Using the remainder theorem, find the remainder,
Question:

Using the remainder theorem, find the remainder, when $p(x)$ is divided by $g(x)$, where $p(x)=2 x^{3}-9 x^{2}+x+15, g(x)=2 x-3 .$

 

Solution:

$p(x)=2 x^{3}-9 x^{2}+x+15$

$g(x)=2 x-3=2\left(x-\frac{3}{2}\right)$

By remainder theorem, when $p(x)$ is divided by $(2 x-3)$, then the remainder $=p\left(\frac{3}{2}\right)$.

Putting $x=\frac{3}{2}$ in $p(x)$, we get

$p\left(\frac{3}{2}\right)=2 \times\left(\frac{3}{2}\right)^{3}-9 \times\left(\frac{3}{2}\right)^{2}+\frac{3}{2}+15=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+15=\frac{27-81+6+60}{4}=\frac{12}{4}=3$

∴ Remainder = 3

Thus, the remainder when p(x) is divided by g(x) is 3.

 

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