Question

Find the remainder (without division) on dividing $f\left(x\right)$ by $(x-2)$, where:
(i) $f\left(x\right)=5x^2-7x+4$
(ii) $f\left(x\right)=2x^3-7x^2+3$.

Answer 1

We know that the remainder theorem states that if a polynomial $f\left(x\right)$ is divided by $(x-a)$, then the remainder is $f\left(a\right)$.

(i) Here, we have the polynomial $f\left(x\right)=5x^2-7x+4$ which is divided by $(x-2)$, therefore, by remainder theorem, the remainder $R$ is:

$R=f\left(2\right)=5(2{)}^2-7(2)+4=(5\times 4)-14+4=20-14+4=24-14=10$.

Hence, the remainder is $10$.

(ii) Here, we have the polynomial $f\left(x\right)=2x^3-7x^2+3$ which is divided by $(x-2)$, therefore, by remainder theorem, the remainder $R$ is:

$R=f\left(2\right)=2(2{)}^3-7(2{)}^2+3=(2\times 8)-(7\times 4)+3=16-28+3=19-28=-9$.

Hence, the remainder is $-9$.