Question

Factor Trinomials of the Form $a{x}^2+bx+c$using the ‘ac’ Method.

$24p^2+160p+96$

Short answer

The solution is $\begin{array}{rcl}& & 8(p+6)(3p+2)\end{array}$.

Step-by-step solution

Step 1. Given information

The given trinomial is $24p^2+160p+96$.

Step 2. Find greatest common factor.

Factor the greatest common factor.

$24p^2+160p+96=8(3p^2+20p+12)$

Step 3. Compare the trinomial 3p2+20p+12 to ax2+bx+c and then find the product of ac. 

The values are as follows:

$\begin{array}{rcl}a& =& 3\\ b& =& 20\\ c& =& 12\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 3\left(12\right)\\ & =& 36\end{array}$

Step 4. Determine two numbers that multiply to ac and add to b. 

Two numbers that multiply to $ac$and add to $b$are as follows:

$\begin{array}{rcl}18·2& =& 36\\ & =& ac\\ 18+2& =& 20\\ & =& b\\ & & \end{array}$

So, two numbers are $18$and localid="1645185478776" $2$.

Now, split the middle term of the trinomial $3p^2+20p+12$as follows:

width="256">3p2+20p+12=3p2+18p+2p+12

Step  5. Factor by grouping.

$\begin{array}{rcl}3p^2+20p+12& =& 3p(p+6)+2(p+6)\\ & =& (p+6)(3p+2)\end{array}$

Step 6. Check by multiplying all the factors 8(p+6)(3p+2).

$\begin{array}{rcl}8(p+6)(3p+2)& =& 8\left(3p^2+18p+2p+12\right)\\ & =& 8\left(3p^2+20p+12\right)\\ & =& 24p^2+\hspace{0.17em}160p\hspace{0.17em}+\hspace{0.17em}96\end{array}$

Hence, the factor is $\begin{array}{rcl}& & 8(p+6)(3p+2)\end{array}$.