A polynomial $f(x)$ with real coefficients leaves the remainder $15$ when divided by $x-3$ and the remainder $2x+1$ when divided by $(x-1)^2$. The remainder when $f(x)$ is divided by $(x-3)(x-1)^2$ is?
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Since function $f(x)$ leaves remainder $15$ when divided by $x−3$, therefore $f(x)$ can be written as $$f(x)=(x−3)l(x)+15\quad ...(1)$$ Also, $f(x)$ leaves remainder $2x+1$ when divided by $(x−1)^2$. Thus, $f(x)$ can also be written as $$f(x)=(x−1)^2m(x)+2x+1\quad ...(2)$$ If $R(x)$ be the remainder when $f(x)$ is divided by $(x−3)(x−1)^2$, then we may write, $$f(x)=(x−3)(x−1)^2n(x)+R(x)\quad ...(3)$$ Since $(x−3)(x−1)^2$ is a polynomial of degree three, the remainder has to be a polynomial of degree less than or equal to two.
Thus let $R(x)=ax^2+bx+c$ From $(1)$ and $(3)$, we have, $$f(3)=15=R(3)\Rightarrow 9a+3b+c=15\quad ...(4)$$ From $(2)$ and $(3)$, we have, $$f(1)=3=R(1)\Rightarrow a+b+c=3\quad ...(5)$$ From $(2)$ and$ (3)$, we have, $$f′(1)=2=R′(1)\Rightarrow 2a+b=2\quad...(6)$$ Solving equation $(4)$, $(5)$ and $(6)$, we get $$a=2, b=−2, c=3$$
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