Chapter 5.6: Factoring Quadratics of Increasing Difficulty

Mike LePine

Chapter 5.6: 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 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 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 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 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 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:

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