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 [latex]y^4 - 81x^4[/latex].
This is a standard difference of squares that can be rewritten as [latex](y^2)^2 - (9x^2)^2[/latex], which factors to [latex](y^2 - 9x^2)(y^2 + 9x^2)[/latex]. This is not completely factored yet, since [latex](y^2 - 9x^2)[/latex] can be factored once more to give [latex](y - 3x)(y + 3x)[/latex].
Therefore, [latex]y^4 - 81x^4 = (y^2 + 9x^2)(y - 3x)(y + 3x)[/latex].
This multiple factoring of an equation is also common in mixing differences of squares with differences of cubes.
Example 7.6.2
Factor [latex]x^6 - 64y^6[/latex].This is a standard difference of squares that can be rewritten as [latex](x^3)^2 + (8x^3)^2[/latex], which factors to [latex](x^3 - 8y^3)(x^3 + 8x^3)[/latex]. This is not completely factored yet, since both [latex](x^3 - 8y^3)[/latex] and [latex](x^3 + 8x^3)[/latex] can be factored again.
[latex](x^3-8y^3)=(x-2y)(x^2+2xy+y^2)[/latex] and
[latex](x^3+8y^3)=(x+2y)(x^2-2xy+y^2)[/latex]
This means that the complete factorization for this is:
[latex]x^6 - 64y^6 = (x - 2y)(x^2 + 2xy + y^2)(x + 2y)(x^2 - 2xy + y^2)[/latex]
Example 7.6.3
A more challenging equation to factor looks like [latex]x^6 + 64y^6[/latex]. 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.
[latex]x^6 + 64y^6 = (x^2)^3 + (4y^2)^3[/latex]
In this form, [latex](x^2)^3+(4y^2)^3[/latex] factors to [latex](x^2+4y^2)(x^4+4x^2y^2+64y^4)[/latex].
Therefore, [latex]x^6 + 64y^6 = (x^2 + 4y^2)(x^4 + 4x^2y^2 + 64y^4)[/latex].
Example 7.6.4
Consider encountering a sum and difference of squares question. These can be factored as follows: [latex](a + b)^2 - (2a - 3b)^2[/latex] factors as a standard difference of squares as shown below:
[latex](a+b)^2-(2a-3b)^2=[(a+b)-(2a-3b)][(a+b)+(2a-3b)][/latex]
Simplifying inside the brackets yields:
[latex][a + b - 2a + 3b] [a + b + 2a - 3b][/latex]
Which reduces to:
[latex][-a + 4b] [3a - 2b][/latex]
Therefore:
[latex](a + b)^2 - (2a - 3b)^2 = [-a - 4b] [3a - 2b][/latex]
Examples 7.6.5
Consider encountering the following difference of cubes question. This can be factored as follows:
[latex](a + b)^3 - (2a - 3b)^3[/latex] factors as a standard difference of squares as shown below:
[latex](a+b)^3-(2a-3b)^3[/latex]
[latex]=[(a+b)-(2a+3b)][(a+b)^2+(a+b)(2a+3b)+(2a+3b)^2][/latex]
Simplifying inside the brackets yields:
[latex][a+b-2a-3b][a^2+2ab+b^2+2a^2+5ab+3b^2+4a^2+12ab+9b^2][/latex]
Sorting and combining all similar terms yields:
[latex]\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] \\ \hline &[-a-2b]&[7a^2+19ab+13b^2] \end{array}[/latex]
Therefore, the result is:
[latex](a + b)^3 - (2a - 3b)^3 = [-a - 2b] [7a^2 + 19ab + 13b^2][/latex]
Questions
Completely factor the following equations.
- [latex]x^4-16y^4[/latex]
- [latex]16x^4-81y^4[/latex]
- [latex]x^4-256y^4[/latex]
- [latex]625x^4-81y^4[/latex]
- [latex]81x^4-16y^4[/latex]
- [latex]x^4-81y^4[/latex]
- [latex]625x^4-256y^4[/latex]
- [latex]x^4-81y^4[/latex]
- [latex]x^6-y^6[/latex]
- [latex]x^6+y^6[/latex]
- [latex]x^6-64y^6[/latex]
- [latex]64x^6+y^6[/latex]
- [latex]729x^6-y^6[/latex]
- [latex]729x^6+y^6[/latex]
- [latex]729x^6+64y^6[/latex]
- [latex]64x^6-15625y^6[/latex]
- [latex](a+b)^2-(c-d)^2[/latex]
- [latex](a+2b)^2-(3a-4b)^2[/latex]
- [latex](a+3b)^2-(2c-d)^2[/latex]
- [latex](3a+b)^2-(a-b)^2[/latex]
- [latex](a+b)^3-(c-d)^3[/latex]
- [latex](a+3b)^3+(4a-b)^3[/latex]
