Q 124. - Intermediate Algebra

Question

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

$24p^2+160p+96$

Step-by-Step Solution

Answer

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

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 3 p 2 + 20 p + 12 to a x 2 + b x + c and then find the product of a c .

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 a c 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 $2$.

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

width="256" style="max-width: none; vertical-align: -4px;" $3p^2+20p+12=3p^2+18p+2p+12$

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