Q. 377 - Intermediate Algebra

Question

In the following exercise, factor.

$3x^2+ 3x - 36$

Step-by-Step Solution

Answer

The factorisation of the given polynomial is: 

$3x^2+3x-36=3(x-3)(x+12)$

Step 1. Given information

The given expression is 

$3x^2+3x-36$

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  $3x^2+3x-36$, we have:

$$\begin{gathered} ac=3\times (-36) \\ =-108 \end{gathered}$$

And we have to think of two numbers whose sum is equal to $3$and the product is equal to $-108$.

Then,

$$\begin{gathered} 12-9=3 \\ 12\times (-9)=-108 \end{gathered}$$

The required numbers are $12\&-9$

Step 3. Perform factorisation

Now,  

$$\begin{gathered} 3x^2+3x-36 \\ 3x^2+12x-9x-36 \\ 3x(x+12)-9(x+12) [taking out 3x and -9 as common] \\ (3x-9)(x+12) [taking out (x+12) as a common] \\ 3(x-3)(x+12) [taking 3 out as a common] \end{gathered}$$

The factorisation of the given polynomial is: 

style="max-width: none; vertical-align: -5px;" $3x^2+3x-36=3(x-3)(x+12)$

Step 4. Check the answer

Multiplying the factors, we get:

$$\begin{gathered} 3x^2+3x-36=3(x-3)(x+12) \\ 3x^2+3x-36=3(x^2+x-12) \\ 3x^2+3x-36=3x^2+3x-36 \end{gathered}$$

This is true.