factoring trinomials: ax2 + bx + c

OBJECTIVE:

find the factors of a trinomial of the form ax2 + bx + c

pp 138-139, text

factoring trinomials: ax2 + bx + c

factors:

Review of past lessons numbers or variables that make up a given product greatest number that could be

found in every set of factors of a given group numbers

GCF:

binomial:

a polynomial of two terms trinomial:

a polynomial of three terms

factoring trinomials: ax2 + bx + c

coefficient:

Review of past lessons the numerical factor next to a variable the small number on the upper

hand of a factor that tells how many times it will used as factor

exponent:

binomial:

a polynomial of two terms trinomial:

a polynomial of three terms

factoring trinomials: ax2 + bx + c

( x + 4)2

Review of past lessons = x2

+ 4x

+ 16 ( b - 3)2

= b2

- 6b

+ 9 ( y - 5)

( y + 3)

= y2

- 2y

-15 ( m - 7)

( m + 7)

= m2

- 49 ( a2+16a+64)

= ( )( )

a

a

+ 8

+ 8

= (a + 8)2

( 4a2+20a+24)

= 4

( )

a2

+5a

+ 6

factoring trinomials: ax2 + bx + c

(4a2+20a+24)

= 4

( )

a2

+5a

+ 6 (a2 + 5a + 6)

= 4

= 4

( )

( )

a

a

+ 3

+ 2

Example 1. Factor 12y2 – y – 6

Find the product of the coefficient of the first term (12) and the last term (–6).

12y2 – y – 6

Find the factors of -72 that will add up to -1.

12(-6) = -72

-72 = -9, 8-9 + 8 = -1

Use the factors -9 and 8 for the coefficient of the middle term (-1)

12y2 + (– 9 + 8)y – 6 Use the DPMoA

12y2 + (– 9y + 8y) – 6 Remove the parenthesis.

(4y – 3) Use the Distributive Property.

3y The factored form of 12y2 – y – 6

12y2 – 9y + 8y – 6 Group terms that have common monomial factors

(12y2 – 9y) + (8y – 6) Factor each binomial using GCF.

3y (4y – 3) + 2

(4y – 3y) ( ) + 2

Example 2. Factor 3x2 + 4x + 1

Find the product of the coefficient of the first term (3) and the last term (1).

3x2 + 4x + 1

Find the factors of 3 that will add up to 4.

3(1) = 3

3 = 3,13 + 1 = 4

Use the factors 3 and 1 for the coefficient of the middle term (4)

3x2 + (3+ 1)x + 1 Use the DPMoA

3x2 + (3x+ x) + 1 Remove the parenthesis.

+ (x+1) Use the Distributive Property.

(x + 1) The factored form of 3x2 + 4x + 1.

3x2 + 3x + x + 1 Group terms that have common monomial factors

(3x2 + 3x) Factor each binomial using GCF.(x + 1)+

(x + 1)3x

3x

( ) + 1

Example 3. Factor completely 21y2 – 35y – 56.

Factor out the GCF.

Factor the new polynomial, if possible. Find the product of 3 and -8. -24

7[3y2 + (– 8 + 3)y – 8] Remove the parenthesis.

21y2 – 35y – 56

7(3y2 – 5y – 8)

Find the factors of -24 that will add up to -5 which is the middle term.

-24 = - 8, 3 Use the -8 + 3 in place of -5 in the middle term.

7[3y2 – 8y + 3y – 8] Group terms that have common monomial factors

Use the Distributive Property.

7

The factored form of 12y2 – y – 6.

Group terms that have common monomial factors

7[(3y2 – 8y) + (3y – 8)] Take out the GCF from the first binomial.

7[3y2 – 8y + 3y – 8]

7[y(3y – 8) + (3y – 8)]

( ) y (3y – 8)

+ 1

factoring trinomials: ax2 + bx + c

Classwork p 163, Practice book

pp 138-139, text

homework p 164, Practice book