Question Regarding Remainder Theorem and Polynomials

Question

Show that when the polynomial $f(x)$ is divided by $(x-a)(x-b)$ where $a \neq b$, the remainder is $ \frac{(x-a)f(a)-(x-a)f(b)}{a-b} $.

Thanks!

Answer 1

Hint:

WLOG $$f(x)=(x-a)(x-b)g(x)+m(x-a)+n(x-b)$$ where $m,n$ are arbitrary constants

Set $x=a,b$ one by one

Or $f(x)=(x-a)(x-b)h(x)+qx+r$

Set $x=a,b$

Answer 2

The remainder should be $\displaystyle \frac{(x-b)f(a)-(x-a)f(b)}{a-b}$.

Let $\displaystyle g(x)=f(x)-\frac{(x-b)f(a)-(x-a)f(b)}{a-b}$.

Note that $\displaystyle g(a)=f(a)-\frac{(a-b)f(a)}{a-b}=0$ and $\displaystyle g(x)=f(b)-\frac{-(b-a)f(b)}{a-b}=0$.

$g(x)$ is divisible by both $x-a$ and $x-b$. It is divisible by $(x-a)(x-b)$.

Answer 3

Note that $$ x-a\mid f(x)-f(a) $$ and that $$ \left.x-b\,\middle|\,\vphantom{\frac{f}{a}}\right.\overbrace{\frac{f(x)-f(a)}{x-a}}^{\large g(x)}-\overbrace{\frac{f(b)-f(a)}{b-a}}^{\large g(b)} $$ Therefore, $$ \left.(x-a)(x-b)\,\middle|\,f(x)-f(a)-(x-a)\frac{f(b)-f(a)}{b-a}\right. $$ Since $f(a)+(x-a)\frac{f(b)-f(a)}{b-a}=\frac{(x-b)f(a)-(x-a)f(b)}{a-b}$, we have $$ \left.(x-a)(x-b)\,\middle|\,f(x)-\frac{(x-b)f(a)-(x-a)f(b)}{a-b}\right. $$