Question

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

• A. $\frac{(x-a)f(a)-(x-b)f(b)}{a-b}$
• B. $\frac{(x-a)f(b)-(x-b)f(a)}{a-b}$
• C. $\frac{(x-a)f(b)-(x-b)f(a)}{b-a}$
• D. $\frac{(x-a)f(a)-(x-b)f(b)}{b-a}$

Answer 1

Answer: (C)

Let $f(x)=(x-a)(x-b)\cdot q(x)+r(x)$
Let $r(x)=\alpha x+\beta [\because d\text{egr}(x)<$ deg. of divisor $]$
$\therefore f(x)=(x-a)(x-b)\cdot q(x)+ax+\beta$
$f(a)=\alpha a+\beta \dots .(i)$
$f(a)=\alpha a+\beta \dots .(ii)$
Subtract Eqs. (i) from (ii)
$f(a)-f(b)=a(a-b)$
$\alpha =\frac{f(a)-f(b)}{a-b}$
$\alpha =\frac{f(b)-f(a)}{b-a}$
Put, $\alpha$ in Eq. (i)
$f(a)=\left(\frac{f(b)-f(a)}{b-a}\right)\times a+\beta$
$f(a)[b-a]=af(b)-af(a)+\beta (b-a)$
$bf(a)-af(a)=af(b)-af(a)+\beta (b-a)$
$\beta =\frac{bf(a)-af(b)}{(b-a)}$
$\therefore r(x)=ax+\beta$
$=\frac{f(b)-f(a)x}{b-a}+\frac{bf(a)-af(b)}{(b-a)}$
$=\frac{xf(b)-xf(a)+bf(a)-af(b)}{b-a}$
$r(x)=\frac{(x-a)f(b)+(b-x)f(a)}{b-a}$
$r(x)=\frac{(x-a)f(b)-(x-b)f(a)}{b-a}$