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I am currently learning how to find zeroes of polynomials. However, 4 things confuse me.

  1. The first question I have is why exactly does the rational zeroes theorem hold true? I get how to use it, but I don't get why it is true. Can someone please explain this?

  2. The second question I have is why is the Descartes rule of signs true?

  3. My third question is how do we find the imaginary roots of polynomial? It can't be through the rational zeroes theorem! But if we don't need to find imaginary roots, why else would we need things like the conjugate pairs theorem?

  4. My last question is of synthetic division in finding the zeroes of polynomials. I don't get why using synthetic division to test if a root is a zero of a polynomial is by taking it, and synthetically dividing it. What I don't get is, why don't you need to change the sign of the zero before synthetically dividing here? Let's say I suspected that the zero of a polynomial is x=-4. Don't I need to change the sign of the root to x+4=0, then synthetically divide by 4? Why is that you don't need to do this?

Can you explain all of this using simple algebra, without complicated techniques? I don't understand any complicated techniques and theorems beyond the quadratic formula. Can you also show and explain your working, so it is easier for me to follow through? I am still a beginner, so that would help very much.

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    $\begingroup$ Each of your questions really belongs in a separate post. $\endgroup$ – Blue Jul 17 '18 at 10:45
  • $\begingroup$ See, for instance, the question "Intuition behind Descartes' Rule of Signs". $\endgroup$ – Blue Jul 17 '18 at 10:54
  • $\begingroup$ Fun fact: The Rule of Signs, which applies to polynomials with real coefficients, has an extension to polynomials with complex coefficients. I call it the "Rule of Sweeps". I described it eight(!) years ago on MathOverflow. (I'm still seeking a direct proof.) $\endgroup$ – Blue Jul 17 '18 at 11:49
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  1. The rational roots or rational zeros theorem only applies to polynomials that have integer coefficients - although a polynomial with rational coefficients can be turned into a polynomial with integer coefficients by multiplying through by the LCM of the coefficients' denominators. The simplest way to see why it is true is to notice that if $\frac{p}{q}$ is a root of the polynomial then $qx-p$ is a factor of the polynomial and so $q$ must be a factor of the coefficient of the highest power term and $p$ must be a factor of the constant term.
  2. There is a simple proof of Descartes rule of signs here: https://www.math.hmc.edu/funfacts/ffiles/20001.1.shtml. It uses induction of the degree of the polynomial, and compares the sign changes in a polynomial with the sign changes in its derivative.
  3. There are explicit formulae for finding the roots of cubic and quartic polynomials (similar to the quadratic formula but more complicated). Finding complex roots of higher degree polynomials is difficult, and usually requires an iterative algorithm such as Newton's method to approximate the roots.
  4. I'm not sure I understand this question. To test whether $-4$ is a root of a polynomial you can just substitiute $x=-4$ to see whether $f(-4)=0$. If you know $-4$ is a root of the polynomial then you can find the quotient after you have removed that root by dividing the polynomial by $x+4$. You know $x+4$ know must be a factor of the polynomial because $f(-4)=0$.
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  • $\begingroup$ Thank you. Final question. For the rational root theorem, will it work if only the coefficient of the largest term is an integer, and if all the other terms are decimals? If not, why? $\endgroup$ – Ethan Chan Jul 17 '18 at 11:50
  • $\begingroup$ If the coefficients are terminating decimals then you can use the rational root theorem because all terminating decimals are rational numbers. If the coefficients are non-terminating decimals than you cannot always use the rational root theorem because the coefficients may be irrational numbers such as $\sqrt{2}$ or $\pi$. Note that the rational root theorem only provides a set of possible rational roots - if the polynomial has irrational roots (e.g. $x^2-2$) then the rational root theorem will not help you find them. $\endgroup$ – gandalf61 Jul 17 '18 at 11:54
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The first question I have is why exactly does the rational zeroes theorem hold true? I get how to use it, but I don't get why it is true. Can someone please explain this?

It is because when you have a normed polynom $p(x)$ with a root $r\in\mathbb{Z}$, then you can rewrite it as

$p(x)=q(x)(x-r)$, since we can just use long division in that case.

When you factor this out how ever you see, that the constant part of $p(x)$ must be divisable by $r$. This is because the constant part c of q(x) multiplied with $(-r)$ is the only way to get the constant part of $p(x)$, wich is $-cr$ then.

Hence if you have an integer root, it has to divide the constant part.

The second question I have is why is the Descartes rule of signs true?

[Edit: I just saw, that the sign rule of decart is more specific about this, so the following does not really match.]

This is called the intermidate value theorem. https://en.wikipedia.org/wiki/Intermediate_value_theorem

It bascically says, that when you have a continuous function $f$ which is negative for some $x_1$ and positiv for some $x_2$, then it has to pass the x-axis at some point between $x_1$ and $x_2$. Hence you have a root $r\in (x_1, x_2)$.

You might draw a picture for that.

My third question is how do we find the imaginary roots of polynomial? It can't be through the rational zeroes theorem! But if we don't need to find imaginary roots, why else would we need things like the conjugate pairs theorem?

This question can not be answered this easily.

About your 4th question, I have never heard about synthetic division before, but you should be correct anyways.

If you have a root $r$ you have to divide by $(x-r)$ for a (helpful) division.

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The reason in the fourth question is that we can always divide one polynomial by another leaving a remainder of lower degree, so that dividing by a linear polynomial gives a constant remainder. Suppose we divide by $(x-a)$ so that $$p(x)=(x-a)q(x)+r$$

If we then set $x=+a$ we get $$p(a)=r$$ and if $a$ is a root we have $r=0$.

The signs work as they do because we need to make the $(x-a)q(x)$ factor disappear - we are not interested in $q(x)$ - and we can make this happen by setting $x-a=0$

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  • $\begingroup$ I'm sorry, I don't follow what you are saying. What has this got to do with synthetic division, and the signs being wrong? Can you please elaborate and explain? $\endgroup$ – Ethan Chan Jul 17 '18 at 12:05
  • $\begingroup$ @EthanChan "Synthetic" division gives you the outcome $p(x)=(x-a)q(x)+r$ - I am trying to show you how a different perspective makes sense of the signs. $\endgroup$ – Mark Bennet Jul 17 '18 at 12:08

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