1. Exponents, Polynomials and Functions3
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2. 3.5 Special Factoring Techniques
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3. Objectives Factor perfect square trinomials.
Factor difference of squares.
Factor difference and sum of cubes.
Factor polynomials that require multiple factoring processes or steps.
Factor trinomials that are quadratic in form.

4. Perfect Square Trinomials

5. Perfect Square Trinomials
A perfect square trinomial comes as a result of squaring a binomial. Using the distributive property and combining like terms, we get the following:
(a + b)2 = (a + b)(a + b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
(a – b)2 = (a – b)(a – b)
= a2 – ab – ab + b2
= a2 – 2ab + b2

6. Perfect Square Trinomials
If we look at the resulting trinomials, we can see a pattern for a perfect square trinomial.
The first and last terms of the trinomial will be perfect squares, and the middle term will be either plus or minus twice the product of the first and last terms that are being squared.
For factoring purposes, the relationship is better seen as follows.
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2

7. Example 1 – Factor perfect square trinomials
Factor the following:
a. x2 + 6x + 9
b. m2 – 14m + 49
c. 4a2 + 20a + 25
d. 9x2 – 24xy + 16y2

8. Example 1(a) – Solution
cont’d
First, we must confirm that x2 + 6x + 9 is a perfect square trinomial. The first term is x squared, the last term is 3 squared, and the middle term is twice the product of these two terms (2  3  x).
This trinomial can be factored into a binomial squared.
x2 + 6x + 9 = (x + 3)2
Check the factorization.
(x + 3)2 = (x + 3)(x + 3)
= x2 + 3x + 3x + 9
= x2 + 6x + 9

9. Example 1(b) – Solution
cont’d
The first term of m2 – 14m + 49 is m squared, the last term is 7 squared, and the middle term is minus twice the product of these two terms (2  7  m).
m2 – 14m + 49 = (m – 7)2
Check the factorization (m – 7)2 = (m – 7)(m – 7)
= m2 – 7m – 7m + 49
= m2 – 14m + 49

10. Example 1(c) – Solution
cont’d
The first term of 4a2 + 20a + 25 is 2a squared, the last term is 5 squared, and the middle term is twice the product of these two terms (2  2a  5).
4a2 + 20a + 25 = (2a + 5)2
Check the factorization.
(2a + 5)2 = (2a + 5)(2a + 5)
= 4a2 + 10a + 10a + 25
= 4a2 + 20a + 25

11. Example 1(d) – Solution
cont’d
The first term of the 9x2 – 24xy + 16y2 is 3x squared, the last term is 4y squared, and the middle term is minus twice the product of these two terms (2  3x  4y).
9x2 – 24xy + 16y2 = (3x – 4y)2
Check the factorization.
(3x – 4y)2 = (3x – 4y)(3x – 4y)
= 9x2 – 12xy – 12xy + 16y2
= 9x2 – 24xy + 16y2

12. Difference of Squares

13. Difference of Squares
A similar pattern that we can use is the difference of two squares.
When multiplying two binomials, one that is the sum of two terms and the other that is the difference of those same two terms, we get the difference of two squares.
(a + b)(a – b) = a2 – ab + ab – b2
= a2 – b2
From this we can see that a binomial made up of the difference of two perfect squares can be factored by following the pattern a2 – b2 = (a + b)(a – b)

14. Difference of Squares
The factorization is the sum of the two terms multiplied by the difference of the two terms.
Although the difference of two squares is factorable, the sum of two squares is not factorable. Many students want to factor the sum of two square as
(a + b)(a + b)
This is not a correct factorization. We can check this by multiplying the factored form out.
(a + b)(a + b) = a2 + ab + ab + b2
= a2 + 2ab + b2

15. Difference of Squares
We can see that the middle term of the final polynomial does not add to zero, so the product does not result in the sum of two squares.

16. Example 2 – Factoring the difference of two squares
Factor the following:
a. x2 – b. 4a2 – 25
c. x d. 9m2 – 16n2
Solution:
a. The expression x2 – 9 is the difference of x squared and squared, so we can factor it using the difference of two squares pattern.
x2 – 9 = x2 – 32
= (x + 3)(x – 3)

17. Example 2 – Solution Check the factorization.
cont’d
Check the factorization.
(x + 3)(x – 3) = x2 – 3x + 3x – 9
= x2 – 9
b. The expression 4a2 – 25 is the difference of 2a squared and 5 squared, so we can factor it using the difference of two squares pattern.
4a2 – 25 = (2a)2 – 52
= (2a + 5)(2a – 5)

18. Example 2 – Solution Check the factorization.
cont’d
Check the factorization.
(2a + 5)(2a – 5) = 4a2 – 10a + 10a – 25
= 4a2 – 25
c. The expression x is the sum of two squares, not the difference of two squares, and is not factorable using the rational numbers.

19. Example 2 – Solution
cont’d
d. The expression 9m2 – 16n2 is the difference of 3m squared and 4n squared, so we can factor it using the difference of two squares pattern.
9m2 – 16n2 = (3m)2 – (4n)2
= (3m – 4n)(3m + 4n)
Check the factorization.
(3m – 4n)(3m + 4n) = 9m2 + 12mn – 12mn – 16n2
= 9m2 – 16n2

20. Difference and Sum of Cubes

21. Difference and Sum of Cubes
Remember that the sum of two squares is not factorable; only the difference of two squares is. Two other patterns that we can use are the sum and difference of two cubes.
The sign changes in these patterns are very important, and we should be very careful when using these patterns to get the signs correct.
Sum of two cubes
Difference of two cubes

22. Example 3 – Factor sum and differences of cubes
Factor the following:
a. x3 + 27
b. m3 – 8
c. 8a b3
d. 2p3 – 54r3

23. Example 3(a) – Solution
The first term of x is x cubed, and the second term is 3 cubed, so we can use the pattern for the sum of two cubes.
x = x3 + 33
= (x + 3)(x2 – 3x + 9)
Check the factorization.
(x + 3) (x2 – 3x + 9) = x2 – 3x2 + 9x + 3x2 – 9x + 27
= x3 + 27

24. Example 3(b) – Solution
cont’d
The first term of m3 – 8 is m cubed, and the second term is 2 cubed, so we can use the pattern for the difference of two cubes.
m3 – 8 = m3 – 23
= (m – 2)(m2 + 2m + 4)
Check the factorization.
(m – 2)(m2 + 2m + 4) = m3 + 2m2 + 4m – 2m2 – 4m – 8
= m3 – 8

25. Example 3(c) – Solution
cont’d
The first term of 8a b3 is 2a cubed, and the second term is 5b cubed, so we can use the pattern for the sum of two cubes.
8a b3 = (2a)3 + (5b)3
= (2a + 5b)(4a2 – 10ab + 25b2)
Check the factorization.
(2a + 5b)(4a2 – 10ab + 25b2)
= 8a3 – 20a2b + 50ab2 + 20a2b – 50ab b3
= 8a b3

26. Example 3(d) – Solution
cont’d
For the expression 2p3 – 54r3, we will first factor out the 2 that is in common and then use the pattern for the difference of two cubes.
2p3 – 54r3 = 2(p3 – 27r3)
= 2(p – 3r)(p2 + 3pr + 9r2)
Check the factorization.
2(p – 3r)(p2 + 3pr + 9r2)
= 2(p3 + 3p2r + 9pr2 – 3p2r – 9pr2 – 27r3)
= 2(p3 – 27r3)
= 2p3 – 54r3
= 2[p3 – (3r)3]

27. Multistep Factorizations

28. Multistep Factorizations
Some polynomials take several steps to factor completely or require slightly different thinking to find the key to factoring them.
Always look for the greatest common factor and factor that out first. Then look for a pattern that you recognize and begin factoring.
Several of the patterns that we have discussed can be found in more complicated expressions.

29. Multistep Factorizations
The sum or difference of two cubes or the difference of two squares can be found in expressions with exponents that are multiples of 2 or 3.
x6 – y6 = (x3)2 – (y3)2
x6 – y6 = (x2)3 – (y2)3
This expression can be looked at as either a difference of two squares or a difference of two cubes.
This leads to two different paths to factoring, but they have the same result in the end. Using the difference of two squares first may make it easier to factor completely.

30. Multistep Factorizations
Difference of squares, then sum and difference of cubes.
x6 – y6 = (x3)2 – (y3)2
= (x3 + y3)(x3 – y3)
= (x + y)(x2 – xy + y2)(x – y)(x2 + xy + y2)
Difference of cubes, then difference of squares and AC method.
x6 – y6 = (x2)3 – (y2)3
= (x2 – y2)(x4 + x2y2 + y4)
= (x + y)(x – y)(x2 – xy + y2)(x2 + xy + y2)

31. Example 4 – Multistep factoring
Factor the following:
a. x4 – 81
b. 3x6 – 192y6
Solution:
a. This expression x4 – 81 is the difference of two squares, since the first term is x2 squared and the second term is squared.
x4 – 81 = (x2)2 – 92
= (x2 + 9)(x2 – 9)
= (x2 + 9)(x + 3)(x – 3)
The second factor is still the difference of two squares, so we can factor again.

32. Example 4 – Solution Check the factorization.
cont’d
Check the factorization.
(x2 + 9)(x + 3)(x – 3) = (x2 + 9)(x2 – 3x + 3x – 9)
= (x2 + 9)(x2 – 9)
= x4 – 9x2 + 9x2 – 81
= x4 – 81
b. We first factor out the 3 that is in common. The remaining expression x6 – 64y6 can be viewed as either the difference of two squares or the difference of two cubes.

33. Example 4 – Solution
cont’d
If we consider it to be the difference of two cubes, we end up needing to factor again, using the difference of two squares to completely finish the factoring.
3x6 – 192y6 = 3(x6 – 64y6)
= 3[(x2)3 – (4y2)3]
= 3(x2 – 4y2)(x4 + 4x2y2 + 16y4)
= 3(x – 2y)(x + 2y)(x4 + 4x2y2 + 16y4)
= 3(x – 2y)(x + 2y)(x2 – 2xy + 4y2)(x2 + 2xy + 4y2)
Factor using the difference of two cubes.
Finish by using the difference of two squares.

34. Example 4 – Solution Check the factorization.
cont’d
Check the factorization.
3(x – 2y)(x + 2y)(x2 – 2xy + 4y2)(x2 + 2xy + 4y2)
= 3(x2 – 4y2)(x4 + 4x2y2 + 16y4)
= 3(x6 + 4x2y2 + 16x2y2 – 4x4y2 – 16x2y4 – 64y6)
= 3(x6 – 64y6)
= 3x6 – 192y6

35. Trinomials in Quadratic Form

36. Trinomials in Quadratic Form
Trinomials of one variable are said to be in quadratic form when the degree of the highest term is twice that of the next term and the final term is a constant.
a(expression)2 + b(expression) + c
These types of situations take practice to see, so consider the following example and pay close attention to the thinking behind each factorization.

37. Example 5 – Multistep factoring
Factor the following:
a. 3a6 – a3 – 10
b. 12w4 + 52w2 + 35
Solution:
a. The trinomial 3a6 – a3 – 10 is quadratic in form because the first term has degree twice that of the second term and the third term is a constant.
We can use a substitution for a3 to make the trinomial appear to be quadratic. Then factor and replace the a3 back into the expression.

38. Example 5 – Solution
cont’d
We can use any variable for this substitution, so for this problem, we will use u. Letting u = a3, we get the following:
3a6 – a3 – 10 = 3(a3)2 – a3 – 10
= 3u2 – u – 10
= (3u + 5)(u – 2)
= (3a3 + 5)(a3 – 2)
Check the factorization.
(3a3 + 5)(a3 – 2) = 3a6 – 6a3 + 5a3 – 10
= 3a6 – a3 – 10
Substitute in u and factor.
u = a3
Replace the u with a3.

39. Example 5 – Solution
cont’d
b. The trinomial 12w4 + 52w is quadratic in form. Use substitution and then factor the remaining quadratic Using u = w2, we get the following:
12w4 + 52w = 12(w2)2 + 52w2 + 35
= 12u2 + 52u2 + 35
= (2u + 7)(6u + 5)
= (2w2 + 7)(6w2 + 5)
Check the factorization.
(2w2 + 7)(6w2 + 5) = 12w4 + 10w2 + 42w2 + 35
= 12w4 + 52w2 + 35
Substitute in u and factor.
u = w2
Replace the u with w2.

40. Choosing a Factoring Method

41. Choosing a Factoring Method
When we are faced with a polynomial to factor, we should choose a method that is best suited to factor the polynomial. We have looked at four methods of factoring:
Factor by grouping
AC method
Trial and error

42. Choosing a Factoring Method
Pattern recognition
Perfect square trinomials
Difference of squares
Difference of cubes
Sum of cubes
Quadratic form

43. Choosing a Factoring Method
All of these methods start by requiring you to take out the greatest common factor of all the terms.
If you do not factor out the GCF, all of these methods become more complicated or impossible.
Determining what method is best depends on the characteristics of the polynomial we are trying to factor.

44. Choosing a Factoring Method
Factor by Grouping
Factoring by grouping is used when we have four terms in the polynomial.
5x2 + 10x – 3x – 6
12a2 – 14a + 30ab – 35b

45. Choosing a Factoring Method
AC method or Trial and Error
Either the AC method or trial and error can be used with quadratics in standard form ax2 + bx + c.
x2 + 5x + 6
4x2 – 2x – 15
Trial and error is easiest when a = 1. After using the AC method, many people start to see more of the common factorization patterns and actually start using more trial and error.

46. Choosing a Factoring Method
Pattern Recognition
Recognizing patterns can help you to factor some polynomials quickly. The following types of polynomials can all be factored by using a pattern.
x2 + 6x x2 – 16
x x3 – x4 – 5x2 – 6
Perfect square trinomial
Difference of squares
Sum of cubes
Difference of cubes
Quadratic form

47. Choosing a Factoring Method
Perfect Square Trinomials
A polynomial in the form of a perfect square trinomial a2 + 2ab + b2 or a2 – 2ab + b2 can be factored by using the patterns
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
Difference of Squares
The difference of two squares a2 – b2 can be factored by using the pattern
a2 – b2 = (a + b)(a – b)

48. Choosing a Factoring Method
Sum and Difference of Cubes
The sum or difference of cubes a3 + b3 or a3 – b3 can be factored by using the patterns
a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab + b2)
Quadratic Form
A polynomial that is in quadratic form, that is, a(expression)2 + b(expression) + c, can be factored by using either the AC method or trial and error.