Q 89 - Intermediate Algebra

Question

In the following exercises, factor completely using trial and error. 

$p^3-8p^2-20p$

Step-by-Step Solution

Answer

The factorization of the expression is

$p^3-8p^2-20p=p(p-10)(p+2)$

Step 1. Given Information

Consider the expression $p^3-8p^2-20p$.

The objective is to factor the expression completely by using the trial and error method.

Step 2. Factor the GCF

The GCF of the three terms in the expression is $p$. So factor the GCF from the expression we get

$\begin{array}{rcl}{p}^3-8p^2-20p& =& p\times p^2-p\times 8p-p\times 20\\ & =& p(p^2-8p-20)\end{array}$

Step 3. Factor using trial and error

The factors of the first term and the last term of the trinomial $p^2-8p-20$ are

$$\begin{gathered} p^2=p·p \\ -20=-10·2 or -5·4 \end{gathered}$$

Consider all the combinations of factors

$p^2-8p-20$
Possible Factors	Product
$(p-10)(p+2)$	$p^2+2p-10p-20=p^2-8p-20$
$(p+10)(p-2)$	$p^2-2p+10p-20=p^2+8p-20$
$(p+5)(p-4)$	$p^2-4p+5p-20=p^2+p-20$
$(p-5)(p+4)$	$p^2+4p-5p-20=p^2-p-20$

So the correct factors is $(p-10)(p+2)$.

Thus, $p^3-8p^2-20p=p(p-10)(p+2)$

Step 4. Check the factors

Multiply the factors to check the solution

$\begin{array}{rcl}p(p-10)(p+2)& =& p(p^2+2p-10p-20)\\ & =& p(p^2-8p-20)\\ & =& p^3-8p^2-20p\end{array}$

And we get the given expression. So the expression is correctly factored.