# 4.11 Roots of polynomials

Any quadratic (which we can write as $ax^{2}+bx+c$) that has roots $\alpha$ and $\beta$ can be equivalently written as $(x-\alpha)(x-\beta)$. This is by the factor theorem (Section 4.6).

When we expand $(a-\alpha)(x-\beta)$ we get a quadratic.

We can then compare the coefficients of Equation 4.120 with those of $ax^{2}+bx+c$.

This gives a set of equations relating the roots and the coefficients of a polynomial.

Something similar is also the case for higher-degree polynomials
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TODO: add a proof for this.

###### Example 4.11.1

A quadratic equation $x^{2}-8x+12=0$ has roots $\alpha$ and $\beta$. Find a quadratic with roots $\alpha^{2}$ and $\beta^{2}$.

Solution 1: To find the coefficients of our new quadratic, we need to find the value of $\alpha^{2}\beta^{2}$ and $\alpha^{2}+\beta^{2}$.

$\displaystyle\alpha^{2}\beta^{2}$ | $\displaystyle=(\alpha\beta)^{2}$ | ||

$\displaystyle=12^{2}$ | |||

$\displaystyle=144$ |

$\displaystyle\alpha^{2}+\beta^{2}$ | $\displaystyle=(\alpha+\beta)^{2}-2(\alpha\beta)$ | ||

$\displaystyle=8^{2}-2(12)$ | |||

$\displaystyle=40$ |

Therefore, a quadratic with roots $\alpha^{2}$ and $\beta^{2}$ is $x^{2}-40x+144=0$.

Solution 2: We can also do this using a substitution. First, note that for our original quadratic, we know that $x=\alpha$ is a root. We want a new polynomial, however, where it is not $x=\alpha$ which is a root, but rather $x=\alpha^{2}$ that is a root. Consider the graph of our function (below)

What we want to do is transform the position of the roots.
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If this is leading to thoughts about transformations of graphs (an
earlier topic in the algebra section) then, I mean, yes! Let’s look at one of
the roots (the one at $x=2$) - it should clearly end up at $2^{2}=4$ (as this is
what the question is asking for). If we call our quadratic $P(x)$, and think
about the point $x$, at the point $x$, we’d like to have a y-value of
$P(\sqrt{x})$, rather than the current $P(x)$.^{24}^{24}
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Because that way
$\sqrt{x}\mapsto x$, which also means that $x\mapsto x^{2}$

Therefore, our new quadratic should be

$\displaystyle P(\sqrt{x})$ | $\displaystyle=(\sqrt{x})^{2}-8\sqrt{x}+12$ | ||

$\displaystyle=x-8\sqrt{x}+12$ |

We’re interested in when this function happens to be zero, so we want

^{25}

^{25}25 To manipulate this equation we use a trick where something is of the form $x_{\text{nasty}}+y_{\text{nice}}+z_{\text{nice}}+...=0$, so we move all the ”nice” stuff over to one side and then apply a function to both sides in order to eliminate the ”nasty” stuff.

$\displaystyle P(\sqrt{x})=0$ | ||

$\displaystyle x-8\sqrt{x}+12=0$ | ||

$\displaystyle x+12=8\sqrt{x}$ | ||

$\displaystyle(x+12)^{2}=64x$ | ||

$\displaystyle x^{2}+24x+144=64x$ | ||

$\displaystyle x^{2}-40x+144=0$ |

This is the same as the other method which relied upon manipulating the roots directly! In general: use whichever method is nicer.

## 4.11.1 A harder example

###### Example 4.11.2

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^{26}26 From https://madasmaths.com/archive/maths_booklets/further_topics/various/roots_of_polynomial_equations.pdf

: the cubic $C$, with roots $\alpha$, $\beta$, and $\gamma$ is given by

The integer function $S_{n}$ is defined as

Find the values of $S_{3}$.

Solution: The easier (in my opinion), way to solve this is by using a substitution. In the case of $S_{3}$ we have our old root $x=x$ and we want to transform it (on the same axis) to the position $(2x+1)^{3}$, which defines a seperate axis, $X$. Therefore, to write the $x$-axis in terms of the $X$-axis, we rearrange

to be in terms of $X$, after which we can then substitute $X$ for $x$ in the polynomial.

$\displaystyle(2x+1)^{3}=X$ | (4.127) | ||

$\displaystyle 2x+1=X^{\frac{1}{3}}$ | (4.128) | ||

$\displaystyle 2x=X^{\frac{1}{3}}-1$ | (4.129) | ||

$\displaystyle x=\frac{X^{\frac{1}{3}}-1}{2}$ | (4.130) |

Using this we can then substitute into the original polynomial

$\displaystyle 8x^{3}+12x^{2}+2x-3=0$ | (4.131) | ||

$\displaystyle\implies 8\left(\frac{X^{\frac{1}{3}}-1}{2}\right)^{3}+12\left(% \frac{X^{\frac{1}{3}}-1}{2}\right)^{2}+2\left(\frac{X^{\frac{1}{3}}-1}{2}% \right)-3=0$ | (4.132) |

This can then be simplified, a lot.

$\displaystyle 8\left(\frac{X^{\frac{1}{3}}-1}{2}\right)^{3}+12\left(\frac{X^{% \frac{1}{3}}-1}{2}\right)^{2}+2\left(\frac{X^{\frac{1}{3}}-1}{2}\right)-3=0$ | (4.133) | ||

$\displaystyle\implies\left(X^{\frac{1}{3}}-1\right)^{3}+3\left(X^{\frac{1}{3}}% -1\right)^{2}+\left(X^{\frac{1}{3}}-1\right)-3=0$ | (4.134) | ||

$\displaystyle\implies\left(X-3X^{\frac{2}{3}}+3X^{\frac{1}{3}}-1\right)+3\left% (X^{\frac{2}{3}}-2X^{\frac{1}{3}}+1\right)+\left(X^{\frac{1}{3}}-1\right)-3=0$ | (4.135) | ||

$\displaystyle\implies\left(X-3X^{\frac{2}{3}}+3X^{\frac{1}{3}}-1\right)+\left(% 3X^{\frac{2}{3}}-6X^{\frac{1}{3}}+3\right)+\left(X^{\frac{1}{3}}-1\right)-3=0$ | (4.136) | ||

$\displaystyle\implies X+\left(-3X^{\frac{2}{3}}+3X^{\frac{2}{3}}\right)+\left(% +3X^{\frac{1}{3}}-6X^{\frac{1}{3}}+X^{\frac{1}{3}}\right)+\left(-1+3-1-3\right% )=0$ | (4.137) | ||

$\displaystyle\implies X-2X^{\frac{1}{3}}-2=0$ | (4.138) |

Here, we pull the familiar trick^{27}^{27}
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Where familiar means ”did it once in
the previous example”. and rearrange

From here, we cube both sides and march onwards

$\displaystyle(X-2)^{3}=\left(2X^{\frac{1}{3}}\right)^{3}$ | (4.140) | ||

$\displaystyle\implies X^{3}+3\left(X^{2}\right)(-2)+(3)(X)\left((-2)^{2}\right% )+(-2)^{3}=8X$ | (4.141) | ||

$\displaystyle\implies X^{3}-6X^{2}+12X-8=8X$ | (4.142) | ||

$\displaystyle\implies X^{3}-6X^{2}+4X-9=0$ | (4.143) |

Because we know this polynomial has the desired roots, and the sum of the roots is equal to $-\frac{b}{a}$, the value of $S_{3}$ is

## 4.11.2 Some further examples

###### Example 4.11.3

The equation $x^{3}+bx^{3}+cx^{2}+dx+e$ has roots $p$, $2p$, $3p$ and $4p$. Show that $125e=\frac{3}{10}b^{4}$.

As always, the first thing to do is to try to orient oneself. What is the question asking for? A relationship between $e$ and $b$ is one way to think about this. A logical next question is whether we can find a simple way to relate $e$ and $b$ directly to one another (I can’t). Perhaps another question which is not unreasonable to ask after this is whether it is possible to express $e$ in terms of $b$ by first expressing $e$ in terms of some third quantity, and then expressing this quantity in terms of $b$. Considering the question gives us the answer; we can express $e$ in terms of $p$, and $b$ in terms of $p$ (which also means we can express $p$ in terms of $b$).

Using Vieta’s formulae we can express $p$ in terms of $e$,

We can also express $p$ in terms of $b$,

Therefore, $p=-\frac{b}{10}$. We can then just substitute this into Equation 4.145,

$\displaystyle e$ | $\displaystyle=24p^{4}$ | (4.147) | ||

$\displaystyle=24\left(-\frac{b}{10}\right)^{4}$ | (4.148) | |||

$\displaystyle=\frac{24}{10,000}b^{4}$ | (4.149) | |||

$\displaystyle=\frac{3}{1250}b^{4}$ | (4.150) |

Therefore, after multiplying by $125$, the end result is that