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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Aicharya · Accepted Answer

Let $f(x)=(x-a)(x-b) \cdot q(x)+r(x)$ Let $r(x)=\alpha x+\beta \quad[\because \operatorname{deg} r(x) < \operatorname{deg}$. of divisor] $\begin{aligned} \therefore \quad f(x) & =(x-a)(x-b) \cdot q(x)+\alpha x+\beta \ f(a) & =\alpha a+\beta \quad \ldots (i) \ f(b) & =\alpha b+\beta \quad \ldots (ii) \end{aligned}$ Subtract Eqs. (i) from (ii) $\begin{aligned} f(a)-f(b) & =\alpha(a-b) \ \alpha & =\frac{f(a)-f(b)}{a-b} \ \alpha & =\frac{f(b)-f(a)}{b-a} \end{aligned}$ Put, $\alpha$ in Eq. (i) $\begin{aligned} f(a) & =\left(\frac{f(b)-f(a)}{b-a}\right) \times a+\beta \ f(a)[b-a] & =a f(b)-a f(a)+\beta(b-a) \ b f(a)-a f(a) & =a f(b)-a f(a)+\beta(b-a) \ \beta & =\frac{b f(a)-a f(b)}{(b-a)} \ r(x) & =\alpha x+\beta \ & =\frac{f(b)-f(a) x}{b-a}+\frac{b f(a)-a f(b)}{(b-a)} \ & =\frac{x f(b)-x f(a)+b f(a)-a f(b)}{b-a} \ r(x) & =\frac{(x-a) f(b)+(b-x) f(a)}{b-a} \ r(x) & =\frac{(x-a) f(b)-(x-b) f(a)}{b-a} \end{aligned}$ Hence, option (c) is correct.

Let \(f(x)\) be a polynomial and \(a, b\) be distinct real numbers. Then the remainder in the division of…

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