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I’ve being practicing polynomial long division for the last week and have built some competence/confidence around the algorithm for performing the operation, but this is stumping me:

Given P($x$) = $(x^3-2x^2-x+2)/(x-k)$ has three values for k in which thee quotient has no remainder, what are the possible k values?

I’m not sure where to begin. The idea of the modulus function comes to mind, but I’m not sure that’s the way either. I’m posting in the hope someone could enlightenment me with the general strategy for this.

duckegg
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    Hint: the remainder when dividing $p(x)$ by $x-k$ is $p(k)$. – Arturo Magidin May 15 '22 at 21:22
  • Thanks @ArturoMagidin. I’ve been doing more research and the answer appears to lie in the direction of the Remainder Theorem you’re hinting at. If that’s the case, then $P(k)=k^3-2k^2-k+2=0$ Or $P(k)=k^3-2k^2-k=-2$. To determine k, some kind of formula similar to the quadratic equation should be invoked. Unless there is some other vector I’m unaware of? – duckegg May 15 '22 at 21:50
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    $k=1$ is an obvious solution, factor it out and you are left with a quadratic. Yes, there is a "cubic" formula (called Cardano's formula), but it is often hard to use in practice. – Arturo Magidin May 15 '22 at 21:58
  • @ArturoMagidin Thanks for your hint. Having ‘discovered’ the Remainder Theorem, the more I look at this, the more it appears to me that I’m going have to teach myself to solve for $x$ in cubic equations. – duckegg May 15 '22 at 21:58
  • @ArturoMagidin Factoring out $k$ and invoking the null factor theorem, I get that $k=0$ or $k=1$. I can sort of ‘cheat’ and use some function in my calculator to see that the other value should be -1. Without recourse to Cardano’s formula, is there a way to deduce that the other answer is -1? – duckegg May 15 '22 at 22:19
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    No. Since $p(1)=0$, you can write $ p(x)=(x-1)q(x)$, with $q(x)$ of degree $2$. Then the remaining roots of $p(x)$ are the roots of $q(x)$. You don't "factor out $k$". You factor out $x-r$ once you know $r$ is a root. – Arturo Magidin May 15 '22 at 22:29

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A useful theorem you can use is the following:

Let $P\in\mathbb{R}\left[X\right]$ be a polynomial over the reals. Then $a$ is a root of $P$ if and only if $\left(x-a\right)$ divides $P(x)$. (This is true not only for the reals but it will suffice in this case).

The fact that $P$ is divisible by a polynomial of the form $x-k$ means that there exists a polynomial $Q$ with $\deg Q =2$ such that $P(x)=(x-k)Q(x)$. So by finding a root of the numerator polynomial, you can represent the numerator like the product above, divide by $(x-k)$, i.e. you found one value you were looking for, and may proceed by finding other real roots if they exist (in this case they do).

Yia War
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After more research, I found a way to solve for x in a cubic. Dividing all of the factors of d by the factors in a and adding to these the set of their selves to the unit negative (-1), thus reducing the trial space to, in this instance, a cardinality of four: -2, -1, 1 and 2. Of these candidate values, -1, 1 and 2 were valid values for k in the original post via either synthetic division or trial and error.

duckegg
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What they are basically trying to ask is: What are the roots of the equation $x^{3}-2x^{2}-x+2$? (k=the roots) This is because they are asking for k such that the quotient has no remainder. This implies that they want you to factor the cubic.

Answer spoiler: (try not to look to the right) k=2,1,-1

But you may be asking how do we factor a cubic.

There are 3 ways in which I know:

Cardano's formula (It's a "cubic" formula, like a quadratic formula)

Lagrange resolvents (Substitution to another variable to convert it to a quadratic)

Factorisation (Encouraged for this example).

I'm not going to take the fun out of factorisation, but here's how you would do it: expand $(x+a)(x+b)(x+c)$, equate coefficients in your example, and solve for the variables, (assuming that a,b,c are integers would be helpful as they are commonly the case).

Maths Addict
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