What will be the roots of $x^4+x^3-16x^2-4x+48=0$ if the product of two of its roots is $6$?

Given:

A polynomial $x^4+x^3-16x^2-4x+48=0$ and the sum of two of its roots is $6$.

As the degree of the polynomial is $4$, so it will have four roots.

Let the roots of the given polynomial be $a, b, c, \mathrm{and} d$ and $a\times b=6$

It means the possibilities are $a=1, b=6 or a=2, b=3 or a=6, b=1 or a=3, b=2$.

Now let $P\left(x\right)=x^4+x^3-16x^2-4x+48$.

For $x=1$,

$$\begin{gathered} P\left(1\right)=1^4+1^3-16{\left(1\right)}^2-4\left(1\right)+48 \\ =1+1-16-4+48 \\ =30 \end{gathered}$$

Now, as $P\left(1\right)\ne 0$, so $1 \mathrm{and} 6$ are not the roots of the given polynomial.

For $x=2$,

$$\begin{gathered} P\left(2\right)=2^4+2^3-16{\left(2\right)}^2-4\left(2\right)+48 \\ =16+8-64-8+48 \\ =0 \end{gathered}$$

Now, as $P\left(2\right)=0$, so $2 \mathrm{and} 3$ are the roots of the given polynomial.

So,

$\frac{x^4+x^3-16x^2-4x+48}{\left(x-2\right)\left(x-3\right)}=x^2+6x+8$

Now for other two roots we will factorise the quadratic equation by splitting the middle term.

$$\begin{gathered} x^2+6x+8 \\ =x^2+2x+4x+8 \\ =x\left(x+2\right)+4\left(x+2\right) \\ =\left(x+4\right)\left(x+2\right) \end{gathered}$$
 

So, the roots are  $3,2,-4,-2$.

Hence, option D is correct.