Q 120. - Intermediate Algebra

Question

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

$6u^2-46u-16$

Step-by-Step Solution

Answer

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

Step 1. Given information

The given trinomial is $6u^2-46u-16$.

Step 2. Find greatest common factor.

Factor the greatest common factor.

$6u^2-46u-16=2(3u^2-23u-8)$

Step 3. Compare the trinomial 3 u 2 - 26 u - 8 to a x 2 + b x + c and then find the product of a c .

The values are as follows:

$\begin{array}{rcl}a& =& 3\\ b& =& -26\\ c& =& -8\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 3(-8)\\ & =& -24\end{array}$

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:

$\begin{array}{rcl}-24\left(1\right)& =& -24\\ & =& ac\\ -24+1& =& -23\\ & =& b\end{array}$

So, two numbers are $-24$ and $1$.

Now, split the middle term of the trinomial $3u^2-23u-8$ as follows:

$3u^2-23u-8=3u^2-24u+u-8$

Step 5. Factor by grouping.

$\begin{array}{rcl}3u^2-23u-8& =& 3u(u-8)+1(u-8)\\ & =& (u-8)(3u+1)\end{array}$

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

$\begin{array}{rcl}2(u-8)(3u+1)& =& 2(u^2-24u+u-8)\\ & =& 2(u^2-23u-8)\\ & =& 2u^2-46u-16\end{array}$

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