Q 90 - Intermediate Algebra

Question

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

$q^3-5q^2-24q$

Step-by-Step Solution

Answer

The factorization of the expression is

$q^3-5q^2-24q=q(q-8)(q+3)$

Step 1. Given Information

Consider the expression $q^3-5q^2-24q$

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

Step 2. Factor out the GCF

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

$\begin{array}{rcl}{q}^3-5q^2-24q& =& q·q^2-q·5q-q·24\\ & =& q(q^2-5q-24)\end{array}$

Step 3. Factor using trial and error

The factors of the first term and the last term of the trinomial $\begin{array}{rcl}& & q^2-5q-24\end{array}$ are

$$\begin{gathered} q^2=q·q \\ -24=-2·12 or -3·8 or -4·6 \end{gathered}$$

Consider all the combinations of factors 

$q^2-5q-24$
Possible Factors	Product
$(q-2)(q+12)$	$q^2+12q-2q-24=q^2+10q-24$
$(q+2)(q-12)$	$q^2-12q+2q-24=q^2-10q-24$
$(q-3)(q+8)$	$q^2+8q-3q-24=q^2+5q-24$
$(q+3)(q-8)$	$q^2-8q+3q-24=q^2-5q-24$
$(q+4)(q-6)$	$q^2-6q+4q-24=q^2-2q-24$
$(q-4)(q+6)$	$q^2+6q-4q-24=q^2+2q-24$

So the correct factors are $(q+3)(q-8)$

Thus, $q^3-5q^2-24q=q(q+3)(q-8)$

Step 4. Check the factors

Multiply the factors to check the solution 

$\begin{array}{rcl}q(q+3)(q-8)& =& q(q^2-8q+3q-24)\\ & =& q(q^2-5q-24)\\ & =& q^3-5q^2-24q\end{array}$

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