7.6 Factoring Quadratics of Increasing Difficulty
Factoring equations that are more difficult involves factoring equations and then checking the answers to see if they can be factored again.
Factor y4−81x4.
This is a standard difference of squares that can be rewritten as (y2)2−(9x2)2, which factors to (y2−9x2)(y2+9x2). This is not completely factored yet, since (y2−9x2) can be factored once more to give (y−3x)(y+3x).
Therefore, y4−81x4=(y2+9x2)(y−3x)(y+3x).
This multiple factoring of an equation is also common in mixing differences of squares with differences of cubes.
Factor x6−64y6.This is a standard difference of squares that can be rewritten as (x3)2+(8x3)2, which factors to (x3−8y3)(x3+8x3). This is not completely factored yet, since both (x3−8y3) and (x3+8x3) can be factored again.
(x3−8y3)=(x−2y)(x2+2xy+y2) and
(x3+8y3)=(x+2y)(x2−2xy+y2)
This means that the complete factorization for this is:
x6−64y6=(x−2y)(x2+2xy+y2)(x+2y)(x2−2xy+y2)
A more challenging equation to factor looks like x6+64y6. This is not an equation that can be put in the factorable form of a difference of squares. However, it can be put in the form of a sum of cubes.
x6+64y6=(x2)3+(4y2)3
In this form, (x2)3+(4y2)3 factors to (x2+4y2)(x4+4x2y2+64y4).
Therefore, x6+64y6=(x2+4y2)(x4+4x2y2+64y4).
Consider encountering a sum and difference of squares question. These can be factored as follows: (a+b)2−(2a−3b)2 factors as a standard difference of squares as shown below:
(a+b)2−(2a−3b)2=[(a+b)−(2a−3b)][(a+b)+(2a−3b)]
Simplifying inside the brackets yields:
[a+b−2a+3b][a+b+2a−3b]
Which reduces to:
[−a+4b][3a−2b]
Therefore:
(a+b)2−(2a−3b)2=[−a−4b][3a−2b]
Consider encountering the following difference of cubes question. This can be factored as follows:
(a+b)3−(2a−3b)3 factors as a standard difference of squares as shown below:
(a+b)3−(2a−3b)3
=[(a+b)−(2a+3b)][(a+b)2+(a+b)(2a+3b)+(2a+3b)2]
Simplifying inside the brackets yields:
[a+b−2a−3b][a2+2ab+b2+2a2+5ab+3b2+4a2+12ab+9b2]
Sorting and combining all similar terms yields:
[−1a+0b][0a2+02ab+00b2][−2a−3b][2a2+05ab+03b2]+[4a2+12ab+09b2][−a−2b][7a2+19ab+13b2]
Therefore, the result is:
(a+b)3−(2a−3b)3=[−a−2b][7a2+19ab+13b2]
Questions
Completely factor the following equations.
- x4−16y4
- 16x4−81y4
- x4−256y4
- 625x4−81y4
- 81x4−16y4
- x4−81y4
- 625x4−256y4
- x4−81y4
- x6−y6
- x6+y6
- x6−64y6
- 64x6+y6
- 729x6−y6
- 729x6+y6
- 729x6+64y6
- 64x6−15625y6
- (a+b)2−(c−d)2
- (a+2b)2−(3a−4b)2
- (a+3b)2−(2c−d)2
- (3a+b)2−(a−b)2
- (a+b)3−(c−d)3
- (a+3b)3+(4a−b)3
Answer Key 7.6