Question

Let p(x) be a quadratic polynomial with constant term 1. Suppose $p\left(x\right)$ when divided by $x-1$ leaves remainder 2 andwhen divided by $x+1$ leaves remainder 4. Then the sum of the roots of $p\left(x\right)=0$ is

• A. $-1$
• B. $1$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Given polynomial: $p\left(x\right)=a{x}^2+bx+1$

If we divide $P\left(x\right)$ by $x-1$, we get remainder as $2$

i.e. $p\left(1\right)=2\Rightarrow a+b+1=2$

If we divide $P\left(x\right)$ by $x+1$, we get remainder as $4$

i.e. $p(-1)=4\Rightarrow a-b+1=4$

Solving above two equations

$\Rightarrow a=2$ and $b=-1$

Now, Sum of roots $=-\frac{b}{a}$

But $-\frac{b}{a}=\frac{1}{2}$

Hence, sum of roots is $\frac{1}{2}$