# Special Factoring Contributed by: 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.
Sec 7.3 - 1
2. Chapter 7
Factoring
Sec 7.3 - 2
3. 7.3
Special Factoring
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.
5. 7.3 Special Factoring
The Difference of Squares
Difference of Squares
x2 – y2 = (x + y)(x – y)
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.
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)
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.
9. 7.3 Special Factoring
Perfect Square Trinomial
Perfect Square Trinomial
x2 + 2xy + y2 = (x + y)2
x2 – 2xy + y2 = (x – y)2
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.
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.
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.
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).
14. 7.3 Special Factoring
Difference of Cubes
Difference of Cubes
x3 – y3 = (x – y)(x2 + xy + y2)
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.
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)
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)
18. 7.3 Special Factoring
Sum of Cubes
Sum of Cubes
x3 + y3 = (x + y)(x2 – xy + y2)
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)
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)
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)