math083 day 1 chapter 6 2013 fall

Math 083 College Algebra

• Fall 2013• Instructor Mr. Bianco

Math 083 Bianco

•Today :–Syllabus–Multiplying–Factoring

Mr. Bianco

• 15 years of teaching High School Math- Algebra to Calculus

• 5 Years at Essex

What you need to know

• BE ON TIME!!!!• Try all homework• Check website(s) if you miss

class- But make sure you inform me first

• Don’t wait to get help!• No electronic devices on during

class

Which number doesn’t belong and be ready to explain why

9 16 25 43

6

POLYNOMIALSPOLYNOMIALS

MULTIPLY:

1. (7) (2x - 5)

(7) (2x) - (7) (5)

14x - 35

7

POLYNOMIALSPOLYNOMIALS

MULTIPLY:

2. (2x2)(5x4 + 7x)(2x2) (5x4)+ (2x2) (7x)

10x6 + 14x3

8

Multiplying by FOILMultiplying by FOIL

(3x + 2) (5x + 4)

F O I L

9

Multiplying by FOILMultiplying by FOIL

(3x + 2) (5x + 4)

F O I L

15x2

F ---> First Terms

10

Multiplying by FOILMultiplying by FOIL

(3x + 2) (5x + 4)

F O I L

O ---> Outer Terms15x2 + 12x

11

Multiplying by FOILMultiplying by FOIL

(3x + 2) (5x + 4)

F O I L

I ---> Inner Terms15x2 + 12x + 10x

12

Multiplying by FOILMultiplying by FOIL

(3x + 2) (5x + 4)

F O I L

L ---> Last Terms15x2 + 12x + 10x + 8

13

Multiplying by FOILMultiplying by FOIL

(3x + 2) (5x + 4)

F O I LCombine Like Terms

15x2 + 22x + 815x2 + 12x + 10x + 8

14

Multiplying by FOILMultiplying by FOIL

(3x + 2) (5x + 4)

F O I L15x2 + 22x + 8

15

Multiplying by FOILMultiplying by FOIL

(5x + 1) (5x + 1)

F O I L

16

Multiplying by FOILMultiplying by FOIL

(5x + 1) (5x + 1)

F O I L

25x2

F ---> First Terms

17

Multiplying by FOILMultiplying by FOIL

(5x + 1) (5x + 1)

F O I L

O ---> Outer Terms 25x2 + 5x

18

Multiplying by FOILMultiplying by FOIL

(5x + 1) (5x + 1)

F O I L

I ---> Inner Terms 25x2 + 5x + 5x

19

Multiplying by FOILMultiplying by FOIL

(5x + 1) (5x + 1)

F O I L

L ---> Last Terms 25x2 + 5x + 5x + 1

20

Multiplying by FOILMultiplying by FOIL

(5x + 1) (5x + 1)

F O I LCombine Like Terms

25x2 + 10x + 125x2 + 5x + 5x + 1

21

Multiplying by FOILMultiplying by FOIL

(5x + 1) (5x + 1)

F O I L25x2 + 10x + 1

22

Multiplying by FOILMultiplying by FOIL

(5x + 1)2

(5x + 1) (5x + 1)

25x2 + 10x + 1

In the process of using the FOIL method on products of certain types of binomials, we see specific patterns that lead to special products.Square of a Binomial

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

Product of the Sum and Difference of Two Terms(a + b)(a – b) = a2 – b2

Special Products

Although you will arrive at the same results for the special products by using the techniques of this section or last section, memorizing these products can save you some time in multiplying polynomials.

Special Products

Factors

Factors (either numbers or polynomials)When an integer is written as a product of integers, each of the integers in the product is a factor of the original number.When a polynomial is written as a product of polynomials, each of the polynomials in the product is a factor of the original polynomial.

Factoring – writing a polynomial as a product of polynomials.

Greatest common factor – largest quantity that is a factor of all the integers or polynomials involved.

Finding the GCF of a List of Monomials1) Find the GCF of the numerical coefficients.2) Find the GCF of the variable factors.3) The product of the factors found in Step 1 and 2 is

the GCF of the monomials.

Greatest Common Factor

Find the GCF of each list of numbers.1) 12 and 8

12 = 2 · 2 · 3 8 = 2 · 2 · 2So the GCF is 2 · 2 = 4.

2) 7 and 20 7 = 1 · 720 = 2 · 2 · 5There are no common prime factors so the GCF is 1.

Greatest Common Factor

Example:

Find the GCF of each list of numbers.1) 6, 8 and 46

6 = 2 · 3 8 = 2 · 2 · 246 = 2 · 23So the GCF is 2.

2) 144, 256 and 300144 = 2 · 2 · 2 · 3 · 3256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2300 = 2 · 2 · 3 · 5 · 5So the GCF is 2 · 2 = 4.

Greatest Common Factor

Example:

1) x3 and x7

x3 = x · x · xx7 = x · x · x · x · x · x · xSo the GCF is x · x · x = x3

• 6x5 and 4x3

6x5 = 2 · 3 · x · x · x4x3 = 2 · 2 · x · x · x So the GCF is 2 · x · x · x = 2x3

Find the GCF of each list of terms.

Greatest Common Factor

Example:

Find the GCF of the following list of terms.

a3b2, a2b5 and a4b7

a3b2 = a · a · a · b · ba2b5 = a · a · b · b · b · b · b a4b7 = a · a · a · a · b · b · b · b · b · b · b

So the GCF is a · a · b · b = a2b2

Notice that the GCF of terms containing variables will use the smallest exponent found amongst the individual terms for each variable.

Greatest Common Factor

Example:

The first step in factoring a polynomial is to find the GCF of all its terms.

Then we write the polynomial as a product by factoring out the GCF from all the terms.

The remaining factors in each term will form a polynomial.

Factoring Polynomials

Factor out the GCF in each of the following polynomials.

1) 6x3 – 9x2 + 12x =3 · x · 2 · x2 – 3 · x · 3 · x + 3 · x · 4 =3x(2x2 – 3x + 4)

2) 14x3y + 7x2y – 7xy =7 · x · y · 2 · x2 + 7 · x · y · x – 7 · x · y · 1 =7xy(2x2 + x – 1)

Factoring out the GCF

Example:

Factor out the GCF in each of the following polynomials.

1) 6(x + 2) – y(x + 2) =6 · (x + 2) – y · (x + 2) =(x + 2)(6 – y)

2) xy(y + 1) – (y + 1) =xy · (y + 1) – 1 · (y + 1) =(y + 1)(xy – 1)

Factoring out the GCF

Example:

Factoring polynomials often involves additional techniques after initially factoring out the GCF.

One technique is factoring by grouping.

Factor xy + y + 2x + 2 by grouping.

Notice that, although 1 is the GCF for all four terms of the polynomial, the first 2 terms have a GCF of y and the last 2 terms have a GCF of 2.

xy + y + 2x + 2 = x · y + 1 · y + 2 · x + 2 · 1 =

y(x + 1) + 2(x + 1) = (x + 1)(y + 2)

Factoring by Grouping

Example:

Factoring a Four-Term Polynomial by Grouping1) Arrange the terms so that the first two terms have a

common factor and the last two terms have a common factor.

2) For each pair of terms, use the distributive property to factor out the pair’s greatest common factor.

3) If there is now a common binomial factor, factor it out.4) If there is no common binomial factor in step 3, begin

again, rearranging the terms differently. • If no rearrangement leads to a common binomial factor, the

polynomial cannot be factored.

Factoring by Grouping

1) x3 + 4x + x2 + 4 = x · x2 + x · 4 + 1 · x2 + 1 · 4 = x(x2 + 4) + 1(x2 + 4) = (x2 + 4)(x + 1)

2) 2x3 – x2 – 10x + 5 = x2 · 2x – x2 · 1 – 5 · 2x – 5 · (– 1) = x2(2x – 1) – 5(2x – 1) = (2x – 1)(x2 – 5)

Factor each of the following polynomials by grouping.

Factoring by Grouping

Example:

Factor 2x – 9y + 18 – xy by grouping.Neither pair has a common factor (other than 1).So, rearrange the order of the factors.

2x + 18 – 9y – xy = 2 · x + 2 · 9 – 9 · y – x · y =2(x + 9) – y(9 + x) =2(x + 9) – y(x + 9) = (make sure the factors are identical)(x + 9)(2 – y)

Factoring by Grouping

Remember that factoring out the GCF from the terms of a polynomial should always be the first step in factoring a polynomial.

This will usually be followed by additional steps in the process.

Factor 90 + 15y2 – 18x – 3xy2.90 + 15y2 – 18x – 3xy2 = 3(30 + 5y2 – 6x – xy2) =3(5 · 6 + 5 · y2 – 6 · x – x · y2) = 3(5(6 + y2) – x (6 + y2)) =3(6 + y2)(5 – x)

Factoring

Example:

Factoring polynomials often involves additional techniques after initially factoring out the GCF.One technique is factoring by grouping.

Factor xy + y + 2x + 2 by grouping.Notice that, although 1 is the GCF for all four terms of the polynomial, the first 2 terms have a GCF of y and the last 2 terms have a GCF of 2.xy + y + 2x + 2 = x · y + 1 · y + 2 · x + 2 · 1 =y(x + 1) + 2(x + 1) = (x + 1)(y + 2)

Factoring by Grouping

Example:

Factoring a Four-Term Polynomial by Grouping1) Arrange the terms so that the first two terms have a

common factor and the last two terms have a common factor.

2) For each pair of terms, use the distributive property to factor out the pair’s greatest common factor.

3) If there is now a common binomial factor, factor it out.4) If there is no common binomial factor in step 3, begin again,

rearranging the terms differently. • If no rearrangement leads to a common binomial factor,

the polynomial cannot be factored.

Factoring by Grouping

§ 5.6

Factoring Trinomials

Factor the polynomial x2 + 13x + 30.Since our two numbers must have a product of 30 and a sum of 13, the two numbers must both be positive.

Positive factors of 30 Sum of Factors1, 30 312, 15 17

3, 10 13

Note, there are other factors, but once we find a pair that works, we do not have to continue searching.

So x2 + 13x + 30 = (x + 3)(x + 10).

Factoring Trinomials of the Form x2 + bx + c

Example:

Factor the polynomial x2 – 11x + 24.Since our two numbers must have a product of 24 and a sum of –11, the two numbers must both be negative.

Negative factors of 24 Sum of Factors – 1, – 24 – 25 – 2, – 12 – 14

– 3, – 8 – 11

So x2 – 11x + 24 = (x – 3)(x – 8).

Factoring Trinomials of the Form x2 + bx + c

Example:

Factor the polynomial x2 – 6x + 10.

Since our two numbers must have a product of 10 and a sum of – 6, the two numbers will have to both be negative.

Negative factors of 10 Sum of Factors – 1, – 10 – 11 – 2, – 5 – 7

Since there is not a factor pair whose sum is – 6, x2 – 6x +10 is not factorable and we call it a prime polynomial.

Prime Polynomials

Example:

You should always check your factoring results by multiplying the factored polynomial to verify that it is equal to the original polynomial.Many times you can detect computational errors or errors in the signs of your numbers by checking your results.

Check Your Result!

Factoring Trinomials of the Form ax2 + bx + c

Returning to the FOIL method, F O I L

(3x + 2)(x + 4) = 3x2 + 12x + 2x + 8 = 3x2 + 14x + 8

To factor ax2 + bx + c into (#1·x + #2)(#3·x + #4), note that a is the product of the two first coefficients, c is the product of the two last coefficients and b is the sum of the products of the outside coefficients and inside coefficients.Note that b is the sum of 2 products, not just 2 numbers, as in the last section.

Factor the polynomial 2x2 + 11x + 5.

Factoring Trinomials of the Form ax2 + bx + c

Example:

Factor the polynomial 2x2 + 11x + 5.

Factoring Trinomials of the Form ax2 + bx + c

Example:

My Fat Dog Sam Stinks

Factor the polynomial 2x2 + 11x + 5.

Factoring Trinomials of the Form ax2 + bx + c

Example:

My Fat Dog Sam Stinks

Multiply

Factor the polynomial 2x2 + 11x + 5.

Factoring Trinomials of the Form ax2 + bx + c

Example:

My Fat Dog Sam Stinks

Factor

Factor the polynomial 2x2 + 11x + 5.

Factoring Trinomials of the Form ax2 + bx + c

Example:

My Fat Dog Sam Stinks

Divide

Factor the polynomial 2x2 + 11x + 5.

Factoring Trinomials of the Form ax2 + bx + c

Example:

My Fat Dog Sam Stinks

Simplify

Factor the polynomial 2x2 + 11x + 5.

Factoring Trinomials of the Form ax2 + bx + c

Example:

My Fat Dog Sam Stinks

Slide

Factor the polynomial 2x2 + 11x + 5.Possible factors of 25x2 are {x, 25x} or {5x, 5x}.Possible factors of 4 are {1, 4} or {2, 2}.

We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors.

Keep in mind that, because some of our pairs are not identical factors, we may have to exchange some pairs of factors and make 2 attempts before we can definitely decide a particular pair of factors will not work.

Factoring Trinomials of the Form ax2 + bx + c

Continued.

Example:

We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 20x.

{x, 25x} {1, 4} (x + 1)(25x + 4) 4x 25x 29x

(x + 4)(25x + 1) x 100x 101x

{x, 25x} {2, 2} (x + 2)(25x + 2) 2x 50x 52x

Factors of 25x2

Resulting Binomials

Product of Outside Terms

Product of Inside Terms

Sum of Products

Factors of 4

{5x, 5x} {2, 2} (5x + 2)(5x + 2) 10x 10x 20x

Factoring Trinomials of the Form ax2 + bx + c

Continued.

Example continued:

Check the resulting factorization using the FOIL method.

(5x + 2)(5x + 2) =

= 25x2 + 10x + 10x + 4

5x(5x)F

+ 5x(2)O

+ 2(5x)I

+ 2(2)L

= 2x2 + 11x + 5

So our final answer when asked to factor 2x2 + 11x + 5 will be (5x + 2)(5x + 2) or (5x + 2)2.

Factoring Trinomials of the Form ax2 + bx + c

Example continued:

Factor the polynomial 21x2 – 41x + 10.

Possible factors of 21x2 are {x, 21x} or {3x, 7x}.

Since the middle term is negative, possible factors of 10 must both be negative: {-1, -10} or {-2, -5}.

We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors.

Factoring Trinomials of the Form ax2 + bx + c

Continued.

Example:

We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to 41x.

Factors of 21x2

Resulting Binomials

Product of Outside Terms

Product of Inside Terms

Sum of Products

Factors of 10

{x, 21x}{1, 10}(x – 1)(21x – 10) –10x 21x – 31x

(x – 10)(21x – 1) –x 210x – 211x

{x, 21x} {2, 5} (x – 2)(21x – 5) –5x 42x – 47x

(x – 5)(21x – 2) –2x 105x – 107x

Factoring Trinomials of the Form ax2 + bx + c

Continued.

Example continued:

Factors of 21x2

Resulting Binomials

Product of Outside Terms

Product of Inside Terms

Sum of Products

Factors of 10

(3x – 5)(7x – 2) 6x 35x 41x

{3x, 7x}{1, 10}(3x – 1)(7x – 10) 30x 7x 37x

(3x – 10)(7x – 1) 3x 70x 73x

{3x, 7x} {2, 5} (3x – 2)(7x – 5) 15x 14x 29x

Factoring Trinomials of the Form ax2 + bx + c

Continued.

Example continued:

Check the resulting factorization using the FOIL method.

(3x – 5)(7x – 2) =

= 21x2 – 6x – 35x + 10

3x(7x)F

+ 3x(-2)O

- 5(7x)I

- 5(-2)L

= 21x2 – 41x + 10

So our final answer when asked to factor 21x2 – 41x + 10 will be (3x – 5)(7x – 2).

Factoring Trinomials of the Form ax2 + bx + c

Example continued: