 # Section A.3 – Polynomials and Factoring

Section Details.
• Definition, properties, and terminology for polynomials
• Operations with polynomials
• Special products of polynomials
• Factoring polynomials

### Practice Problems

Directions. The following are review problems for the section. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it.
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, $$4x^2y^2-12xy+9=(2xy-3)^2$$

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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
$(5x + 1)(2x - 1)$

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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}

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4. Factor the following expression $$-6z^2+17z+3$$

$$(6z+1)(-z+3)$$

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