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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:
\(\begin{array}{rrl}
&[\phantom{-1}a+\phantom{0}b]&[\phantom{0}a^2+\phantom{0}2ab+\phantom{00}b^2] \\
&[-2a-3b]&[2a^2+\phantom{0}5ab+\phantom{0}3b^2] \\
+&&[4a^2+12ab+\phantom{0}9b^2] \\
\midrule
&[-a-2b]&[7a^2+19ab+13b^2]
\end{array}\)
Therefore, the result is:
\((a + b)^3 – (2a – 3b)^3 = [-a – 2b] [7a^2 + 19ab + 13b^2]\)
Questions
Completely factor the following equations.
- \(x^4-16y^4\)
- \(16x^4-81y^4\)
- \(x^4-256y^4\)
- \(625x^4-81y^4\)
- \(81x^4-16y^4\)
- \(x^4-81y^4\)
- \(625x^4-256y^4\)
- \(x^4-81y^4\)
- \(x^6-y^6\)
- \(x^6+y^6\)
- \(x^6-64y^6\)
- \(64x^6+y^6\)
- \(729x^6-y^6\)
- \(729x^6+y^6\)
- \(729x^6+64y^6\)
- \(64x^6-15625y^6\)
- \((a+b)^2-(c-d)^2\)
- \((a+2b)^2-(3a-4b)^2\)
- \((a+3b)^2-(2c-d)^2\)
- \((3a+b)^2-(a-b)^2\)
- \((a+b)^3-(c-d)^3\)
- \((a+3b)^3+(4a-b)^3\)
Answers to odd questions
1. \((x^2-4y^2)(x^2+4y^2)\)
\((x-2y)(x+2y)(x^2+4y^2)\)
3. \((x^2-16y^2)(x^2+16y^2)\)
\((x-4y)(x+4y)(x^2+16y^2)\)
5. \((9x^2-4y^2)(9x^2+4y^2)\)
\((3x+2y)(3x-2y)(9x^2+4y^2)\)
7. \((25x^2-16y^2)(25x^2+16y^2)\)
\((5x-4y)(5x+4y)(25x^2+16y^2)\)
9. \((x^3-y^3)(x^3+y^3)\)
\((x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)\)
11. \((x^3-8y^3)(x^3+8y^3)\)
\((x-2y)(x^2+2xy+4y^2)(x+2y)(x^2-2xy+4y^2)\)
13. \((27x^3-y^3)(27x^3+y^3)\)
\((3x-y)(9x^2+3xy+y^2)(3x+y)(9x^2-3xy+y^2)\)
15. \((9x^2)^3+(4y^2)^3\)
\((9x^2+4y^2)(81x^4-36x^2y^2+16y^4)\)
17. \([(a+b)-(c-d)][(a+b)+(c-d)]\)
\([a+b-c+d][a+b+c-d]\)
19. \([(a+3b)-(2c-d)][(a+3b)+(2c-d)]\)
\([a+3b-2c+d][a+3b+2c-d]\)
21. \([(a+b)-(c-d)][(a+b)^2+(a+b)(c-d)+(c-d)^2]\)
\([a+b-c+d][a^2+2ab+b^2+ac-ad+bc-bd+c^2-2cd+d^2]\)
