Q. 380 - Intermediate Algebra

Question

In the following exercise, factor.

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

Step-by-Step Solution

Answer

The factorisation of the given polynomial is: 

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

Step 1. Given information

The given expression is  

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

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   $\mathrm{b}$.

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 $-8\&4$

Step 3. Perform factorisation

Now, 

$$\begin{gathered} 3n^2(n^2-4n-32) \\ 3n^2[n^2-8n+4n-32] \\ 3n^2\left[n\right(n-8)+4(n-8\left)\right] [taking out n and 4 as common] \\ 3n^2(n-8)(n+4) [taking out (n-8) as a common] \end{gathered}$$

The factorisation of the given polynomial is: 

$3n^4- 12n^3 - 96n^2=3n^2(n-8)(n+4)$ 

Step 4. Check the answer

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.