Special Factoring

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Sharp Tutor Mon, Jan 17, 2022 06:55 PM UTC

OBJECTIVES:
1. Factor a difference of squares. 2. Factor a perfect square trinomial. 3. Factor a difference of cubes. 4. Factor a sum of cubes.

1. Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 7.3 - 1

2. Chapter 7
Factoring
Copyright © 2010 Pearson Education, Inc. All rights reserved
Sec 7.3 - 2

3. 7.3
Special Factoring
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Sec 7.3 - 3

4. 7.3 Special Factoring
Objectives
1. Factor a difference of squares.
2. Factor a perfect square trinomial.
3. Factor a difference of cubes.
4. Factor a sum of cubes.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 4

5. 7.3 Special Factoring
The Difference of Squares
Difference of Squares
x2 – y2 = (x + y)(x – y)
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 5

6. 7.3 Special Factoring
EXAMPLE 1 Factoring Differences of Squares
Factor each polynomial.
(a) 2n2 – 50
There is a common factor of 2.
2n2 – 50 = 2(n2 – 25) Factor out the common factor.
= 2(n + 5)(n – 5) Factor the difference of squares.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 6

7. 7.3 Special Factoring
EXAMPLE 1 Factoring Differences of Squares
Factor each polynomial.
(b) 9g2 – 16
9g2 – 16 = (3g)2 – (4)2 = (3g + 4)(3g – 4)
A2 – B2 (A + B)(A – B)
(c) 4h2 – (w + 5)2
4h2 – (w + 5)2 = (2h)2 – (w + 5)2 = (2h + w + 5)(2h – [w + 5])
= (2h + w + 5)(2h – w – 5)
A2 – B2 (A + B) (A – B)
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 7

8. 7.3 Special Factoring
Caution
Assuming no greatest common factor except 1, it is not possible to
factor (with real numbers) a sum of squares, such as x2 + 16.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 8

10. 7.3 Special Factoring
EXAMPLE 2 Factoring Perfect Square Trinomials
Factor each polynomial.
(a) 9g2 – 42g + 49
Here 9g2 = (3g)2 and 49 = 72. The sign of the middle term is –, so if
9g2 – 42g + 49 is a perfect square trinomial, the factored form will
have to be
(3g – 7)2.
Take twice the product of the two terms to see if this is correct.
2(3g)(–7) = –42g
This is the middle term of the given trinomial, so
9g2 – 42g + 49 = (3g – 7)2.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 10

11. 7.3 Special Factoring
EXAMPLE 2 Factoring Perfect Square Trinomials
Factor each polynomial.
(b) 25x2 + 60xy + 64y2
If this is a perfect square trinomial, it will equal (5x + 8y)2. By the
pattern described earlier, if multiplied out, this squared binomial has a
middle term of 2(5x)(8y), which does not equal 60xy. Verify that this
trinomial cannot be factored by the methods of the previous section
either. It is prime.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 11

12. 7.3 Special Factoring
EXAMPLE 2 Factoring Perfect Square Trinomials
Factor each polynomial.
(c) (n – 4)2 + 18(n – 4) + 81 = [ (n – 4) + 9 ]2
= (n + 5)2,
since 2(n – 4)9 = 18(n – 4), the middle term.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 12

13. 7.3 Special Factoring
EXAMPLE 2 Factoring Perfect Square Trinomials
Factor each polynomial.
(d) c2 – 6c + 9 – h2
Since there are four terms, we will use factoring by grouping. The first
three terms here form a perfect square trinomial. Group them together,
and factor as follows.
(c2 – 6c + 9) – h2 = (c – 3)2 – h2
The result is the difference of squares. Factor again to get
= (c – 3 + h)(c – 3 – h).
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 13

15. 7.3 Special Factoring
EXAMPLE 3 Factoring Difference of Cubes
Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).
(a) a3 – 125 = a3 – 53
= (a – 5)(a2 + 5a + 52)
= (a – 5)(a2 + 5a + 25)
a3 –125
Check: = (a – 5)(a2 + 5a + 25)
–5a
Opposite of the product of the cube
roots gives the middle term.
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 15

16. 7.3 Special Factoring
EXAMPLE 3 Factoring Difference of Cubes
Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).
(b) 8g3 – h3 = (2g)3 – h3
= (2g – h) [ (2g)2 + (2g)(h) + h2) ]
= (2g – h)(4g2 + 2gh + h2)
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 16

17. 7.3 Special Factoring
EXAMPLE 3 Factoring Difference of Cubes
Factor each polynomial. Recall, x3 – y3 = (x – y)(x2 + xy + y2).
(c) 64m3 – 27n3 = (4m)3 – (3n)3
= (4m – 3n) [ (4m)2 + (4m)(3n) + (3n)2 ]
= (4m – 3n)(16m2 + 12mn + 9n2)
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 17

19. 7.3 Special Factoring
Note on Signs
The sign of the second term in the binomial factor of a sum or difference
of cubes is always the same as the sign in the original polynomial.
In the trinomial factor, the first and last terms are always positive;
the sign of the middle term is the opposite of the sign of the second term
in the binomial factor.
Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2)
Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 19

20. 7.3 Special Factoring
EXAMPLE 4 Factoring Sums of Cubes
Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).
(a) n 3 + 8 = n3 + 23
= (n + 2)(n2 – 2n + 22)
= (n + 2)(n2 – 2n + 4)
(b) 64v3 + 27g3 = (4v)3 + (3g)3
= (4v + 3g) [ (4v)2 – (4v)(3g) + (3g)2 ]
= (4v + 3g) (16v2 – 12gv + 9g2)
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 20

21. 7.3 Special Factoring
EXAMPLE 4 Factoring Sums of Cubes
Factor each polynomial. Recall, x3 + y3 = (x + y)(x2 – xy + y2).
(c) 2k3 + 250 = 2(k3 + 125)
= 2(k3 + 53)
= 2(k + 5)(k2 – 5k + 25)
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 21

22. 7.3 Special Factoring
Factoring Summary
Special Types of Factoring (Memorize)
Difference of Squares x2 – y2 = (x + y)(x – y)
Perfect Square Trinomial x2 + 2xy + y2 = (x + y)2
x2 – 2xy + y2 = (x – y)2
Difference of Cubes x3 – y3 = (x – y)(x2 + xy + y2)
Sum of Cubes x3 + y3 = (x + y)(x2 – xy + y2)
Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 7.3 - 22

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