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$f(x)$ 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?

Michael Rozenberg
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emil
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  • what should be the remainder when $f(x)$ is divided by $(x-a)$ ? – J. W. Tanner Jun 24 '19 at 17:58
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    Note $f(x)=Q(x)(x-a)(x-b)(x-c)+R(x)$. What can you say on $R(a), R(b), R(c)$ ? The degree of $R$ is less than $2$. So you can use Lagrange interpolation. There is no need to make an explicit calculation of $A$, $B$, $C$. – Dlem Jun 24 '19 at 18:01
  • Note that the question doesn't ask you to derive that expression for the remainder. Rather it only asks to prove that it equals the remainder, which is very easy - see my answer. – Bill Dubuque Jun 24 '19 at 18:57

3 Answers3

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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 $C$.

Qurultay
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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).

Michael Rozenberg
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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.

Bill Dubuque
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  • e.g. $\ r(a) = f(a)\dfrac{(a!-!b)(a!-!c)}{(a!-!b)(a!-!c)} = f(a).\ $ So $,r(b) = f(b),\ ,r(c) = f(c)\ $ similarly (or by symmetry). – Bill Dubuque Jun 24 '19 at 18:34