56 7.5 Factoring Special Products
Now transition from multiplying special products to factoring special products. If you can recognize them, you can save a lot of time. The following is a list of these special products (note that a2 + b2 cannot be factored):
\(\begin{array}{lll}
a^2-b^2&=&(a+b)(a-b) \\
(a+b)^2&=&a^2+2ab+b^2 \\
(a-b)^2&=&a^2-2ab+b^2 \\
a^3-b^3&=&(a-b)(a^2+ab+b^2) \\
a^3+b^3&=&(a+b)(a^2-ab+b^2) \\
\end{array}\)
The challenge is therefore in recognizing the special product.
Example 7.5.1
Factor \(x^2 - 36\).
This is a difference of squares. \((x - 6)(x + 6)\) is the solution.
Example 7.5.2
Factor \(x^2 - 6x + 9\).
This is a perfect square. \((x - 3)(x - 3)\) or \((x - 3)^2\) is the solution.
Example 7.5.3
Factor \(x^2 + 6x + 9\).
This is a perfect square. \((x + 3)(x + 3)\) or \((x + 3)^2\) is the solution.
Example 7.5.4
Factor \(4x^2 + 20xy + 25y^2\).
This is a perfect square. \((2x + 5y)(2x + 5y)\) or \((2x + 5y)^2\) is the solution.
Example 7.5.5
Factor \(m^3 - 27\).
This is a difference of cubes. \((m - 3)(m^2 + 3m + 9)\) is the solution.
Example 7.5.6
Factor \(125p^3 + 8r^3\).
This is a difference of cubes. \((5p + 2r)(25p^2 - 10pr + 4r^2)\) is the solution.
Questions
Factor each of the following polynomials.
- \(r^2-16\)
- \(x^2-9\)
- \(v^2-25\)
- \(x^2-1\)
- \(p^2-4\)
- \(4v^2-1\)
- \(3x^2-27\)
- \(5n^2-20\)
- \(16x^2-36\)
- \(125x^2+45y^2\)
- \(a^2-2a+1\)
- \(k^2+4k+4\)
- \(x^2+6x+9\)
- \(n^2-8n+16\)
- \(25p^2-10p+1\)
- \(x^2+2x+1\)
- \(25a^2+30ab+9b^2\)
- \(x^2+8xy+16y^2\)
- \(8x^2-24xy+18y^2\)
- \(20x^2+20xy+5y^2\)
- \(8-m^3\)
- \(x^3+64\)
- \(x^3-64\)
- \(x^3+8\)
- \(216-u^3\)
- \(125x^3-216\)
- \(125a^3-64\)
- \(64x^3-27\)
- \(64x^3+27y^3\)
- \(32m^3-108n^3\)
