Q 121. - Intermediate Algebra

Question

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

$48z^3-102z^2-45z$

Step-by-Step Solution

Answer

The solution is $\begin{array}{rcl}& & 3z(2z-5)(8z+3)\end{array}$.

Step 1. Given information

The given trinomial is $48z^3-102z^2-45z$.

Step 2. Find greatest common factor.

Factor the greatest common factor.

$48z^3-102z^2-45z=3z(16z^2-34z-15)$

Step 3. Compare the trinomial 16 z 2 - 34 z - 15 to a x 2 + b x + c and then find the product of a c .

The values are as follows:

$\begin{array}{rcl}a& =& 16\\ b& =& -34\\ c& =& -15\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 16(-15)\\ & =& -240\end{array}$

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:

$\begin{array}{rcl}-40\left(6\right)& =& -240\\ & =& ac\\ -40+6& =& -34\\ & =& b\end{array}$

So, two numbers are $-40$ and $6$.

Now, split the middle term of the trinomial $16z^2-34z-15$ as follows:

$\begin{array}{rcl}16z^2-34z-15& =& 16z^2-40z+6z-15\end{array}$

Step 5. Factor by grouping.

$\begin{array}{rcl}16z^2-34z-15& =& 8z(2z-5)+3(2z-5)\\ & =& (2z-5)(8z+3)\end{array}$

Step 6. Check by multiplying all the factors 3 z ( 2 z - 5 ) ( 8 z + 3 ) .

$\begin{array}{rcl}3z(2z-5)(8z+3)& =& 3z(16z^2-40z+6z-15)\\ & =& 3z(16z^2-34z-15)\\ & =& 48z^3-102z-45z\end{array}$

Hence, the factor is $\begin{array}{rcl}& & 3z(2z-5)(8z+3)\end{array}$.