Let f(x) be a polynomial and a, b be distinct | vedclass

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(C) Let $f(x) = (x-a)(x-b)q(x) + r(x)$.Since the divisor is of degree $2$,the remainder $r(x)$ must be of degree at most $1$. Let $r(x) = \alpha x + \

Vedclass · Accepted Answer

$\frac{(x-a) f(b)-(x-b) f(a)}{b-a}$

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(C) Let $f(x) = (x-a)(x-b)q(x) + r(x)$.Since the divisor is of degree $2$,the remainder $r(x)$ must be of degree at most $1$. Let $r(x) = \alpha x + \beta$.Then $f(x) = (x-a)(x-b)q(x) + \alpha x + \beta$.Substituting $x = a$ and $x = b$:$f(a) = \alpha a + \beta$  $(i)$$f(b) = \alpha b + \beta$  $(ii)$Subtracting $(ii)$ from $(i)$:$f(a) - f(b) = \alpha(a - b) \implies \alpha = \frac{f(a) - f(b)}{a - b} = \frac{f(b) - f(a)}{b - a}$.Substituting $\alpha$ into $(i)$:$\beta = f(a) - \alpha a = f(a) - \left(\frac{f(b) - f(a)}{b - a}\right)a = \frac{f(a)(b - a) - a(f(b) - f(a))}{b - a} = \frac{b f(a) - a f(a) - a f(b) + a f(a)}{b - a} = \frac{b f(a) - a f(b)}{b - a}$.Thus,$r(x) = \alpha x + \beta = \left(\frac{f(b) - f(a)}{b - a}\right)x + \frac{b f(a) - a f(b)}{b - a} = \frac{x f(b) - x f(a) + b f(a) - a f(b)}{b - a}$.Rearranging terms:$r(x) = \frac{f(b)(x - a) - f(a)(x - b)}{b - a}$.Therefore,the correct option is $C$.

Let $f(x)$ be a polynomial and $a, b$ be distinct real numbers. Then the remainder in the division of $f(x)$ by $(x-a)(x-b)$ is

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