Q 94 - Intermediate Algebra

Question

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

$6y^4+12y^3-48y^2$

Step-by-Step Solution

Answer

The factorization of the expression is  $6y^4+12y^3-48y^2=6y^2(y-2)(y+4)$

Step 1. Given Information

Consider the expression $6y^4+12y^3-48y^2$

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

Step 2. Factor out GCF

The GCF of the three terms in the expression is $6y^2$. So factor the GCF from the expression we get 

$\begin{array}{rcl}6y^4+12y^3-48y^2& =& 6y^2·y^2+6y^2·2y-6y^2·8\\ & =& 6y^2(y^2+2y-8)\end{array}$

Step 3. Factor using trial and error

The factors of the first term and the last term of the trinomial $y^2+2y-8$are

$$\begin{gathered} y^2=y·y \\ -8=-2·4 \end{gathered}$$

Consider all the combinations

Possible Factors	Product
$(y+2)(y-4)$	$y^2-4y+2y-8=y^2-2y-8$
$(y-2)(y+4)$	$y^2+4y-2y-8=y^2+2y-8$

So the other factors are $(y-2)(y+4)$.

Thus, $6y^4+12y^3-48y^2=6y^2(y-2)(y+4)$.

Step 4. Check the factors

Multiply the factors to check the solution 

$\begin{array}{rcl}6y^2(y-2)(y+4)& =& 6y^2(y^2+4y-2y-8)\\ & =& 6y^2(y^2+2y-8)\\ & =& 6y^4+12y^3-48y^2\end{array}$

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