Q 108. - Intermediate Algebra

Question

Factor Trinomials of the Form $a{x}^2+bx+c$ Using Trial and Error.

$-81a^2+153a+18$

Step-by-Step Solution

Answer

The solution is $\begin{array}{rcl}& & -9(a-2)(9a+1)\end{array}$.

Step 1. Given information

Consider the trinomial.

$-81a^2+153a+18$

Step 2. Write the trinomial in descending order and then find the greatest common factor.

The trinomial $-81a^2+153a+18$ is given in descending order.

There is a greatest common factor in the given trinomial.

Factor the greatest common factor.

$-81a^2+153a+18=-9(9a^2-17a-2)$

Step 3. Find the factors of the first term and the last term of the trinomial 9 a 2 - 17 a - 2 .

The first term of the trinomial $9a^2-17a-2$ is $9a^2$.

So, the factors of $9a^2$ are as follows:

$\begin{array}{rcl}9a^2& =& a·9a\\ 9a^2& =& 3a·3a\end{array}$

The last term of the trinomial $9a^2-17a-2$ is $-2$.

Since the last term of the polynomial  $9a^2-17a-2$ is negative, the factors of the last term will have opposite signs. 

The factors of $-2$ are as follows:

$\begin{array}{rcl}-2& =& 1·(-2)\\ -2& =& (-1)·2\end{array}$

Step 4. Make a table for all the combination of factors of 9 a 2 - 17 a - 2 .

Possible factors	Product
$(a+1)(9a-2)$	$9a^2+7a-2$
$(a-2)(9a+1)$	$9a^2-17a-2$
$(a-1)(9a+2)$	$9a^2-7a-2$
$(a+2)(9a-1)$	$9a^2+17a-2$
$(3a+1)(3a-2)$	$9a^2-3a-2$
$(3a-1)(3a+2)$	$9a^2+3a-2$

From the table, conclude that the combination $(a-2)(9a+1)$ is correct.

Step 5. Check by multiplying all the factors - 9 ( a - 2 ) ( 9 a + 1 ) .

$\begin{array}{rcl}-9(a-2)(9a+1)& =& -9(9a^2+a-18a-2)\\ & =& -9(9a^2-17a-2)\\ & =& -81a^2+153a+18\end{array}$

Hence, the factor is $\begin{array}{rcl}& & -9(a-2)(9a+1)\end{array}$.