If I have a polynomial $f(x)$ and is divided by $(x- 3)$ and $(x + 6)$ the respective remainders are $7$ and $22$, what is the remainder when $f(x)$ is divided by $(x-3)(x + 6)$?
I tried it by doing:
$$f(x) =(x-3)(x+6)q(x) + ax+b $$ And, $a$ and $b$ comes out to be $-\dfrac53$ and $12$ respectively.
But I'm not sure how to solve any further. And kindly explain exactly how it's done
A slick way via $\ {ab\bmod ac = a(b\bmod c)} = $ mod Distributive Law, $ $ is to add $\ f(3) = 7\ $ to
$\quad f(x)\!-\!f(3)\bmod (x\!-\!3)(x\!+\!6)\, =\, (x\!-\!3)\Bigg[\dfrac{\overbrace{f(x)\!-\!f(3)}^{\Large\color{#c00}{ 22\ \ -\ \ 7}\ }}{\underbrace{x - 3}_{\large\color{#c00}{\Large -6\ -\ 3\ }\ \ }}\underbrace{\bmod_{\phantom{1_{1_1{1_{1_{1_1}}}}}}\!\!\!\!\!\!\!\!\!\!\! x\!+\!6}_{\Large\Rightarrow\ \ x\ \equiv\ \color{#c00}{-6}\!}\Bigg] =\, \color{#c00}{-\dfrac{5}3}(x\!-\!3) $
$f(x)=(x-3)a(x)+7\Rightarrow f(3)=7$
If you can write $f$ is of the form $f(x) =(x-3)(x+6)q(x) + ax+b$.
Solution is so easy:
$$f(3)=3a+b=7$$ $$f(-6)=-6a+b=22$$
Hence, $a=-\dfrac53$ and $b=12$
Let $$f(x)=Q1(x-3)+7$$ and $$f(x)=Q2(x+6)+22$$ (where Q1 and Q2 are some functions of x, since degree of f can be 2)
So, if you substitute x=3 and x=-6 in above equations respectively, you get f(3)=7 and f(-6)=22
Now, let $$f(x)=Q3(x-3)(x+6)+(ax+b)$$ where (ax+b) is remainder
(remainder may be a constant or a linear in x because of degree restriction over f(x))
Substitute x=3 $$f(3)=0+(3a+b) $$
Substitute x=-6 $$f(-6)=0+(-6a+b)$$
So, 3a+b=7 and -6a+b=22
Solving, $$a= \frac{-5}{3}$$ and $$b=12 $$ Remainder is $$ \frac{-5x}{3}+12$$
