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
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