Question

Let $p\left(x\right)$ be a quadratic polynomial with constant term $1$. Suppose $p\left(x\right)$, when divided by $x-1$ leaves remainder $2$ and when 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

The correct option is D $\frac{1}{2}$

Let $p\left(x\right)=a{x}^2+bx+c.......\left(i\right)$

Given constant term '$c$'$=1$
$\therefore$ $p\left(x\right)=a{x}^2+bx+1=\mathrm{0......}\left(ii\right)$

Now by given condition $p\left(1\right)=2$ (remainder)
$\Rightarrow$ $a+b+1=2$
$\Rightarrow$ $a+b=\mathrm{1.......}\left(iii\right)$

and $p(-1)=4$
$\Rightarrow$ $a-b+1=4$
$\Rightarrow$ $a-b=\mathrm{3.....}\left(iv\right)$

On adding eqs. $\left(iii\right)$ and $\left(iv\right)$ we get
$2a=4$ $\Rightarrow$ $a=2$ form eqs. $\left(iii\right)$ $b=-1$

On putting the values of $a$ and $b$ in eq $\left(ii\right)$ we get
$p\left(x\right)=2x^2-x+1=0$
$\therefore$ Sum of the roots $=-\frac{\text{Coefficient of}x}{\text{Coefficient of}{x}^2}=\frac{-(-1)}{2}$ $=\frac{1}{2}$