Question

Question 14
By Remainder theorem, find the remainder when p(x) is divided by g(x).

(i) $p\left(x\right)=x^3–2x^2–4x–1,g\left(x\right)=x+1$
(ii) $p\left(x\right)=x^3–3x^2+4x+50,g\left(x\right)=x–3$
(iii) $p\left(x\right)=4x^3–12x^2+14x–3,g\left(x\right)=2x–1$
(iv) $p\left(x\right)=x^3–6x^2+2x-4,g\left(x\right)=1-\frac{3}{2}x$

Answer 1

(i) $p\left(x\right)=x^3–2x^2–4x–1,g\left(x\right)=x+1$
Here, zero of g(x) is -1.
When we divide p(x) by g(x) using remainder theorem, we get the remainder = p(-1).
$p(-1)=(-1{)}^3–2(-1{)}^2–4(-1)–1$
$=-1–2+4–1$
$=4–4=0$
Hence, remainder is 0.

(ii) $p\left(x\right)=x^3–3x^2+4x+50,g\left(x\right)=x–3$
Here, zero of g(x) is 3.
When we divide p(x) by g(x) using remainder theorem, we get the remainder = p(3).
$p\left(3\right)=(3{)}^3–3(3{)}^2+4\left(3\right)+50$
$=27–27+12+50=62$
Hence, remainder is 62.

(iii) $p\left(x\right)=4x^3–12x^2+14x–3andg\left(x\right)=2x–1$
Here, zero of g(x) is $\frac{1}{2}$.
When we divide p(x) by g(x) using remainder theorem, we get the remainder = $p\left(\frac{1}{2}\right)$.
$\therefore p\left(\frac{1}{2}\right)=4{\left(\frac{1}{2}\right)}^3-12{\left(\frac{1}{2}\right)}^2+14\left(\frac{1}{2}\right)-3=4\times \frac{1}{8}-12\times \frac{1}{4}+14\times \frac{1}{2}-3$
$=\frac{1}{2}-3+7-3=\frac{1}{2}+1=\frac{1+2}{2}=\frac{3}{2}$
Hence, remainder is $\frac{3}{2}$.

(iv) Given, $p\left(x\right)=x^3–6x^2+2x–4$ and $g\left(x\right)=1-\frac{3}{2}x$.
Here, zero of g(x) is $\frac{2}{3}$.
When we divide p(x) by g(x) using remainder theorem, we get the remainder = $p\left(\frac{2}{3}\right)$.
$=\frac{8}{27}-6\times \frac{4}{9}+2\times \frac{2}{3}-4=\frac{8}{27}-\frac{24}{9}+\frac{4}{3}-4$
$p\left(\frac{2}{3}\right)=\frac{8-72+36-108}{27}=\frac{-136}{27}$
Hence, remainder is $\frac{-136}{27}$.