Q 123. - Intermediate Algebra

Question

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

$16s^2+40s+24$

Step-by-Step Solution

Answer

The solution is $\begin{array}{rcl}& & 8(s+1)(2s+3)\end{array}$.

Step 1. Given information

The given trinomial is $16s^2+40s+24$.

Step 2. Find greatest common factor.

Factor the greatest common factor.

$16s^2+40s+24=8(2s^2+5s+3)$

Step 3. Compare the trinomial 2 s 2 + 5 s + 3 to a x 2 + b x + c and then find the product of a c .

The values are as follows:

$\begin{array}{rcl}a& =& 2\\ b& =& 5\\ c& =& 3\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 2·3\\ & =& 6\end{array}$

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:

$\begin{array}{rcl}2·3& =& 6\\ & =& ac\\ 2+3& =& 5\\ & =& b\end{array}$

So, two numbers are $2$ and $3$.

Now, split the middle term of the trinomial $\begin{array}{rcl}& & 2s^2+5s+3\end{array}$ as follows:

$\begin{array}{rcl}2s^2+5s+3& =& 2s^2+2s+3s+3\end{array}$

Step 5. Factor by grouping.

$\begin{array}{rcl}2s^2+5s+3& =& 2s(s+1)+3(+1)\\ & =& (s+1)(2s+3)\end{array}$

Step 6. Check by multiplying all the factors 8 ( s + 1 ) ( 2 s + 3 ) .

$\begin{array}{rcl}8(s+1)(2s+3)& =& 8\left(2s^2+2s+3s+3\right)\\ & =& 16s^2+40s+24\end{array}$

Hence, the factor is $\begin{array}{rcl}& & 8(s+1)(2s+3)\end{array}$.