Q 112. - Intermediate Algebra

Question

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

$8w^2+25w+3$

Step-by-Step Solution

Answer

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

Step 1. Given information

The given trinomial is $8w^2+25w+3$.

Step 2. Find greatest common factor.

There is no greatest common factor.

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

The values are as follows:

$\begin{array}{rcl}a& =& 8\\ b& =& 25\\ c& =& 3\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 8·3\\ & =& 24\end{array}$

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

Since all the terms in the given polynomial is positive, both factors will be positive.

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

$\begin{array}{rcl}24·1& =& 24\\ & =& ac\\ 24+1& =& 25\\ & =& b\end{array}$

So, two numbers are $24$ and $1$.

Now, split the middle term of the given trinomial $8w^2+25w+3$ as follows:

$8w^2+25w+3=8w^2+24w+w+3$

Step 5. Factor by grouping.

$\begin{array}{rcl}8w^2+25w+3& =& 8w(w+3)+1(w+3)\\ & =& (8w+1)(w+3)\end{array}$

Step 6. Check by multiplying ( 8 w + 1 ) ( w + 3 ) .

$\begin{array}{rcl}(8w+1)(w+3)& =& 8w^2+24w+w+3\\ & =& 8w^2+25w+3\end{array}$

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