A Polynomial Remainder

As $f(x)$ leaves a remainder of $2x+1$ when divided by $\left( x-1\right)^{2}$ we may assume

$f(x)=\left( x-1\right)^{2}g(x)+(2x+1) ....................(\text{a})$

Substituting $x=1$ in $(\text{a})$ we get:

$f(1)=3$

Now differentiating both sides of $(\text{a})$ w.r.t $x$ we get:

$f'(x)=(x-1)\left( (x-1)g'(x)+2g(x)\right)+2$

Substituting $x=1$ in the above equation we get:

$f'(1)=2$

Now since $f(x)$ also leaves a remainder of $15$ on division by $x-3$ we may assume

$f(x)=\left( x-1\right)^{2}(x-3)h(x)+ax^{2}+bx+c ....................(\text{b})$

Substituting $x=1$ in $(\text{b})$

$a+b+c=3 ....................(1)$

Substituting $x=3$ in $(\text{b})$

$9a+3b+c=15 ....................(2)$

Now differentiating both sides of $(\text{b})$ w.r.t. $x$ we get:

$f'(x)=(x-1)\left( (x-1)(x-3)h'(x)+(x-1)h(x)+2(x-3)h(x)\right)+2ax+b$

Substituting $x=1$ in the above equation we get:

$2a+b=2 ....................(3)$

Solving $(1)$ , $(2)$ and $(3)$ we get:

$a=2$ , $b=-2$ and $c=3$

Thus the required answer is

$a\times b\times (-c)=2\times (-2)\times (-3)=\boxed{12}$

The first condition is that $f(3) = 15$ . Now we can write $f(x) = (x-1)^2g(x) + 2x+1$ , and plugging in $x = 3$ gives $g(3) = 2$ . So write $g(x) = (x-3)h(x) + 2$ and simplify to get $f(x) = (x-1)^2(x-3)h(x) + 2(x-1)^2+2x+1,$ so the remainder upon division by $(x-1)^2(x-3)$ is $2x^2-2x+3$ , so the answer is $\fbox{12}$ .