c. Any quadratic of form ax2+bx+c
is expressible in the product of two linear polynomials:
ax2+bx+c= (a1x+b1)(a2x+b2)
where a is the product of a1 and a2, c is the product of b1 and b2 and b
is the sum of the product of a1b2 and a2b1. Consider one more
example,
(3x+2)(x+4)= 3x2+ 2x +12x +8 = 3x2+14x+ 8
Here, 3 = 3 × 1 = a1 × a2 = a in general form. 14 = 2 × 1 + 4 × 3 =
(a1b2) × (a2b1) = b in general form. 8 = 4 × 2 = b1 × b2= c in general
form. Example: Factorize 3x2 – 5x – 2
● Step 1: find factors of 3 × – 2 = -6.
● Step 2: Product of factors of -6 are = -3 × 2; -2 × 3; 6 × -1; -1 ×
6. Here, the possible pair of factor gives a summation of -1 is
(-6,1).
● Step 3: Break the middle term as the summation of two
numbers such that its product is equal to -6. Calculated above
such two numbers are -6 and 1.
● Step 4: Breaking the middle term: x2-6x+x-2
● Step 5: Making pairs of terms: (3x2-6x)+(x-2)
● Step 6: Factor out GCD for both the pairs: 3x(x-2)+1(x-2).
Note: 1 is factored out from the second bracket as there was no
other common factor.
● Step 7: Factor out GCD from the resulting summation of
products: =3x(x-2)+1(x-2) =(x-2)(3x+1)
Factorization of Polynomials by Grouping
We can factorize polynomial containing even number of terms by
forming pairs of terms.
● Step 1: Form pairs out of given even number of terms.
● Step 2: Try factoring out GCD from all the pairs separately.
● Step 3: Lastly, factor out the remaining common factor from
the products formed.
Note: In case you do not get common factors for the pairs formed, try
rearranging the terms and follow the same procedure again. Example:
Factorize x2+ 4xy+4y+x
= (x2 + 4xy) + (4y + x)
=x(x + 4y) + 1( 4y + x)
=(x + 4y)(x + 1)
Factorization of Polynomials by Perfect Squares
A trinomial which can be factored, such that both the factors are same.
Then, it will form a perfect square trinomial. For example, x2+2x+1 =
(x + 1)(x + 1) = (x + 1)2. There are certain identities which are
important for perfect square trinomials are as follows:
(a+b)2=(a2+2ab+b2)
Example:Do the factorization of polynomials, 4x2+12x+9
Solution: 4x2+12x+9 = (2x)2 + 2(2x)(3x) + 32 = (2x + 3)2
It can be seen from the identities and example above, that a trinomial
with first and last terms as perfect squares and the middle term can be
written as twice of the product of roots of first and last terms then the
trinomial can be expressed as a perfect square.
Factorization of Polynomials by Difference of Squares
This applies mainly to the pair of two polynomial terms which are a
perfect square and expressed as the difference between them.
a2-b2 = (a – b)(a + b)
Example: 9x2 – 4
Solution: 9x2 – 4 = (3x+2)(3x-2) { Using Identity }
Solved Example for You
Question: Factorise x2+11x+18
Solution: Consider the quadratic polynomial, x2+11x+18
=x2 + 9x + 2x + 18
=(x2 + 9x) + (2x + 18)
=x(x + 9) + 2(x + 9)
=(x + 9)(x + 2)
Remainder Theorem
When you divide one polynomial by another the process can be very
long. The Remainder and Factor Theorems help us avoid this long
division process by providing certain rules. We will learn about the
Remainder Theorem in this article.
Introduction
Divide 15 by 6. What answer do you get? By using the simple division
process, we find that the quotient is 2 and the remainder is 3. Hence,
we write 15 = (2 x 6) + 3. Note: The remainder ‘3’ is less than the
divisor ‘6’. On the other hand, when we divide 12 by 6, we get a
quotient of 2 and remainder 0. In this case, we say that 6 is a factor of
12 OR 12 is a multiple of 6.
Dividing One Polynomial by Another
Let’s start by dividing a polynomial with a monomial as follows:
Dividend Polynomial: 2x3 + x2 + x
Divisor Monomial: x
We have, (2x3 + x2 + x) / x = (2x3)/x + x2/x + x/x = 2x2 + x + 1
Observe that ‘x’ is common to each term of the dividend polynomial.
We can also write the dividend as, x(2x2 + x + 1). Hence, ‘x’ and ‘2x2
+ x + 1’ are factors of ‘2x3 + x2 + x’.
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Let’s look at another example.
Dividend polynomial: 3x2 + x + 1
Divisor Monomial: x
We have, (3x2 + x + 1) / x = (3x2)/x + x/x + 1/x = 3x + 1 + 1/x
Here, 1 is not divisible by x. So, we stop the division here and note
that 1 is the remainder. Hence, we have 3x2 + x + 1 = {x(3x + 1)} + 1.
So, the result of the division is,
Quotient: 3x + 1
Remainder: 1
Note that since the remainder is not zero, x is not a factor of 3x2 + x +
1.
Dividing a Polynomial by Any Non-zero Polynomial
Let’s do the following division: Dividend: x + 3x2 – 1 and Divisor: 1
+ x
Step 1.
We arrange the terms on the descending order of their degrees. Hence,
we have dividend: 3x2 + x – 1 and divisor: x + 1
Step 2.
Divide the first term of the dividend by the first term of the divisor:
3x2/x = 3x. This is the first term of the quotient.
Step 3.
Multiply the divisor by the first term of the quotient and subtract this
product from the dividend,
{3x2 + x – 1} – {3x(x + 1)}
= {3x2 + x – 1} – {3x2 + 3x}
= – 2x – 1
‘– 2x – 1’ is the remainder.
Step 4.
Now, the new dividend is ‘– 2x – 1’ and the divisor is still the same.
Repeat step 2 to get the next term of the quotient. Divide the first term
of the new dividend by the first term of the divisor: (– 2x)/x = – 2 =
the second term of the quotient
Step 5.
Multiply the divisor by the second term of the quotient and subtract
this product from the new dividend,
(– 2x – 1) – {– 2(x + 1)} = (– 2x – 1) – (– 2x – 2)
= – 2x – 1 + 2x + 2 = 1 = Remainder
Remember: This process continues until the degree of the new
dividend is less than the degree of the divisor.
Step 6.
Hence, we have quotient: 3x – 2 and remainder: 1. It is important to
note here that, 3x2 + x – 1 = (x + 1) (3x – 2) + 1 or Dividend =
(Divisor × Quotient) + Remainder
Therefore, we can conclude that, if p(x) and g(x) are two polynomials
such that degree of p(x) ≥ degree of g(x) and g(x) ≠ 0, then we can
find polynomials q(x) and r(x) such that:
p(x) = g(x)q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of
g(x). Here we say that p(x) divided by g(x), gives q(x) as quotient and
r(x) as remainder.
Next, let’s try to find a link between the remainder and the dividend.
Let’s find the value of the dividend polynomial at x = -1
p(x) = 3x2 + x – 1
p(-1) = 3(- 1)2 + (- 1) – 1
= 3 – 1 – 1
= 1 … which is the remainder!
So, the remainder of (3×2 + x – 1) / (x + 1) = Value of (3×2 + x – 1) at
x = – 1 (or the zero of the divisor [x + 1]). In other words, the
remainder obtained on dividing a polynomial by another is the same as
the value of the dividend polynomial at the zero of the divisor
polynomial. This brings us to the first theorem of this article.
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Remainder Theorem
Let p(x) be any polynomial of degree greater than or equal to one and
let ‘a’ be any real number. If p(x) is divided by the linear polynomial
(x – a), then the remainder is p(a).
Proof: p(x) is a polynomial with degree greater than or equal to one. It
is divided by a polynomial (x – a), where ‘a’ is a real number. Let’s
assume that the quotient is q(x) and the remainder is r(x). So, we can
write,
p(x) = (x – a)q(x) + r(x)
Now, the degree of (x – a) is 1. Also, since r(x) is the remainder, its
degree is less than the degree of the divisor: (x – a). Therefore, the
degree of r(x) = 0. In other words, r(x) is a constant. Let’s call the
constant ‘r’. Hence, for all values of ‘x’, r(x) = r. Therefore,
p(x) = (x – a) q(x) + r
Now, let’s find p(a) or the value of p(x) at x = a.
p(a) = (a – a) q(a) + r
= (0)q(a) + r
= r
We see that when a polynomial p(x) of a degree greater than or equal
to one is divided by a linear polynomial (x – a), where a is a real
number, then the remainder is r which is also equal to p(a). This
proves the Remainder Theorem.
For example, check whether the polynomial q(t) = 4t3 + 4t2 – t – 1 is a
multiple of 2t+1.
Solution: q(t) will be a multiple of 2t + 1 only, if 2t + 1 divides q(t)
with remainder zero. Let’s find the zero of the divisor polynomial:
2t + 1 = 0 Or t = – ½
Next, let’s find the value of q(t) at t= – ½
q(- ½) = 4(- ½)3 + 4(- ½)2 – (- ½) – 1
= 4(- 1/8) + 4(- ¼) + ½ – 1
= – ½ + 1 + ½ – 1 = 0.
Hence, we can conclude that the remainder obtained on dividing q(t)
by 2t + 1 is 0. And, (2t + 1) is a factor of ‘4t3 + 4t2 – t – 1’.
Solved Examples for You
Question: Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by
(x – 1).
Solution: Dividend Polynomial = p(x) = x4 + x3 – 2x2 + x + 1 and
Divisor Polynomial = x – 1
Zero of the divisor polynomial is x – 1 = 0 or, x = 1.
Therefore, p(1) = (1)4 + (1)3 – 2(1)2 + 1 + 1 = 1 + 1 – 2 + 1 + 1 = 2.
So, by Remainder Theorem, the remainder is 2.
Question: Find the remainder when (x3 – ax2 + 6x – a) is divided by (x
– a).
Solution: Dividend Polynomial = p(x) = x3 – ax2 + 6x – a and Divisor
Polynomial = x – a
Zero of the divisor polynomial is x – a = 0 or, x = a.
Therefore, p(a) = (a)3 – a(a)2 + 6a – a = a3 – a3 + 6a – a = 5a
So, by Remainder Theorem, the remainder is 5a.
Factor Theorem
In this part, we will look at the Factor Theorem, which uses the
remainder theorem and learn how to factorise polynomials. Further,
we will be covering the splitting method and the factor theorem
method.
Factor Theorem
If p(x) is a polynomial of degree n > 1 and a is any real number, then
● x – a is a factor of p(x), if p(a) = 0, and
● p(a) = 0, if x – a is a factor of p(x).
Let’s look at an example to understand this theorem better.
Example:
Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6.
Solution: To begin with, we know that the zero of the polynomial (x +
2) is –2. Let p(x) = x3 + 3x2 + 5x + 6
Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 = –8 + 12 – 10 + 6 = 0
According to the factor theorem, if p(a) = 0, then (x – a) is a factor of
p(x). In this example, p(a) = p(- 2) = 0
Therefore, (x – a) = {x – (-2)} = (x + 2) is a factor of ‘x3 + 3x2 + 5x +
6’ or p(x).
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Factorisation of polynomials
You can factorise polynomials by splitting the middle term as follows:
to begin with, consider a polynomial ax2 + bx + c with factors (px + q)
and (rx + s). Therefore, we have ax2 + bx + c = (px + q) (rx + s). So,
ax2 + bx + c = prx2 + (ps + qr) x + qs
If we compare the coefficients of x2, we get a = pr. Also, on
comparing the coefficients of x, we get b = ps + qr. Finally, on
comparing the constants, we get c = qs. Hence, b is the sum of two
numbers ‘ps’ and ‘qr’, whose product is (ps)(qr) = (pr)(qs) = ac.
Therefore, to factorise ax2 + bx + c, we have to write b as the sum of
two numbers whose product is ‘ac’. Let’s look at an example to
understand this clearly.
Example
Factorise 6×2 + 17x + 5 by splitting the middle term.
Solution 1 (By splitting method): As explained above, if we can find
two numbers, ‘p’ and ‘q’ such that, p + q = 17 and pq = 6 x 5 = 30,
then we can get the factors.
After looking at the factors of 30, we find that numbers ‘2’ and ‘15’
satisfy both the conditions, i.e. p + q = 2 + 15 = 17 and pq = 2 x 15 =
30. So,
6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5
= 6x2 + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)
Therefore, the factors of (6x2 + 17x + 5) are (3x + 1) and (2x + 5) with
a remainder, zero.
Example
Factorise y2 – 5y + 6 by using the Factor Theorem.
Solution: Let p(y) = y2 – 5y + 6. Now, if p(y) = (y – a) (y – b), the
constant term will be ab as can be seen below,
p(y) = (y – a)(y – b)
= y2 – by – ay + ab
On comparing the constants, we get ab = 6. Next, the factors of 6 are
1, 2 and 3. Now, p(2) = 22– (5 × 2) + 6 = 4 – 10 + 6 = 0. So, (y – 2) is
a factor of p(y). Also, p(3) = 32 – (5 × 3) + 6 = 9 – 15 + 6 = 0. So, (y –
3) is also a factor of y2 – 5y + 6. Therefore, y2 – 5y + 6 = (y – 2)(y –
3)
More Solved Examples for You
Question: Factorise x3 – 23x2 + 142x – 120
Solution: Let p(x) = x3 – 23x2 + 142x – 120. To begin with, we will
start finding the factors of the constant ‘– 120’, which are:
±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60
and ±120
Further, by trial, we find that p(1) = 0. Hence, we conclude that (x – 1)
is a factor of p(x). Also, we see that
[x3 – 23x2 + 142x – 120] = x3 – x2 – 22x2 + 22x + 120x – 120
So, by removing the common factors, we have x3 – 23x2 + 142x – 120
= x2(x –1) – 22x(x – 1) + 120(x – 1)
Further, taking ‘x – 1’ common, we get x3 – 23x2 + 142x – 120 = (x –
1) (x2 – 22x + 120)
Therefore, x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
Also, note that if we divide p(x) by ‘x – 1’, then the result will be (x2 –
22x + 120)
Going on, x2 – 22x + 120 can be factorised further. So, by splitting the
middle term, we get:
x2 – 22x + 120 = x2 – 12x – 10x + 120 … [ (– 12 – 10 = – 22) and
{(–12)( –10) = 120}]
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
Therefore, we have x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)
Question: Factorise x3 – 2x2 – x + 2
Solution: Let p(x) = x3 – 2x2 – x + 2. To begin with, we will start
finding the factors of the constant ‘2’, which are: 1, 2
By trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a
factor of p(x).
So, by removing the common factors, we have x3 – 2x2 – x + 2 = x2(x
– 2) – (x – 2) = (x2 – 1)(x – 2)
= (x + 1)(x – 1)(x – 2) … [Using the identity (x2 – 1) = (x + 1)(x – 1)]
Therefore, the factors of x3 – 2x2 – x + 2 are (x + 1), (x – 1) and (x –
2)
Zeroes of Polynomial
We already know that a polynomial is an algebraic term with one or
many terms. Zeroes of Polynomial are the real values of the variable
for which the value of the polynomial becomes zero. So, real numbers,
‘m’ and ‘n’ are zeroes of polynomial p(x), if p(m) = 0 and p(n) = 0.
Understanding Zeroes of Polynomial
Example 1
Let’s look at the polynomial, p(x) = 5x3 – 2x2 + 3x -2. Now, let’s find
the value of the polynomial(x) at x = 1, p(1) = 5(1)3 – 2(1)2 + 3(1) – 2
= 5 – 2 + 3 – 2 = 4. Therefore, we can say that the value of the
polynomial p(x) at x = 1 is 4.
Next, let’s find the value of the polynomial(x) at x = 0, p(0) = 5(0)3 –
2(0)2 + 3(0) – 2 = 0 – 0 + 0 – 2 = – 2. Therefore, we can say that the
value of the polynomial p(x) at x = 0 is – 2.
Example 2
Let’s look at another polynomial now, p(x) = x – 1. Let’s find the
value of the polynomial at x = 1, p(1) = 1 – 1 = 0. So, the value of the
polynomial p(x) at x = 1 is 0. Therefore, 1 is the zero of the
polynomial p(x).
Similarly, for the polynomial q(x) = x – 2, the zero of the polynomial
is 2. To summarize, zeroes of polynomial p(x) are numbers c and d
such that p(c) = 0 and p(d) = 0.
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for Class 10 Maths
Calculating Zeroes of polynomial
When we calculated zeroes of polynomial p(x) = x – 1, we equated it
to 0, x – 1 = 0 or, x = 1. Hence, we say that p(x) = 0 is the Polynomial
Equation. 1 is the Root of the Polynomial equation p(x) = 0. OR 1 is
the Zero of the Polynomial equation p(x) = x – 1 = 0.
Now, let’s look at a constant polynomial ‘5’. You can write this as
5x0. What is the Root of this constant polynomial? The answer is a
Non-zero constant polynomial has no zero. Also, every real number is
a zero of the Zero Polynomial.
Let’s look at the following linear polynomial to understand the
calculation of the roots or ‘zeroes of polynomial’: p(x) = ax + b …
where a ≠ 0. To find a zero, we must equate the polynomial to 0. [p(x)
= 0]. Hence, ax + b = 0 … where a ≠ 0. So, ax = – b or, x = – b/a.
Therefore, ‘- b/a’ is the only zero of p(x) = ax + b. We can also say
that a linear polynomial has only one zero.
Observations
● A zero of a polynomial need not be 0.
● 0 may be a zero of a polynomial.
● Every linear polynomial has one and only one zero.
● A polynomial can have more than one zero.
Learn Degree of Polynomials in detail.
More Solved Examples for You
Question: Find p(0), p(1) and p(2) for each of the following
polynomials:
● p(y) = y2 – y + 1
● p(t) = 2 + t + 2t2 – t3
● p(x) = x3
● p(x) = (x – 1) (x + 1)
Solution:
p(y) = y2 – y + 1
● p(0) = 02 – 0 + 1 = 1
● p(1) = 12 – 1 + 1 = 1
● p(2) = 22 – 2 + 1 = 3
p(t) = 2 + t + 2t2 – t3
● p(0) = 2 + 0 + 2(0)2 – (0)3 = 2 + 0 + 0 – 0 = 2
● p(1) = 2 + 1 + 2(1)2 – (1)3 = 2 + 1 + 2 – 1 = 4
● p(2) = 2 + 2 + 2(2)2 – (2)3 = 2 + 2 + 8 – 8 = 4
p(x) = x3
● p(0) = (0)3 = 0
● p(1) = (1)3 = 1
● p(2) = (2)3 = 8
p(x) = (x – 1) (x + 1)
● p(0) = (0 – 1)(0 + 1) = (-1)(1) = – 1
● p(1) = (1 – 1)(1 + 1) = (0)(2) = 0
● p(2) = (2 – 1)(2 + 1) = (1)(3) = 3
Verify the following
Question: Verify whether the following are zeroes of polynomials
indicated against them.
● p(x) = 3x + 1, x = – 1/3
● p(x) = 5x – π, x = 4/5
● p(x) = (x + 1) (x – 2), x = – 1, 2
● p(x) = 3x2 – 1, x = – 1/√3 , 2/√3
Solution:
● p(x) = 3x + 1, x = – 1/3
p(-1/3) = 3(-1/3) + 1 = – 1 + 1 = 0.
Hence, x = -1/3 is a zero of polynomial 3x + 1.
● p(x) = 5x – π, x = 4/5
p(4/5) = 5(4/5) – π = 4 – π ≠ 0.
Hence, x = 4/5 is not a zero of polynomial 5x – π.
● p(x) = (x + 1) (x – 2), x = – 1, 2
p(-1) = (– 1 + 1)(- 1 – 2) = (0)(- 3) = 0.
Hence, x = – 1 is a zero of polynomial (x + 1) (x – 2).
p(2) = (2 + 1)(2 – 2) = (3)(0) = 0.
Hence, x = 2 is a zero of polynomial (x + 1) (x – 2).
● p(x) = 3x2 – 1, x = – 1/√3 , 2/√3
p(- 1/√3) = 3(- 1/√3)2 – 1 = 3 (1/3) – 1 = 1 – 1 = 0.
Hence, x = – 1/√3 is a zero of polynomial 3x2 – 1.
p(2/√3) = 3(2/√3) – 1 = 3(4/3) – 1 = 4 – 1 = 3 ≠ 0.
Hence, x = 2/√3 is not a zero of polynomial 3x2 – 1.
Geometrical Representation of Zeroes of a Polynomial
In the study of polynomials, you are aware that a real number ‘k’ is a
zero of the polynomial p(x) if p(k) = 0. Remember, zero of a
polynomial is different from a zero polynomial. We have seen the
Remainder theorem use the concept of zeroes of a polynomial too. In
order to understand their importance, we will look at the geometrical
representations of linear and quadratic polynomials and the
geometrical meaning of their zeroes.
Suggested Videos
Standard form of Polynomial H
Factorisation of Polynomials by Common Factor Method
Cyclic Expressions, Cyclic Polynomials H
Linear Polynomial
Let’s look at a linear polynomial ax + b, where a ≠ 0. You have
already studied that the graph of y = ax + b is a straight line. Let’s
look at the graph of y = 2x + 3.
x -2 2
y = 2x + 3 -1 7
The straight line y = 2x + 3 will pass through the points (- 2, – 1) and
(2, 7). Here is how the graph looks like:
Fig. 1
From the Fig.1 above, you can see that the graph of y = 2x + 3
intersects the x-axis at the point (- 3/2, 0). Now, the zero of (2x + 3) is
(- 3/2). Therefore, the zero of the linear polynomial (2x + 3) is the
x-coordinate of the point where the graph of y = 2x + 3 intersects the
x-axis. Hence, we can say,
For a linear polynomial ax + b, a ≠ 0, the graph of y = ax + b is a
straight line which intersects the x-axis at exactly one point, namely, (-
b/a, 0). Also, this linear polynomial has only one zero which is the
x-coordinate of the point where the graph of y = ax + b intersects the
x-axis.
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Quadratic Polynomial
Let’s look at a quadratic polynomial, x2 – 3x – 4. To look at the graph
of y = x2 – 3x – 4, let’s list some values:
x – 2 – 1 0 1 2 3 4 5
y = x2 – 3x – 4 6 0 – 4 – 6 – 6 – 4 0 6
The graph of y = x2 – 3x – 4 will pass through (- 2, 6), (- 1, 0), (0, –
4), (1, – 6), (2, – 6), (3, – 4), (4, 0) and (5, 6). Here is how the graph
looks:
Fig. 2
For that matter, for any quadratic polynomial y = ax2 + bx + c, a ≠ 0,
the graph of y = ax2 + bx + c has either one of these two shapes:
● If a > 0, then it is open upwards like the one shown in the graph
above
● If a < 0, then it is open downwards.
These curves are parabolas. A quick look at the table above shows that
(-1) and (4) are zeroes of the quadratic polynomial. From the Fig. 2
above, you can see that (-1) and (4) are the x-coordinates of the points
where the graph of y = x2 – 3x – 4 intersects the x-axis. Therefore, we
can say,
The zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0, are
precisely the x-coordinates of the points where the parabola
representing y = ax2 + bx + c intersects the x-axis.
As far as the shape of the graph is concerned, the following three
cases are possible:
Case (i)
The graph cuts x-axis at two distinct points A and A′, where the
x-coordinates of A and A′ are the two zeroes of the quadratic
polynomial ax2 + bx + c, as shown below:
Fig. 3
Case (ii)
The graph intersects the x-axis at only one point, or at two coincident
points. Also, the x-coordinate of A is the only zero for the quadratic
polynomial ax2 + bx + c, as shown below:
Fig. 4
Case (iii)
The graph is either
● Completely above the x-axis or
● Completely below the x-axis.
So, it does not cut the x-axis at any point. Hence, the quadratic
polynomial ax2 + bx + c has no zero, as shown below:
Fig. 5
To summarize, a quadratic polynomial can have either:
● Two distinct zeroes (as shown in Case i)
● Two equal zeroes (or one zero as shown in Case ii)
● No zero (as shown in Case iii)
It can also be summarized by saying that a polynomial of degree 2 has
a maximum of 2 zeroes.
Cubic Polynomial
Let’s look at a cubic polynomial, x3 – 4x. Next, let’s list a few values
to plot the graph of y = x3 – 4x.
x – 2 – 1 0 1 2
y = x3 – 4x 0 3 0 – 3 0
The graph of y = x3 – 4x will pass through (- 2, 0), (- 1, 3), (0, 0), (1, –
3), and (2, 0). Here is how the graph looks like:
Fig. 6
From the table above, we can see that 2, 0 and – 2 are the zeroes of the
cubic polynomial x3 – 4x. You can also observe that the graph of y =
x3 – 4x intersects the x-axis at 2, 0 and – 2. Let’s take a quick look at
some examples:
Let’s plot the graph of the following two cubic polynomials:
1. x3
2. x3 – x2
The graphs of y = x3 and y = x3 – x2 look as follows:
Fig. 7
From the first graph, you can observe that 0 is the only zero of the
polynomial x3, since the graph of y = x3 intersects the x-axis only at 0.
Similarly, the polynomial x3 – x2 = x2(x – 1) has two zeroes, 0 and 1.
From the second diagram, you can see that the graph of y = x3 – x2
intersects the x-axis at 0 and 1.
Hence, we can conclude that there is a maximum of three zeroes for
any cubic polynomial. Or, any polynomial with degree 3 can have
maximum 3 zeroes. In general,
Given a polynomial p(x) of degree n, the graph of y = p(x) intersects
the x-axis at a maximum of n points. Therefore, a polynomial p(x) of
degree n has a maximum of n zeroes.
Solved Examples for You
Question: The graphs of y = p(x) are given in the figure below, for
some polynomials p(x). Find the number of zeroes of p(x), in each
case.
Fig. 8
Solution: The number of zeroes in each of the graphs above are:
● 0, since the line is not intersecting the x-axis.
● 1, since the line intersects the x-axis once.
● 2, since the line intersects the x-axis twice.
● 2, since the line intersects the x-axis twice.
● 4, since the line intersects the x-axis four times.
● 3, since the line intersects the x-axis thrice.
Geometrical Representation of Zeroes of a Polynomial
In the study of polynomials, you are aware that a real number ‘k’ is a
zero of the polynomial p(x) if p(k) = 0. Remember, zero of a
polynomial is different from a zero polynomial. We have seen the
Remainder theorem use the concept of zeroes of a polynomial too. In
order to understand their importance, we will look at the geometrical
representations of linear and quadratic polynomials and the
geometrical meaning of their zeroes.
Linear Polynomial
Let’s look at a linear polynomial ax + b, where a ≠ 0. You have
already studied that the graph of y = ax + b is a straight line. Let’s
look at the graph of y = 2x + 3.
x -2 2
y = 2x + 3 -1 7
The straight line y = 2x + 3 will pass through the points (- 2, – 1) and
(2, 7). Here is how the graph looks like:
Fig. 1
From the Fig.1 above, you can see that the graph of y = 2x + 3
intersects the x-axis at the point (- 3/2, 0). Now, the zero of (2x + 3) is
(- 3/2). Therefore, the zero of the linear polynomial (2x + 3) is the
x-coordinate of the point where the graph of y = 2x + 3 intersects the
x-axis. Hence, we can say,
For a linear polynomial ax + b, a ≠ 0, the graph of y = ax + b is a
straight line which intersects the x-axis at exactly one point, namely, (-
b/a, 0). Also, this linear polynomial has only one zero which is the
x-coordinate of the point where the graph of y = ax + b intersects the
x-axis.
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Quadratic Polynomial
Let’s look at a quadratic polynomial, x2 – 3x – 4. To look at the graph
of y = x2 – 3x – 4, let’s list some values:
x – 2 – 1 0 1 2 3 4 5
y = x2 – 3x – 4 6 0 – 4 – 6 – 6 – 4 0 6
The graph of y = x2 – 3x – 4 will pass through (- 2, 6), (- 1, 0), (0, –
4), (1, – 6), (2, – 6), (3, – 4), (4, 0) and (5, 6). Here is how the graph
looks:
Fig. 2
For that matter, for any quadratic polynomial y = ax2 + bx + c, a ≠ 0,
the graph of y = ax2 + bx + c has either one of these two shapes:
● If a > 0, then it is open upwards like the one shown in the graph
above
● If a < 0, then it is open downwards.
These curves are parabolas. A quick look at the table above shows that
(-1) and (4) are zeroes of the quadratic polynomial. From the Fig. 2
above, you can see that (-1) and (4) are the x-coordinates of the points
where the graph of y = x2 – 3x – 4 intersects the x-axis. Therefore, we
can say,
The zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0, are
precisely the x-coordinates of the points where the parabola
representing y = ax2 + bx + c intersects the x-axis.
As far as the shape of the graph is concerned, the following three
cases are possible:
Case (i)
The graph cuts x-axis at two distinct points A and A′, where the
x-coordinates of A and A′ are the two zeroes of the quadratic
polynomial ax2 + bx + c, as shown below:
Fig. 3
Case (ii)
The graph intersects the x-axis at only one point, or at two coincident
points. Also, the x-coordinate of A is the only zero for the quadratic
polynomial ax2 + bx + c, as shown below:
Fig. 4
Case (iii)
The graph is either
● Completely above the x-axis or
● Completely below the x-axis.
So, it does not cut the x-axis at any point. Hence, the quadratic
polynomial ax2 + bx + c has no zero, as shown below:
Fig. 5
To summarize, a quadratic polynomial can have either:
● Two distinct zeroes (as shown in Case i)
● Two equal zeroes (or one zero as shown in Case ii)
● No zero (as shown in Case iii)
It can also be summarized by saying that a polynomial of degree 2 has
a maximum of 2 zeroes.
Cubic Polynomial
Let’s look at a cubic polynomial, x3 – 4x. Next, let’s list a few values
to plot the graph of y = x3 – 4x.
x – 2 – 1 0 1 2
y = x3 – 4x 0 3 0 – 3 0
The graph of y = x3 – 4x will pass through (- 2, 0), (- 1, 3), (0, 0), (1, –
3), and (2, 0). Here is how the graph looks like:
Fig. 6
From the table above, we can see that 2, 0 and – 2 are the zeroes of the
cubic polynomial x3 – 4x. You can also observe that the graph of y =
x3 – 4x intersects the x-axis at 2, 0 and – 2. Let’s take a quick look at
some examples:
Let’s plot the graph of the following two cubic polynomials:
1. x3
2. x3 – x2
The graphs of y = x3 and y = x3 – x2 look as follows:
Fig. 7
From the first graph, you can observe that 0 is the only zero of the
polynomial x3, since the graph of y = x3 intersects the x-axis only at 0.
Similarly, the polynomial x3 – x2 = x2(x – 1) has two zeroes, 0 and 1.
From the second diagram, you can see that the graph of y = x3 – x2
intersects the x-axis at 0 and 1.
Hence, we can conclude that there is a maximum of three zeroes for
any cubic polynomial. Or, any polynomial with degree 3 can have
maximum 3 zeroes. In general,
Given a polynomial p(x) of degree n, the graph of y = p(x) intersects
the x-axis at a maximum of n points. Therefore, a polynomial p(x) of
degree n has a maximum of n zeroes.
Solved Examples for You
Question: The graphs of y = p(x) are given in the figure below, for
some polynomials p(x). Find the number of zeroes of p(x), in each
case.
Fig. 8
Solution: The number of zeroes in each of the graphs above are:
● 0, since the line is not intersecting the x-axis.
● 1, since the line intersects the x-axis once.
● 2, since the line intersects the x-axis twice.
● 2, since the line intersects the x-axis twice.
● 4, since the line intersects the x-axis four times.
● 3, since the line intersects the x-axis thrice.