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A polynomial $f(x)$ is such that upon division by $(x-3)$ it leaves a remainder of $15$ and upon division by $(x-1)^2$, it leaves a remainder of $2x+1$. What is the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$?

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closed as off-topic by Lazy Lee, user91500, kingW3, Hippalectryon, Trevor Gunn Jun 13 '17 at 22:08

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Hint. We have that $$f(x)=A(x)(x-3)+15=B(x)(x-1)^2+2x+1=C(x)(x-3)(x-1)^2+r(x)$$ where $r(x)=ax^2+bx+c$ (because the degree of $(x-3)(x-1)^2$ is $3$) and $A,B,C$ are polynomials.

In order to find $a$, $b$, and $c$, let $x=1$ and $x=3$. Consider also the derivative of $f$ at $x=1$. Therefore we obtain the equations: $$r(3)=15,\quad r(1)=3,\quad r'(1)=2.$$ Are you able to find $a,b,c$ now?

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  • $\begingroup$ Any way to do this without differentiating? $\endgroup$ – Archimedesprinciple Jun 13 '17 at 10:35

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