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8-6
8-6 Choosing
Choosinga aFactoring
FactoringMethod
Method
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
11
8-6 Choosing a Factoring Method
Warm Up
Factor each trinomial.
1. x2 + 13x + 40 (x + 5)(x + 8)
2. 5x2 – 18x – 8 (5x + 2)(x – 4)
3. Factor the perfect-square trinomial
16x2 + 40x + 25 (4x + 5)(4x + 5)
4. Factor 9x2 – 25y2 using the difference of
two squares. (3x + 5y)(3x – 5y)
Holt Algebra 1
8-6 Choosing a Factoring Method
Objectives
Choose an appropriate method for
factoring a polynomial.
Combine methods for factoring a
polynomial.
Holt Algebra 1
8-6 Choosing a Factoring Method
Recall that a polynomial is in its fully factored
form when it is written as a product that cannot
be factored further.
GCF
4 Terms --> Grouping
3 Terms --> unfoiling
2 Terms --> Difference of Squares
if a2 - b2
Holt Algebra 1
8-6 Choosing a Factoring Method
If none of the factoring methods work, the polynomial
is said to be unfactorable.
Helpful Hint
For a polynomial of the form ax2 + bx + c, if
there are no numbers whose sum is b and whose
product is ac, then the polynomial is
unfactorable.
Holt Algebra 1
8-6 Choosing a Factoring Method
Caution
x2 + 4 is a sum of squares, and cannot be
factored.
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1A: Factoring
Factor 10x2 + 48x + 32 completely. Check
your answer.
10x2 + 48x + 32
2(5x2 + 24x + 16)
Factor out the GCF.
2(5x + 4)(x + 4)
Factor remaining trinomial.
Check 2(5x + 4)(x + 4) = 2(5x2 + 20x + 4x + 16)
= 10x2 + 40x + 8x + 32
= 10x2 + 48x + 32
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1B: Factoring
Factor 8x6y2 – 18x2y2 completely. Check your
answer.
8x6y2 – 18x2y2
2x2y2(4x4 – 9)
Factor out the GCF. 4x4 – 9 is a
perfect-square trinomial of
the form a2 – b2.
2x2y2(2x2 – 3)(2x2 + 3) a = 2x, b = 3
Check 2x2y2(2x2 – 3)(2x2 + 3) = 2x2y2(4x4 – 9)
= 8x6y2 – 18x2y2 
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1C
Factor each polynomial completely. Check
your answer.
4x3 + 16x2 + 16x
4x3 + 16x2 + 16x
4x(x2 + 4x + 4)
4x(x + 2)2
Factor out the GCF. x2 + 4x + 4 is a
perfect-square trinomial of the
form a2 + 2ab + b2.
a = x, b = 2
Check 4x(x + 2)2 = 4x(x2 + 2x + 2x + 4)
= 4x(x2 + 4x + 4)
= 4x3 + 16x2 + 16x 
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1D
Factor each polynomial completely. Check
your answer.
2x2y – 2y3
2y(x2 – y2)
Factor out the GCF. 2y(x2 – y2) is a
perfect-square trinomial of the
form a2 – b2.
2y(x + y)(x – y)
a = x, b = y
2x2y – 2y3
Check 2y(x + y)(x – y) = 2y(x2 + xy – xy – y2)
= 2x2y +2xy2 – 2xy2 – 2y3
= 2x2y – 2y3
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1E: Factoring
Factor each polynomial completely.
9x2 + 3x – 2
9x2 + 3x – 2
( x + )( x +
)
Factors of 9 Factors of 2
1 and –2
1 and 9
1 and –2
3 and 3
–1 and 2
3 and 3
(3x – 1)(3x + 2)
Holt Algebra 1
The GCF is 1 and there is no
pattern.
a = 9 and c = –2;
Outer + Inner = 3
Outer + Inner
1(–2) + 1(9) = 7
3(–2) + 1(3) = –3
3(2) + 3(–1) = 3 

8-6 Choosing a Factoring Method
Example 1F: Factoring
Factor each polynomial completely.
12b3 + 48b2 + 48b The GCF is 12b; (b2 + 4b + 4)
is a perfect-square
12b(b2 + 4b + 4)
trinomial in the form of
(x + )(x + )
a2 + 2ab + b2.
Factors of 4 Sum
1 and 4
5
2 and 2
4 a = 2 and c = 2
12b(b + 2)(b + 2)
12b(b + 2)2
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1G: Factoring
Factor each polynomial completely.
4y2 + 12y – 72
4(y2 + 3y – 18)
(y +
)(y +
)
Factor out the GCF. There is no
pattern. b = 3 and c = –18;
look for factors of –18 whose
sum is 3.
Factors of –18 Sum
–1 and 18
17 
–2 and 9
7
–3 and 6
3  The factors needed are –3
and 6
4(y – 3)(y + 6)
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1H: Factoring
Factor each polynomial completely.
x4 – x2
Holt Algebra 1
x2(x2 – 1)
Factor out the GCF.
x2(x + 1)(x – 1)
x2 – 1 is a difference of two
squares.
8-6 Choosing a Factoring Method
Example 1I
Factor each polynomial completely.
3x2 + 7x + 4
3x2 + 7x + 4
( x + )( x +
)
Factors of 3 Factors of 4
1 and 4
3 and 1
2 and 2
3 and 1
4 and 1
3 and 1
(3x + 4)(x + 1)
Holt Algebra 1
a = 3 and c = 4;
Outer + Inner = 7
Outer + Inner
3(4) + 1(1) = 13
3(2) + 1(2) = 8
3(1) + 1(4) = 7



8-6 Choosing a Factoring Method
Example 1J
Factor each polynomial completely.
2p5 + 10p4 – 12p3
2p3(p2 + 5p – 6)
(p +
)(p +
)
Factor out the GCF. There is no
pattern. b = 5 and c = –6;
look for factors of –6 whose
sum is 5.
Factors of – 6 Sum
– 1 and 6
5 The factors needed are –1
and 6
2p3(p + 6)(p – 1)
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1K
Factor each polynomial completely.
9q6 + 30q5 + 24q4
3q4(3q2
(
q+
Factor out the GCF. There is no
pattern.
+ 10q + 8)
)(
q+
)
Factors of 3 Factors of 8
1 and 8
3 and 1
2 and 4
3 and 1
4 and 2
3 and 1
3q4(3q + 4)(q + 2)
Holt Algebra 1
a = 3 and c = 8;
Outer + Inner = 10
Outer + Inner
3(8) + 1(1) = 25
3(4) + 1(2) = 14
3(2) + 1(4) = 10



8-6 Choosing a Factoring Method
Example 1L
Factor each polynomial completely.
2x4 + 18
2(x4 + 9)
Factor out the GFC.
x4 + 9 is the sum of squares
and that is not factorable.
2(x4 + 9) is completely factored.
Holt Algebra 1
8-6 Choosing a Factoring Method
Holt Algebra 1
8-6 Choosing a Factoring Method
Lesson Quiz
Tell whether the polynomial is completely
factored. If not, factor it.
1. (x + 3)(5x + 10)
2. 3x2(x2 + 9)
no; 5(x+ 3)(x + 2)
completely factored
Factor each polynomial completely. Check
your answer.
3. x3 + 4x2 + 3x + 12 4. 4x2 + 16x – 48
4(x + 6)(x – 2)
(x + 4)(x2 + 3)
5. 18x2 – 3x – 3
3(3x + 1)(2x – 1)
6. 18x2 – 50y2
2(3x + 5y)(3x – 5y)
7. 5x – 20x3 + 7 – 28x2 (1 + 2x)(1 – 2x)(5x + 7)
Holt Algebra 1