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.

x^{2}-(\alpha+\beta)x+\alpha\beta

(4.120)

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

x^{2}-(\alpha+\beta)x+\alpha\beta=x^{2}+\frac{b}{a}x+\frac{c}{a}

(4.121)

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

-(\alpha+\beta)=\frac{b}{a}\iff\alpha+\beta=-\frac{b}{a}

(4.122)

\alpha\beta=\frac{c}{a}

(4.123)

Something similar is also the case for higher-degree polynomials ²²²² 22 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. ²³²³ 23 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 $P(x)$ .²⁴²⁴ 24 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

²⁵²⁵ 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

²⁶²⁶ 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

8x^{3}+12x^{2}+2x-3=0

(4.124)

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

S_{n}=(2\alpha+1)^{n}+(2\beta+1)^{n}+(2\gamma+1)^{n}

(4.125)

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

X=(2x+1)^{3}

(4.126)

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 trick²⁷²⁷ 27 Where familiar means ”did it once in the previous example”. and rearrange

X-2=2X^{\frac{1}{3}}

(4.139)

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

-\frac{-6}{1}=6

(4.144)

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 $e$ ,

(p)(2p)(3p)(4p)=24p^{4}=e

(4.145)

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

p+2p+3p+4p=10p=-b

(4.146)

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

125e=\frac{3}{10}b^{4}

(4.151)