Section A.3 – Polynomials and Factoring
- Definition, properties, and terminology for polynomials
- Operations with polynomials
- Special products of polynomials
- Factoring polynomials
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Factor the expression \(4x^2y^2-12xy+9\)
Answer: \((2xy-3)^2\)
Solution: This is an example of a binomial square.
\(4x^2y^2=(2xy)^2 \text{, } 9=(3)^2\text{, and } 12xy=2(2xy)(3)\)
Therefore, \(4x^2y^2-12xy+9=(2xy-3)^2\)
2. Factor the expression \(10y^2-3y-1\)
Answer: \((5x+1)(2x-1)\)
Solution: Need factors of \(10\cdot(-1)\) that add up to \(-3\).
Those factors are \(-5\) and \(2\), so we need to place them in parentheses in such a way that the inner terms multiply to \(2x\) and the outer terms multiply to \(-5x\):
\[ ( \quad \quad \qquad )( \quad \quad \quad )\]
\(10x^2\) has factors of \(5x\) and \(2x\), which we can place in our parentheses like below.
\[ (5x\quad \quad )(2x\quad \quad )\]
The outer terms should still multiply to \(-5x\) and the inner terms to \(2x\). \(1\) obviously has factors of \(1\) and \(1\), so...
\[ (5x \quad 1)(2x \quad 1)\]
Since the product of the outside two numbers needs to be \(-5x\) and the inner two numbers need to have a product of \(2x\), we get
\[ (5x + 1)(2x - 1)\]
3. Factor the expression \((x-1)(x^2+3)+(x-1)(x^2-5)\)
Answer: \(2(x-1)^2(x+1)\)
Solution: This type of problem is best factored using factoring by grouping.
Since each term has a factor of \((x-1)\), we can factor \((x-1)\) out.
\[\begin{align}
(x-1)(x^2+3)+(x-1)(x^2-5)
&=(x-1)[(x^2+3)+(x^2-5)]\\
&=(x-1)[2x^2-2]\\
&=2(x-1)(x^2-1)\\
&=2(x-1)(x+1)(x-1)\\
&=2(x-1)^2(x+1)
\end{align}\]