2

Suppose I am given that $(x - \alpha)$ is a factor of $p(x) = a_0x^n + a_1x^{n-1} + .... a_n$

That obviously means $(x - \alpha)$ is a factor of $p(x)$ i.e $p(x) = (x - \alpha)q(x)$. My question is, if I want to find $q(x)$, and therefore the other roots, can I just multiply $(x - \alpha)$ by $q(x)$ and compare coefficients? Because theoretically, $q(x)$ being of degree $n - 1$ would have $n$ coefficients and if I compare them to $p(x)$ with $n+1$ coefficients, I would have n variables and n+1 equations, meaning a solution isn't guaranteed. Is this correct? If so, what does this mean, and how do I find the other roots if this happens?

MNIShaurya
  • 862
  • 3
    Polynomial division might be relevant. – Klaus May 21 '21 at 18:46
  • I presume that you meant to write 'Suppose I am given that $\alpha$ is a root of...' in your first sentence. – Joe May 21 '21 at 18:53
  • I don't exactly understand your question, but if $p(x)$ is a polynomial of degree $n$, then $q(x)$ is a polynomial of degree $n-1$, meaning that there are $n-1$ unknowns you have to solve for. E.g. if $p(x)=2x^3-3x^2+4x+3$, and we know $(2x+1)$ is a factor, then we could write $$ (2x^3-3x^2+4x+3) \equiv (2x+1)(Ax^2+Bx+C) $$ and notice that there are three unknowns ($A$, $B$, and $C$), but $4$ equations that result when we expand the RHS. – Joe May 21 '21 at 19:01
  • It sounds like you are asking how to dividie a polynomial by another polynomial. You can explicitly perform polynomial long division. Or you can set up a system of equations, if you want (and solving it would morally replicate the steps of polynomial long division). Your numerology is off, though --- there is a uniquely defined quotient. I suggest doing an explicit example (perhaps with something you know well, like $(x+1)(x+2)$ divided by $(x+1)$) to see your intuition break. – davidlowryduda May 21 '21 at 19:01
  • @Joe yes, so 3 unknowns and 4 equations will not necessary yield a solution. What happens when there isn't a solution? – MNIShaurya May 21 '21 at 19:28
  • 1
    @MNIShaurya: I take it that you are worried that the system of equations might not be consistent. If you look at Theorem 3.4 from this page, then you will see that you can pretty much always write a polynomial in the form $p(x)=(x-\alpha)q(x) + r(x)$, where $r(x)$ is the remainder term. This remainder term obviously is $0$ if $\alpha$ is a solution to the equation $p(x)=0$. – Joe May 21 '21 at 19:37
  • @MNIShaurya: Hence, given that such a polynomial exists, we know the system of equations always has a solution. I'm afraid that I do not know or understand a proof of this theorem, but hopefully this assuages your fears. – Joe May 21 '21 at 19:38
  • @Joe I think I've understood what you're saying. Is a solution always possible because the remainder polynomial(in this case a constant) is in fact the $n+1$th unknown? Or is it something completely different? – MNIShaurya May 22 '21 at 04:35
  • @MNIShaurya: That's not what I'm saying. Don't worry—I'll elaborate in an answer. – Joe May 22 '21 at 09:13

1 Answers1

1

Theorem 3,4 from this page comes in handy. The theorem is quite general, but we actually only need a special case to answer your question:

Suppose $p(x)$ is a polynomial of degree $n\geq1$ and $d(x)$ is a linear function of the form $d(x)=ax+b$. Then, there exists a unique polynomial $q(x)$ of degree $n-1$ such that $$ p(x)=q(x)d(x)+r \, , $$ where $r$, the remainder, is a constant.

Let's see this theorem in action. Suppose $p(x)=2x^3-3x^2-3x+7$ and $d(x)=x-2$. Then, by Theorem 3.4, we know that there exists a unique polynomial $q(x)$ such that $$ 2x^3-3x^2-3x+7=q(x)(x-2)+r \, . $$ Since the degree of $p(x)$ is $3$, the degree of $q(x)$ is $2$, and so $q(x)$ must be of the form $Ax^2+Bx+C$. Hence, $$ 2x^3-3x^2-3x+7=(Ax^2+Bx+C)(x-2)+r \, . $$ If we expand the RHS, then we get $$ 2x^3-3x^2-3x+7 \equiv Ax^3+(B-2A)x^2 + (C-2B)x + (-2C+r) \, . $$ From this obtain \begin{align} A &= 2 \\ B-2A &= -3 \\ C-2B &= -3 \\ -2C + r &= 7 \end{align} So that's four equations with four unknowns. Note however that not every system of equations with four equations and four unknowns has a solution. Consider the system \begin{align} w+x+2y+5z &= 2 \\ w+x+2y+5z &= 3 \\ 4w+2x+4y+10z &= 6 \\ 7w+4x-5y+z &= -11 \, . \end{align} Since the first two equations contradict each other, there are no solutions. If we change the second equation to $2w+x+2y+5z = 3$, then we get the opposite problem: since equations $2$ and $3$ are multiples of each other, there are infinitely many solutions.

However, in our case Theorem 3.4 guarantees that there is a unique polynomial $q(x)=Ax^2+Bx+C$ such that $$ 2x^3-3x^2-3x+7=(Ax^2+Bx+C)(x-2)+r \, . $$ If the solutions for $A$, $B$, and $C$ didn't exist, then $q(x)$ wouldn't exist, contradicting Theorem 3.4. Likewise, if $A$, $B$, and $C$ had infinitely many solutions, then $q(x)$ wouldn't be unique, also contradicting Theorem 3.4.

I'll end this by saying that it is good that you were sceptical about whether or not a solution was guaranteed. As Marc van Leeuwen puts it in this post,

If you take a docile attitude and accept the statement to be proved from the onset, you will never be able to understand what all this reasoning was needed for in the first place.

However, in this particular case, you have nothing to worry about.

Joe
  • 19,636