Question

What must be subtracted from $16x^3-8x^2-4x+7$ so that the resulting expression has $2x+1$ as a factor?

Answer 1

If any function let, $f\left(x\right)$ is a polynomial and another function is $g\left(x\right)$ polynomial having less degree than the $f\left(x\right)$. Now if $g\left(x\right)$ is a factor of $f\left(x\right)$ then $g\left(x\right)$ should divide the $f\left(x\right)$ leaving no remainder. This means :

$f\left(x\right)=g\left(x\right).q\left(x\right)$

where $q\left(x\right)$ is the quotient .

Now $2x+1$ is a factor of $16x^3-8x^2-4x+7$ then, it should divide by $2x+1$ having zero remainder.

Now on dividing :

$\frac{16x^3-8x^2-4x+7}{2x+1}=8x^2-8x+2$ with having remainder $5$ .

$\Rightarrow 16x^3-8x^2-4x+7=(8x^2-8x+2)(2x+1)+5$

Now if we subtract $5$ from the given polynomial then

$\Rightarrow (16x^3-8x^2-4x+7)-5$

$\Rightarrow 16x^3-8x^2-4x+2$

$\Rightarrow \frac{16x^3-8x^2-4x+2}{2x+1}=8x^2-8x+2$

Now $2x+1$ is a factor of our new polynomial.

So, we have to subtract $5$ from $16x^3-8x^2-4x+7$ for having $2x+1$ as a factor.