Q. 379 - Intermediate Algebra

Question

In the following exercise, factor.

$18a^2- 57a- 21$

Step-by-Step Solution

Answer

The factorisation of the given polynomial is: 

$18a^2-57a-21=3(2a-7)(3a+1)$

Step 1. Given information

The given expression is  

$18a^2-57a- 21$

Step 2. Use ac method for factorisation

To factorise the polynomial $a{x}^2+bx+c$ by $ac$  method, we need to think of two numbers whose product is equal to $ac$ and the sum is equal to  $b$.

For polynomial $18a^2- 57a- 21$, we will take out $3$ as a common.

$3(6a^2-19a-7)$

Then

$$\begin{gathered} ac=6\times (-7) \\ =-42 \end{gathered}$$

And we have to think of two numbers whose sum is equal to $-19$and the product is equal to $-42$.

Then,

$$\begin{gathered} -21+2=-19 \\ -21\times 2=-42 \end{gathered}$$

The required numbers are $-21 \&2$

Step 3. Perform factorisation

Now,

$$\begin{gathered} 3(6a^2-19a-7) \\ 3(6a^2-21a+2a-7) \\ 3\left[3a\right(2a-7)+1(2a-7\left)\right] [taking out 3a and 1 as common] \\ 3(2a-7)(3a+1) [taking (2a-7) as a common] \end{gathered}$$

The factorisation of the given polynomial is: 

$18a^2-57a-21=3(2a-7)(3a+1)$

Step 4. Check the answer

Multiplying the factors, we get:

$$\begin{gathered} 18a^2-57a-21=3(2a-7)(3a+1) \\ 18a^2-57a-21=3(6a^2+2a-21a-7) \\ 18a^2-57a-21=18a^2-57a-21 \end{gathered}$$

This is true.