In the following exercise, factor.

$3n^4-12n^3+96n^2$

The given expression is

$3n^4-12n^3+96n^2$

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 role="math" localid="1644650981219" $\mathrm{b}$.

For polynomial $3n^4-12n^3-96n^2$, we will take out $n^2$as a common.

$3n^2(n^2-4n-32)$

Then

$$\begin{gathered} ac=1\times (-32) \\ =-32 \end{gathered}$$

And we have to think of two numbers whose sum is equal to $-12$and the product is equal to $-96$.

Then,

$$\begin{gathered} -8+4=-4 \\ -8\times 4=-32 \end{gathered}$$

The required numbers are$-8\&4$

Now,

$$\begin{gathered} 3n^2(n^2-4n-32) \\ 3n^2[n^2-8n+4n-32] \\ 3n^2\left[n\right(n-8)+4(n-8\left)\right][takingoutnand4ascommon] \\ 3n^2(n-8)(n+4)[takingout(n-8)asacommon] \end{gathered}$$

The factorisation of the given polynomial is:

localid="1644680812044" $3n^4-12n^3-96n^2=3n^2(n-8)(n+4)$

Multiplying the factors, we get:

$$\begin{gathered} 3n^4-12n^3-96n^2=3n^2(n-8)(n+4) \\ 3n^4-12n^3-96n^2=3n^2(n^2+4n-8n-32) \\ 3n^4-12n^3-96n^2=3n^4-12n^3-96n^2 \end{gathered}$$

This is true.