Dividing polynomial $f(x)$ by $(x-a)(x-b)(x-c)$

Question

$f(x)$ is a polynomial with a degree greater than 3. When $f(x)$ is divided by $(x-a)(x-b)(x-c)$, prove remainder is $$\frac{f(a)(x-b)(x-c)}{(a-b)(a-c)}+ \frac{f(b)(x-a)(x-c)}{(b-c)(b-a)} +\frac{f(c)(x-a)(x-b)}{(c-b)(c-a)}$$

My Try

I tried this using the conventional method,$$f(x)=Q(x)(x-a)(x-b)(x-c)+Ax^2+Bx+C$$

But then I got long answers for coefficients $A$, $B$ & $C$. Is there a better way to solve this? Can anyone give me a hint to work this?

Answer 1

We have: $$f(x)=Q(x)(x-a)(x-b)(x-c)+Ax^2+Bx+C$$ as you mentioned. Now note that $$f(a)=Aa^2+Ba+C\\ f(b)=Ab^2+Bb+C\\ f(c)=Ac^2+Bc+C$$ Now it is easy to solve this system of equations for $A$, $B$ and $C$.

Answer 2

Because $$f(a)=Aa^2+Ba+C,$$ $$f(b)=Ab^2+Bb+C$$ and $$f(c)=Ac^2+Bc+C.$$ Now, work with the given remainder (it's also a polynomial of the second degree).

Answer 3

Call the purported remainder $\,r(x).$ $\,f(x)-r(x)$ has roots $\,x = a,b,c\,$ so by the Factor Theorem $\,f(x)-r(x) = (x\!-\!a)(x\!-\!b)(x\!-\!c) q(x),\,$ so $\deg r < 3\,$ $\Rightarrow\ r(x) =$ remainder by its uniqueness.