Repeated root remainder

Since $(x-1)^2$ is a factor of $f(x)$ then, $f(x) =(x-1)^2(x-d)$ , where $d$ is a real constant. Therefore,

$\begin{aligned} f(x) & = (x-1)^2(x-d) \\ \Rightarrow x^3+bx+c & = (x-1)^2(x-d) \\ & = (x^2-2x+1)(x-d) \\ & = x^3 - (d+2)x^2 + (2d+1)x - d \end{aligned}$

Equating coefficients of both sides, we have, $d+2= 0\quad \Rightarrow d = -2$ . Therefore, we have:

$\begin{aligned} f(x) & = (x-1)^2(x-d) \\ \Rightarrow f(2) & = (2-1)^2[2-(-2)] = \boxed{4} \end{aligned}$

I did this by method of "forcing the factor":

$f(x)=x^3+bx+c \\ \Rightarrow f(x)=x^3-2x^2+x+2x^2-x+bx+c \\\Rightarrow f(x)=x(x^2-2x+1)+2x^2+x(b-1)+c \\\Rightarrow f(x)=x(x-1)^2+2\left(x^2-x\left(\dfrac{b-1}{2}\right)+\dfrac{c}{2}\right)$

Since $(x-1)^2 \ | \ f(x)$ we have:

$x^2-x\left(\dfrac{b-1}{2}\right)+\dfrac{c}{2}=x^2-2x+1 \\ \Rightarrow \left(\dfrac{b-1}{2}\right)=-2 \ , \ \dfrac{c}{2}=1 \\ \Rightarrow \boxed{b=-3 \ , \ c=2}$

So we have $f(x)=x^3-3x+2$ .By Remainder theorem ,

$f(2)=2^3-3(2)+2=8-6+2=2+2=\huge\boxed{4}$

Remainder when we divide $f(x)$ by $(x-1)^{2}$ is zero. Therefore $-1=b+c$ .

Now remainder when $f(x)$ is divided by $x-2$ is $8+2b+c$ i.e $7+b$ .

Now we know by factor theorem that $f(x)=(x-1)^{2}*q(x)$ .

$q(x)$ must be linear so let $q(x)=x-z$ .

Therefore $f(x)=(x-1)^{2}(x-z)$

Now $x^{3}+bx+x=x^{3}-2x^{2}+x-zx^{2}+2xz-z$

Comparing quadratic terms coef on both sides we get $-(2+z)=0$ i.e $z=-2$ .

Now we compare linear terms we get $b=1+2z=1+2*(-2)=1-4=-3$

Therefore our remainder is $7+b=7-3=4$ .

I just divided x-1 with f(x) to get quotient x^2+x+(b+1) now we know this is also divisible by x-1 so we put x=1 in that to get b=-3 and we also know b+c=-1 so we get c=2 and we substitute x=2 in f(x) to get 4

We can use the fact that f(1) as well as derivative of f(1) are 0

Since x=1 is repeated root of the polynomial f(x), therefore, f(x)=0 and f'(x)=0 at x=1. Solving the two equations f(1)=0 and f'(1) =0 we get b=-3 and c=2. substituting these values we have, f(x)= x^3-3x+2. Apply remainder theorem to get the required remainder which is f(2)=4

Super easy solution :