Question

What is the remainder if we divide $6x^3+x^2–2x+4$ by $x-2$ ?

Answer 1

Now, we know that If the value of the polynomial $g\left(x\right)$ becomes equal to 0 when we substitute $x=a$, then 𝒂 is a zero of the polynomial $g\left(x\right)$, and vice-versa.

Therefore, we have
$g\left(x\right)=x-2$

For, $g\left(x\right)=0,\Rightarrow x-2=0\Rightarrow x=2$

Therefore, 2 is the zero of the polynomial f(x).

Now to get the value of the dividend at the zero of the divisor, we will put x = zero of the divisor in $f\left(x\right)$.

Therefore,

$f\left(x\right)=6x^3+x^2-2x+4$ (given)

Putting $x=2$ in $f\left(x\right)$

we get,
$f\left(2\right)=6(2{)}^3+(2{)}^2-2.2+4=6.8+4-4+4=48+4=52$

Hence, $f\left(2\right)=52$

Therefore, 52 is the remainder here.