91Ó°ÊÓ

The given expression is

$18a^2−57a−21$

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

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

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

Then,

role="math" localid="1644649872658" $$\begin{gathered} -21+2=-19 \\ -21×2=-42 \end{gathered}$$

The required numbers are$-21\&2$

Now,

$$\begin{gathered} 3(6a^2-19a-7) \\ 3(6a^2-21a+2a-7) \\ 3\left[3a\right(2a-7)+1(2a-7\left)\right][takingout3aand1ascommon] \\ 3(2a-7)(3a+1)[taking(2a-7)asacommon] \end{gathered}$$

The factorisation of the given polynomial is:

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

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.