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8-4
++bx
8-4 Factoring
Factoringax
ax
bx++cc
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
11
8-4 Factoring ax2 + bx + c
Warm Up
Find each product.
1. (x – 2)(2x + 7)
2x2 + 3x – 14
2. (3y + 4)(2y + 9) 6y2 + 35y + 36
3. (3n – 5)(n – 7) 3n2 – 26n + 35
Find each trinomial.
4. x2 +4x – 32 (x – 4)(x + 8)
5. z2 + 15z + 36 (z + 3)(z + 12)
6. h2 – 17h + 72 (h – 8)(h – 9)
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Objective
Factor quadratic trinomials of the form
ax2 + bx + c.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
In the previous lesson you factored
trinomials of the form x2 + bx + c.
Now you will factor trinomials of the
form ax + bx + c, where a ≠ 0.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
When you multiply (3x + 2)(2x + 5), the coefficient
of the x2-term is the product of the coefficients of
the x-terms. Also, the constant term in the trinomial
is the product of the constants in the binomials.
(3x + 2)(2x + 5) = 6x2 + 19x + 10
Holt Algebra 1
8-4 Factoring ax2 + bx + c
To factor a trinomial like ax2 + bx + c into its
binomial factors, write two sets of parentheses
( x + )( x + ).
Write two numbers that are factors of a next to the
x’s and two numbers that are factors of c in the
other blanks. Multiply the binomials to see if you are
correct.
(3x + 2)(2x + 5) = 6x2 + 19x + 10
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 1: Factoring ax2 + bx + c by Guess and Check
Factor 6x2 + 11x + 4 by guess and check.
(
+
)(
+
)
Write two sets of parentheses.
2
( x + )( x + ) The first term is 6x , so at least
one variable term has a
coefficient other than 1.
The coefficient of the x2 term is 6. The constant term in
the trinomial is 4.
(2x + 4)(3x + 1) = 6x2 + 14x + 4
Try factors of 6 for the
2

(1x + 4)(6x + 1) = 6x + 25x + 4
coefficients and
(1x + 2)(6x + 2) = 6x2 + 14x + 4
factors of 4 for the
(1x + 1)(6x + 4) = 6x2 + 10x + 4
constant terms.
(3x + 4)(2x + 1) = 6x2 + 11x + 4
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 1 Continued
Factor 6x2 + 11x + 4 by guess and check.
(
+
)(
+
)
( x + )( x + )
Write two sets of parentheses.
The first term is 6x2, so at least
one variable terms has a
coefficient other than 1.
The factors of 6x2 + 11x + 4 are (3x + 4) and (2x + 1).
6x2 + 11x + 4 = (3x + 4)(2x + 1)
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 1a
Factor each trinomial by guess and check.
6x2 + 11x + 3
( + )( + )
Write two sets of parentheses.
( x + )( x + )
The first term is 6x2, so at least
one variable term has a
coefficient other than 1.
The coefficient of the x2 term is 6. The constant term in
the trinomial is 3.
(2x + 1)(3x + 3) = 6x2 + 9x + 3  Try factors of 6 for the
coefficients and
(1x + 3)(6x + 1) = 6x2 + 19x + 3
factors of 3 for the
2
(1x + 1)(6x + 3) = 6x + 9x + 3 
constant terms.
(3x + 1)(2x + 3) = 6x2 + 11x + 3
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 1a Continued
Factor each trinomial by guess and check.
6x2 + 11x + 3
(
+
)(
+
)
( x + )( x + )
Write two sets of parentheses.
The first term is 6x2, so at least
one variable term has a
coefficient other than 1.
The factors of 6x2 + 11x + 3 are (3x + 1)(2x + 3).
6x2 + 11x + 3 = (3x + 1)(2x +3)
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 1b
Factor each trinomial by guess and check.
3x2 – 2x – 8
(
+
)(
+
)
Write two sets of parentheses.
2, so at least
The
first
term
is
3x
( x + )( x + )
one variable term has a
coefficient other than 1.
The coefficient of the x2 term is 3. The constant term in
the trinomial is –8.
(1x – 1)(3x + 8) = 3x2 + 5x – 8  Try factors of 3 for the
coefficients and
2

(1x – 4)(3x + 2) = 3x – 10x – 8
factors of 8 for the
(1x – 8)(3x + 1) = 3x2 – 23x – 8  constant terms.
(1x – 2)(3x + 4) = 3x2 – 2x – 8 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 1b
Factor each trinomial by guess and check.
3x2 – 2x – 8
(
+
)(
+
)
( x + )( x + )
Write two sets of parentheses.
The first term is 3x2, so at least
one variable term has a
coefficient other than 1.
The factors of 3x2 – 2x – 8 are (x – 2)(3x + 4).
3x2 – 2x – 8 = (x – 2)(3x + 4)
Holt Algebra 1
8-4 Factoring ax2 + bx + c
So, to factor a2 + bx + c, check the factors of a and
the factors of c in the binomials. The sum of the
products of the outer and inner terms should be b.
Product = c
Product = a
(
X+
)(
x+
) = ax2 + bx + c
Sum of outer and inner products = b
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Since you need to check all the factors of a and the
factors of c, it may be helpful to make a table. Then
check the products of the outer and inner terms to
see if the sum is b. You can multiply the binomials
to check your answer.
Product = c
Product = a
(
X+
)(
x+
) = ax2 + bx + c
Sum of outer and inner products = b
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 2A: Factoring ax2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
2x2 + 17x + 21
(
x+
)(
x+
a = 2 and c = 21,
) Outer + Inner = 17.
Factors of 2 Factors of 21
1 and 21
1 and 2
21 and 1
1 and 2
3 and 7
1 and 2
7 and 3
1 and 2
Outer + Inner
1(21) + 2(1) = 23
1(1) + 2(21) = 43
1(7) + 2(3) = 13
1(3) + 2(7) = 17 
Use the Foil method.
(x + 7)(2x + 3)
Check (x + 7)(2x + 3) = 2x2 + 3x + 14x + 21
= 2x2 + 17x + 21 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Remember!
When b is negative and c is positive, the factors
of c are both negative.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 2B: Factoring ax2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
3x2 – 16x + 16
(
x+
)(
x+
)
a = 3 and c = 16,
Outer + Inner = –16.
Factors of 3 Factors of 16 Outer + Inner
1 and 3
–1 and –16 1(–16) + 3(–1) = –19
1 and 3
– 2 and – 8 1( – 8) + 3(–2) = –14
– 4 and – 4 1( – 4) + 3(– 4)= –16
1 and 3



(x – 4)(3x – 4)
Use the Foil method.
Check (x – 4)(3x – 4) = 3x2 – 4x – 12x + 16
= 3x2 – 16x + 16 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 2a
Factor each trinomial. Check your answer.
6x2 + 17x + 5
(
x+
)(
x+
Factors of 6 Factors of 5
1 and 5
1 and 6
1 and 5
2 and 3
1 and 5
3 and 2
)
a = 6 and c = 5,
Outer + Inner = 17.
Outer + Inner
1(5) + 6(1) = 11
2(5) + 3(1) = 13
3(5) + 2(1) = 17 


Use the Foil method.
(3x + 1)(2x + 5)
Check (3x + 1)(2x + 5) = 6x2 + 15x + 2x + 5
= 6x2 + 17x + 5 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 2b
Factor each trinomial. Check your answer.
9x2 – 15x + 4
( x + )( x +
)
a = 9 and c = 4,
Outer + Inner = –15.
Factors of 9 Factors of 4 Outer + Inner
3 and 3
–1 and – 4 3(–4) + 3(–1) = –15
3 and 3
– 2 and – 2 3(–2) + 3(–2) = –12 
– 4 and – 1 3(–1) + 3(– 4)= –15
3 and 3

(3x – 4)(3x – 1)
Use the Foil method.
Check (3x – 4)(3x – 1) = 9x2 – 3x – 12x + 4
= 9x2 – 15x + 4
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 2c
Factor each trinomial. Check your answer.
3x2 + 13x + 12
(
x+
)(
x+
)
Factors of 3 Factors of 12
1 and 3
1 and 12
2 and 6
1 and 3
3 and 4
1 and 3
(x + 3)(3x + 4)
a = 3 and c = 12,
Outer + Inner = 13.
Outer + Inner
1(12) + 3(1) = 15
1(6) + 3(2) = 12 
1(4) + 3(3) = 13 
Use the Foil method.
Check (x + 3)(3x + 4) = 3x2 + 4x + 9x + 12
= 3x2 + 13x + 12 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
When c is negative, one factor of c will be
positive and the other factor will be negative.
Only some of the factors are shown in the
examples, but you may need to check all of the
possibilities.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 3A: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
3n2 + 11n – 4
(
n+
)(
n+
)
a = 3 and c = – 4,
Outer + Inner = 11 .
Factors of 3 Factors of 4 Outer + Inner
1 and 3
–1 and 4 1(4) + 3(–1) = 1
1(2) + 3(–2) = – 4 
1 and 3
–2 and 2
–4 and 1
1(1) + 3(–4) = –11 
1 and 3
4 and –1
1(–1) + 3(4) = 11
1 and 3

(n + 4)(3n – 1)
Use the Foil method.
Check (n + 4)(3n – 1) = 3n2 – n + 12n – 4
= 3n2 + 11n – 4
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 3B: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
2x2 + 9x – 18
(
x+
)(
x+
)
a = 2 and c = –18,
Outer + Inner = 9.
Factors of 2 Factors of – 18 Outer + Inner
1(– 1) + 2(18) = 35
1 and 2
18 and –1
1(– 2) + 2(9) = 16 
1 and 2
9 and –2
6 and –3
1(– 3) + 2(6) = 9
1 and 2

(x + 6)(2x – 3)
Use the Foil method.
Check (x + 6)(2x – 3) = 2x2 – 3x + 12x – 18
= 2x2 + 9x – 18 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 3C: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
4x2 – 15x – 4
(
x+
)(
x+
Factors of 4 Factors of – 4
1 and 4
–1 and 4
1 and 4
–2 and 2
–4 and 1
1 and 4
(x – 4)(4x + 1)
)
a = 4 and c = –4,
Outer + Inner = –15.
Outer + Inner
1(4) – 1(4) = 0 
1(2) – 2(4) = –6 
1(1) – 4(4) = –15 
Use the Foil method.
Check (x – 4)(4x + 1) = 4x2 + x – 16x – 4
= 4x2 – 15x – 4 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 3a
Factor each trinomial. Check your answer.
6x2 + 7x – 3
(
x+
)(
x+
)
a = 6 and c = –3,
Outer + Inner = 7.
Factors of 6 Factors of – 3 Outer + Inner
6(–3) + 1(1) = –17
6 and 1
1 and –3
6(–1) + 1(3) = – 3 
6 and 1
3 and –1
3(–3) + 2(1) = – 7 
3 and 2
1 and –3
3(–1) + 2(3) = 3 
3 and 2
3 and –1
2(–3) + 3(1) = – 3
2 and 3
1 and –3
1(–2) + 3(3) = 7 
2 and 3
3 and –1


(3x – 1)(2x + 3)
Use the Foil method.
Check (3x – 1)(2x + 3) = 6x2 + 9x – 2x – 3
= 6x + 7x – 3
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 3b
Factor each trinomial. Check your answer.
4n2 – n – 3
(
n+
)(
n+
)
a = 4 and c = –3,
Outer + Inner = –1.
Factors of 4 Factors of –3 Outer + Inner
1 and 4
–1(3) + 1(4) = 1 
1 and –3
1(3) –1(4) = – 1 
1 and 4
–1 and 3
(4n + 3)(n – 1)
Use the Foil method.
Check (4n + 3)(n – 1) = 4n2 – 4n + 3n – 3
= 4n2 – n – 3
Holt Algebra 1
8-4 Factoring ax2 + bx + c
When the leading coefficient is negative,
factor out –1 from each term before using
other factoring methods.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Caution
When you factor out –1 in an early step, you
must carry it through the rest of the steps.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 4A: Factoring ax2 + bx + c When a is
Negative
Factor –2x2 – 5x – 3.
–1(2x2 + 5x + 3)
–1(
x+
)(
x+
)
Factors of 2 Factors of 3
Outer + Inner
1 and 2
3 and 1
1(1) + 3(2) = 7 
1 and 2
1 and 3
1(3) + 1(2) = 5
(x + 1)(2x + 3)
–1(x + 1)(2x + 3)
Holt Algebra 1
Factor out –1.
a = 2 and c = 3;
Outer + Inner = 5
8-4 Factoring ax2 + bx + c
Check It Out! Example 4a
Factor each trinomial. Check your answer.
–6x2 – 17x – 12
–1(6x2 + 17x + 12)
–1(
x+
)(
x+
)
Factors of 6 Factors of 12
Outer + Inner

17 
2 and 3
4 and 3
2(3) + 3(4) = 18
2 and 3
3 and 4
2(4) + 3(3) =
(2x + 3)(3x + 4)
–1(2x + 3)(3x + 4)
Holt Algebra 1
Factor out –1.
a = 6 and c = 12;
Outer + Inner = 17
8-4 Factoring ax2 + bx + c
Check It Out! Example 4b
Factor each trinomial. Check your answer.
–3x2 – 17x – 10
–1(3x2 + 17x + 10)
–1(
x+
)(
x+
)
Factors of 3 Factors of 10
Outer + Inner

17 
1 and 3
2 and 5
1(5) + 3(2) = 11
1 and 3
5 and 2
1(2) + 3(5) =
(3x + 2)(x + 5)
–1(3x + 2)(x + 5)
Holt Algebra 1
Factor out –1.
a = 3 and c = 10;
Outer + Inner = 17)
8-4 Factoring ax2 + bx + c
Lesson Quiz
Factor each trinomial. Check your answer.
1. 5x2 + 17x + 6
(5x + 2)(x + 3)
2. 2x2 + 5x – 12
(2x– 3)(x + 4)
3. 6x2 – 23x + 7
(3x – 1)(2x – 7)
4. –4x2 + 11x + 20 (–x + 4)(4x + 5)
5. –2x2 + 7x – 3
(–2x + 1)(x – 3)
6. 8x2 + 27x + 9
(8x + 3)(x + 3)
Holt Algebra 1