In the following exercise, factor.

$3x^2+3x-36$

The given expression is

$3x^2+3x-36$

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:

role="math" localid="1644646529752" $$\begin{gathered} ac=3\times (-36) \\ =-108 \end{gathered}$$

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$12\&-9$

Now,

$$\begin{gathered} 3x^2+3x-36 \\ 3x^2+12x-9x-36 \\ 3x(x+12)-9(x+12)[takingout3xand-9ascommon] \\ (3x-9)(x+12)[takingout(x+12)asacommon] \\ 3(x-3)(x+12)[taking3outasacommon] \end{gathered}$$

The factorisation of the given polynomial is:

localid="1644646935961">3x2+3x−36=3(x-3)(x+12)

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.