Question

• By remainder theorem , find the remainder when...

p(x)=4x^3-3x^2+2x-5 is divided by g(x)=1-2x.​

Answer 1

The answer is given above.

Hope it helps you.

Answer 2

$\huge\tt{\red{\underline{Given:}}}$

★$p(x) =4x^{3}-3x^{2}+2x-5=0$

★$g(x) =1-2x$

$\huge\tt{\red{\underline{To\:\:Find:}}}$

★The remainder when g(x) divides p(x).

$\huge\tt{\red{\underline{Concept\:\:Used:}}}$

★We will be using 'Remainder Theorem'.

$\huge\tt{\red{\underline{Answer:}}}$

We have,

$\longrightarrow p(x) =4x^{3}-3x^{2}+2x-5=0$

$\longrightarrow g(x) =1-2x$

So, by reminder theorem we know that ,

Let x be a polynomial of degree one or more and let $\alpha$be any real number if p( x) is divided by $(x-\alpha)$then the remainder is p($\alpha$) .

______________________________________

On equating g(x) with 0,

$\implies 1-2x=0$

$\implies -2x=-1$

${\underline{\boxed{.°.\red{ x =\dfrac{1}{2}}}}}$

______________________________________

Now ,the remainder will be $p(\dfrac{1}{2})$

Let remainder be r.

$\implies p(x) =4x^{3}-3x^{2}+2x-5=r$

$\implies p(\dfrac{1}{2}) =4\times (\dfrac{1}{2})^{3}-3\times(\dfrac{1}{2})^{2}+2\times(\dfrac{1}{2})-5 =r$

$\implies r = 4\times\dfrac{1}{8}-3\times\dfrac{1}{4}+2\times\dfrac{1}{2}-5$

$\implies r = \cancel{4}\times\dfrac{1}{\cancel{8}^{2}}-3\times\dfrac{1}{4}+\cancel{2}\times\dfrac{1}{\cancel{2}}-5$

$\implies r = \dfrac{1}{2}- \dfrac{3}{4}-4$

$\implies r= \dfrac{(2-3-16)}{4}$

${\underline{\boxed{.°.\purple{r=\dfrac{-17}{4}}}}}$