Finding remainder

Question

Okay I saw this one on a test so here it goes: A polynomial of (degree > 3) when divided by $ (x-1)^2$ and $x-3$ leaves the remainder $2x+1$ and $15$ respectively. The remainder when it is divided by $(x-1)^2(x-3)$ is $ax^2+bx+c$ then find: $a+2b$;$a-3c-2b$; and the nature of it's roots.. I tried but no luck

Answer 1

Let the polynomial be $f(x)$

We have $$f(x)=(x-1)^2Q_1(x)+2x+1$$ and $$f(x)=(x-3)Q_2(x)+15$$ We also have $$f(x)=(x-1)^2(x-3)Q_3(x)+ax^2+bx+c$$

putting $x=1,$ $$f(1)=3=a+b+c$$ putting $x=3,$ $$f(3)=15=9a+3b+c$$ Therefore eliminating $c$, we have $4a+b=6$

Furthermore, $$f'(x)=2(x-1)Q_1+(x-1)^2Q'_1+2\Rightarrow f'(1)=2$$ and also, $$f'(x)=2(x-1)(x-3)Q_3+(x-1)^2Q_3+(x-1)^2(x-3)Q'_3+2ax+b$$

Hence$$f'(1)=2=2a+b$$ Now you can solve to get $a=2, b=-2, c=3$

Answer 2

HINT:

Let $$f(x)=A(x-1)^2(x-3)+B(x-1)^2+C(x-1)(x-3)+D(x-3)$$

$$15=f(3)=B(3-1)^2\iff B=?$$

$$2x+1\equiv C(x-1)(x-1-2)+D(x-3)\pmod{(x-1)^2}$$ $$\equiv D(x-3)-2C=Dx-(2C+3D)$$

which needs to be $2x+1$

Answer 3

So the remainder when $ax^2+bx+c$ is divided by $(x-1)^2=x^2-2x+1$ is $2x+1$ and $$ax^2+bx+c-a(x^2-2x+1)=(b+2a)x+(c-a)=2x+1$$ So that $2a+b=2$ and $c-a=1$

Then set $x=3$ to obtain $9a+3b+c=15$.

Now solve for $a,b,c$