By remainder theorem, find the remainder when p(x) is divided by g(x)
(i) p(x) = x3-2x2-4x-1, g(x)=x + 1
(ii) p(x) = x3 -3x2 + 4x + 50, g(x)= x – 3
(iii) p(x) = x3 – 12x2 + 14x -3, g(x)= 2x – 1 – 1
(iv) ‘p(x) = x3-6x2+2x-4, g(x) = 1 -(3/2) x
(i) Given, $p(x)=x^{3}-2 x^{2}-4 x-1$ and $g(x)=x+1$
Here, zero of $g(x)$ is $-1$.
When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p(-1)$
$\therefore \quad p(-1)=(-1)^{3}-2(-1)^{2}-4(-1)-1$
$=-1-2+4-1$
$=4-4=0$
Hence, remainder is $0 .$
(ii) Given, $p(x)=x^{3}-3 x^{2}+4 x+50$ and $g(x)=x-3$
Here, zero of $g(x)$ is 3 .
When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p(3)$.
$\therefore \quad p(3)=(3)^{3}-3(3)^{2}+4(3)+50$
$=27-27+12+50=62$
Hence, remainder is 62 .
(iii) Given, $p(x)=4 x^{3}-12 x^{2}+14 x-3$ and $g(x)=2 x-1$
Here, zero of $g(x)$ is $\frac{1}{2}$.
When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p\left(\frac{1}{2}\right)$.
$\therefore \quad p\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^{3}-12\left(\frac{1}{2}\right)^{2}+14\left(\frac{1}{2}\right)-3=4 \times \frac{1}{8}-12 \times \frac{1}{4}+14 \times \frac{1}{2}-3$
$=\frac{1}{2}-3+7-3=\frac{1}{2}+1=\frac{1+2}{2}=\frac{3}{2}$
Hence, remainder is $\frac{3}{2}$.
(iv) Given, $p(x)=x^{3}-6 x^{2}+2 x-4$ and $g(x)=1-\frac{3}{2} x$.
Here, zero of $g(x)$ is $\frac{2}{3}$.
When we divide $p(x)$ by $g(x)$ using remainder theorem, we get the remainder $p\left(\frac{2}{3}\right)$
$\because$ $=\frac{8}{27}-6 \times \frac{4}{9}+2 \times \frac{2}{3}-4=\frac{8}{27}-\frac{24}{9}+\frac{4}{3}-4$
$=\frac{8-72+36-108}{27}=\frac{-136}{27}$
Hence, remainder is $\frac{-136}{27}$.
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