Question

A polynomial $f\left(x\right)$ leaves remainder $15$ when divided by $(x-3)$ and $(2x+1)$ when divided by $(x-1{)}^2$. When $f$ is divided by $(x-3)(x-1{)}^2,$ the remainder is

• A. $2x^2+2x+3$
• B. $2x^2-2x-3$
• C. $2x^2-2x+3$
• D. none of these

Answer 1

Answer: (C)

$2x^2-2x+3$

Since function $f\left(x\right)$ leaves remainder $15$ when divided by $x-3$, therefore $f\left(x\right)$ can be written as

$f\left(x\right)=(x-3)l\left(x\right)+15$ ...(1)

Also, $f\left(x\right)$ leaves remainder $2x+1$ when divided by $(x-1{)}^2.$

Thus, $f\left(x\right)$ can also be written as

$f\left(x\right)={(x-1)}^{2}m\left(x\right)+2x+1$ ...(2)

If $R\left(x\right)$ be the remainder when $f\left(x\right)$ is divided by $(x-3)(x-1{)}^2,$ then we may write

$f\left(x\right)=(x-3){(x-1)}^{2}n\left(x\right)+R\left(x\right)$ ...(3)

Since $(x-3)(x-1{)}^2$ is a polynomial of degree three, the remainder has to be a polynomial of degree less than or equal to two.

Thus let $R\left(x\right)=a{x}^2+bx+c$

From (1) and (3), we have

$f\left(3\right)=15=R\left(3\right)\Rightarrow 9a+3b+c=15$ ...(4)

From (2) and (3), we have

$f\left(1\right)=3=R\left(1\right)\Rightarrow a+b+c=3$ ...(5)

From (2) and (3), we have

$f^{\prime }\left(1\right)=2=R^{\prime }\left(1\right)\Rightarrow 2a+b=2$ ...(6)

Solving equation (4),(5) and (6), we get

$a=2,b=-2,c=3$