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:


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 $Q(x)$


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 $(a+b)c = 2(-2) = -4$.

  • $\begingroup$ Man math is beautiful! Could you explain just the part of $x^3$ where you get $\beta$ in more detail? $\endgroup$ – Aleksa Nov 11 '18 at 11:35
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    $\begingroup$ @Aleksa I agree! :) What I wanted to add is that, for remainder and factor theorem (and polynomial division in general), I often find it useful to go back to the basic theory because it's quite obvious and often very easy when you write it out. Better not to learn too many things by rote. $\endgroup$ – Deepak Nov 11 '18 at 11:37
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    $\begingroup$ Sorry, missed your edit. There are only two "ways" of getting the $x^3$ term of $P(x)$ on the LHS from the expression on the RHS - multiplying the $x^3$ term of $D(x)$ by the constant term of $Q(x)$ and multiplying the $x^2$ term of $D(x)$ by the $x$ term of $Q(x)$. Adding these (signed) terms up on both sides gives you a quick linear equation you can solve for $\beta$. $\endgroup$ – Deepak Nov 11 '18 at 11:39
  • $\begingroup$ Thanks for your feedback! $\endgroup$ – Aleksa Nov 11 '18 at 11:46
  • $\begingroup$ Sure, glad to help! :) $\endgroup$ – Deepak Nov 11 '18 at 11:46

Hint: Compute the remainder of $x^4-x^3$ when divided by $x^3+2x^2-3x+1$.

  • $\begingroup$ I've got $ax^2+bx+c=(x^3+2x^2-3x+1)[Q(x)-(x+3)]-6x^2+8x+4$ now, what do I do from this point? $\endgroup$ – Aleksa Nov 11 '18 at 10:59
  • $\begingroup$ The remainder was $9x^2-10x-3$, so I moved $(x^3-2x^2-3x+1)(x-3)+9x^2-10x-3$ on the other side to get that $\endgroup$ – Aleksa Nov 11 '18 at 11:00

Maybe if you go through the steps for polynomial long division you will get an answer for the remainder in terms of a and b. Then you can equate the x terms with the same order, to determine a and b.


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