Let p(x) be a quadratic polynomial with const | vedclass

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(D) Let $p(x) = ax^2 + bx + c$. Given, the constant term $c = 1$. Therefore, $p(x) = ax^2 + bx + 1$. By the Remainder Theorem, $p(1) = 2$ and $p(-1) =

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$\frac{1}{2}$

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(D) Let $p(x) = ax^2 + bx + c$. Given, the constant term $c = 1$. Therefore, $p(x) = ax^2 + bx + 1$. By the Remainder Theorem, $p(1) = 2$ and $p(-1) = 4$. Substituting $x=1$: $a(1)^2 + b(1) + 1 = 2 \implies a + b = 1$ (Equation $I$). Substituting $x=-1$: $a(-1)^2 + b(-1) + 1 = 4 \implies a - b = 3$ (Equation $II$). Adding Equation $I$ and Equation $II$: $(a+b) + (a-b) = 1 + 3 \implies 2a = 4 \implies a = 2$. Substituting $a=2$ into Equation $I$: $2 + b = 1 \implies b = -1$. Thus, the polynomial is $p(x) = 2x^2 - x + 1$. The sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is given by $-\frac{b}{a}$. Sum of the roots $= -\frac{-1}{2} = \frac{1}{2}$.

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

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