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.

 

Example 7.6.1

Factor y^4 - 81x^4.

This is a standard difference of squares that can be rewritten as (y^2)^2 - (9x^2)^2, which factors to (y^2 - 9x^2)(y^2 + 9x^2). This is not completely factored yet, since (y^2 - 9x^2) can be factored once more to give (y - 3x)(y + 3x).

Therefore, y^4 - 81x^4 = (y^2 + 9x^2)(y - 3x)(y + 3x).

This multiple factoring of an equation is also common in mixing differences of squares with differences of cubes.

Example 7.6.2

Factor x^6 - 64y^6.This is a standard difference of squares that can be rewritten as (x^3)^2 + (8x^3)^2, which factors to (x^3 - 8y^3)(x^3 + 8x^3). This is not completely factored yet, since both (x^3 - 8y^3) and (x^3 + 8x^3) can be factored again.

(x^3-8y^3)=(x-2y)(x^2+2xy+y^2) and
(x^3+8y^3)=(x+2y)(x^2-2xy+y^2)

This means that the complete factorization for this is:

x^6 - 64y^6  = (x - 2y)(x^2 + 2xy + y^2)(x + 2y)(x^2 - 2xy + y^2)

Example 7.6.3

A more challenging equation to factor looks like x^6 + 64y^6. 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.

x^6 + 64y^6 = (x^2)^3 + (4y^2)^3

In this form, (x^2)^3+(4y^2)^3 factors to (x^2+4y^2)(x^4+4x^2y^2+64y^4).

Therefore, x^6 + 64y^6 = (x^2 + 4y^2)(x^4 + 4x^2y^2 + 64y^4).

Example 7.6.4

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]

Examples 7.6.5

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][a^2+2ab+b^2+2a^2+5ab+3b^2+4a^2+12ab+9b^2]

Sorting and combining all similar terms yields:

\begin{array}{rrl} &[\phantom{-1}a+\phantom{0}b]&[\phantom{0}a^2+\phantom{0}2ab+\phantom{00}b^2] \\ &[-2a-3b]&[2a^2+\phantom{0}5ab+\phantom{0}3b^2] \\ +&&[4a^2+12ab+\phantom{0}9b^2] \\ \midrule &[-a-2b]&[7a^2+19ab+13b^2] \end{array}

Therefore, the result is:

(a + b)^3 - (2a - 3b)^3  =  [-a - 2b] [7a^2 + 19ab + 13b^2]

Questions

Completely factor the following equations.

  1. x^4-16y^4
  2. 16x^4-81y^4
  3. x^4-256y^4
  4. 625x^4-81y^4
  5. 81x^4-16y^4
  6. x^4-81y^4
  7. 625x^4-256y^4
  8. x^4-81y^4
  9. x^6-y^6
  10. x^6+y^6
  11. x^6-64y^6
  12. 64x^6+y^6
  13. 729x^6-y^6
  14. 729x^6+y^6
  15. 729x^6+64y^6
  16. 64x^6-15625y^6
  17. (a+b)^2-(c-d)^2
  18. (a+2b)^2-(3a-4b)^2
  19. (a+3b)^2-(2c-d)^2
  20. (3a+b)^2-(a-b)^2
  21. (a+b)^3-(c-d)^3
  22. (a+3b)^3+(4a-b)^3

Answer Key 7.6

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Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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