If the polynomial $$x^4-x^3+ax^2+bx+c$$ divided by the polynomial $$x^3+2x^2-3x+1$$ gives the remainder $$3x^2-2x+1$$ then how much is (a+b)c?
So what I know, and how I solved these problems before, I can write this down like this:
$x^4-x^3+ax^2+bx+c=Q(x)(x^3+2x^2-3x+1)+3x^2-2x+1$
And from here I should get the roots of $x^3+2x^2-3x+1$ so I could use the remainder only since I don't know $Q(x)$
$x^3+2x^2-3x+1=0$
but I can't find the roots of this by hand, so I used WolframAlpha to get the roots, but they are not whole numbers so I couldn't write this in the form of $(x-x_1)(x-x_2)(x-x_3)$ and then just use $P(x_1)$, $P(x_2)$, $P(x_3)$ to get $a,b,$ and $c$
I tried regular dividing on these polynomials to get $Q(x)$ but I couldn't divide them because of the $a,b$ and $c$ present
Let (original) polynomial $P(x) = x^4 - x^3 + ax^2 + bx + c$ and divisor polynomial $D(x) = x^3 + 2x^2 - 3x + 1$. The remainder polynomial $R(x) = 3x^2 - 2x + 1$. We can also define quotient polynomial $Q(x) = \alpha x + \beta$.
We have that $P(x) = Q(x)\cdot D(x) + R(x)$.
By inspection of the coefficient of $x^4$ on both sides, $\alpha = 1$, i.e. $Q(x)$ is monic.
Considering the coefficient of $x^3$ on both sides, we have that $-x^3 = 2x^3 + \beta x^3$, giving $\beta = -3$. So $Q(x) = (x-3)$.
By now substituting convenient values for $x$, we can solve for $a, b, c$ quite easily. But we don't have to go all the way on that, since we only need to find $(a+b)c$.
We can find $c$ by putting $x = 0$, giving $c = (-3)(1) + 1 = -2$.
We can find $a+b$ by putting $x = 1$, giving $1-1 + a + b - 2 = (-2)(1+2-3+1) + 3-2+1$, giving $a+b = 2$.
So $(a+b)c = 2(-2) = -4$.