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A polynomial function $f(x)$ with real coefficients leaves the remainder $15$ when divided by $x-3$, and the remainder $2x+1$ when divided by $(x-1)^2$. Then the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$ is?
What I have thought-
The remainder must be of the form $ax^2+bx+c$. Now applying the remainder theorem, I am able to find $2$ equations in $a,b,c$ . eg
Let $f(x)$=$(x-3)$$h(x)$+$15$ ...................................(1)
Also let $f(x)=(x-3)(x-1)^2g(x)+ax^2+bx+c$...........................(2)
Put $x=3$ and using (1) we get $15=9a+3b+c$
Similarly I can get another equation using the other information given. But I am only able to get 2 equations in 3 variables. From where do I get the 3rd equation in $a,b,c$ and hence the remainder?

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    $\begingroup$ Calculate derivatives both members $\endgroup$ – medicu Mar 24 '14 at 15:59
  • $\begingroup$ Sorry to mislead you. I know calculus. I have removed the tag. And I think @medicu comment will help $\endgroup$ – idpd15 Mar 24 '14 at 16:18
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Hint $\ f(x) = 2x\!+\!1+ (x\!-\!1)^2 (c + (x-3)g(x))\,$ and $\,15 = f(3) = 7+4c\ $ so $\ c = \,\ldots$

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  • $\begingroup$ I am not able to figure out what you did.. $\endgroup$ – idpd15 Mar 25 '14 at 5:56
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    $\begingroup$ @Mridul Since $f(x)$ has remainder $2x+1$ when divided by $(x-1)^2$ we can write $f(x) = 2x+1 + (x-1)^2 h(x).\,$ Now write $h(x) = c + (x-3)g(x)$ then evaluate at $\,x=3$ to solve for $c.\ \ $ $\endgroup$ – Bill Dubuque Mar 25 '14 at 12:38
  • $\begingroup$ okay... thanks! $\endgroup$ – idpd15 Mar 26 '14 at 19:37
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Note that $$ \frac{1}{4}(x-1)^2-\frac{1}{4}(x+1)(x-3)=1. $$ You can get this by dividing $(x-1)^2$ by $x-3$. So take $$ f(x)=15\frac{1}{4}(x-1)^2-(2x+1)\frac{1}{4}(x+1)(x-3). $$ Then the remainder when $f(x)$ is divided by $x-3$ is 15 and by $(x-1)^2$ is $2x+1$ since $(x+1)(x-3)=[(x-1)+2][(x-1)-2]=(x-1)^2-4$.

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