4.11 Roots of polynomials

Any quadratic (which we can write as ax2+bx+cax^{2}+bx+c) that has roots α\alpha and β\beta can be equivalently written as (xα)(xβ)(x-\alpha)(x-\beta). This is by the factor theorem (Section 4.6).

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

x2(α+β)x+αβx^{2}-(\alpha+\beta)x+\alpha\beta (4.120)

We can then compare the coefficients of Equation 4.120 with those of ax2+bx+cax^{2}+bx+c.

x2(α+β)x+αβ=x2+bax+cax^{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.

(α+β)=baα+β=ba-(\alpha+\beta)=\frac{b}{a}\iff\alpha+\beta=-\frac{b}{a} (4.122)
αβ=ca\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 x28x+12=0x^{2}-8x+12=0 has roots α\alpha and β\beta. Find a quadratic with roots α2\alpha^{2} and β2\beta^{2}.

Solution 1: To find the coefficients of our new quadratic, we need to find the value of α2β2\alpha^{2}\beta^{2} and α2+β2\alpha^{2}+\beta^{2}.

α2β2\displaystyle\alpha^{2}\beta^{2} =(αβ)2\displaystyle=(\alpha\beta)^{2}
α2+β2\displaystyle\alpha^{2}+\beta^{2} =(α+β)22(αβ)\displaystyle=(\alpha+\beta)^{2}-2(\alpha\beta)

Therefore, a quadratic with roots α2\alpha^{2} and β2\beta^{2} is x240x+144=0x^{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=αx=\alpha is a root. We want a new polynomial, however, where it is not x=αx=\alpha which is a root, but rather x=α2x=\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=2x=2) - it should clearly end up at 22=42^{2}=4 (as this is what the question is asking for). If we call our quadratic P(x)P(x), and think about the point xx, at the point xx, we’d like to have a y-value of P(x)P(\sqrt{x}), rather than the current P(x)P(x).2424 24 Because that way xx\sqrt{x}\mapsto x, which also means that xx2x\mapsto x^{2}

Therefore, our new quadratic should be

P(x)\displaystyle P(\sqrt{x}) =(x)28x+12\displaystyle=(\sqrt{x})^{2}-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 xnasty+ynice+znice+=0x_{\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.
P(x)=0\displaystyle P(\sqrt{x})=0
x8x+12=0\displaystyle x-8\sqrt{x}+12=0
x+12=8x\displaystyle x+12=8\sqrt{x}
x2+24x+144=64x\displaystyle x^{2}+24x+144=64x
x240x+144=0\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 CC, with roots α\alpha, β\beta, and γ\gamma is given by

8x3+12x2+2x3=08x^{3}+12x^{2}+2x-3=0 (4.124)

The integer function SnS_{n} is defined as

Sn=(2α+1)n+(2β+1)n+(2γ+1)nS_{n}=(2\alpha+1)^{n}+(2\beta+1)^{n}+(2\gamma+1)^{n} (4.125)

Find the values of S3S_{3}.

Solution: The easier (in my opinion), way to solve this is by using a substitution. In the case of S3S_{3} we have our old root x=xx=x and we want to transform it (on the same axis) to the position (2x+1)3(2x+1)^{3}, which defines a seperate axis, XX. Therefore, to write the xx-axis in terms of the XX-axis, we rearrange

X=(2x+1)3X=(2x+1)^{3} (4.126)

to be in terms of XX, after which we can then substitute XX for xx in the polynomial.

(2x+1)3=X\displaystyle(2x+1)^{3}=X (4.127)
2x+1=X13\displaystyle 2x+1=X^{\frac{1}{3}} (4.128)
2x=X131\displaystyle 2x=X^{\frac{1}{3}}-1 (4.129)
x=X1312\displaystyle x=\frac{X^{\frac{1}{3}}-1}{2} (4.130)

Using this we can then substitute into the original polynomial

8x3+12x2+2x3=0\displaystyle 8x^{3}+12x^{2}+2x-3=0 (4.131)
8(X1312)3+12(X1312)2+2(X1312)3=0\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.

8(X1312)3+12(X1312)2+2(X1312)3=0\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)
(X131)3+3(X131)2+(X131)3=0\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)
(X3X23+3X131)+3(X232X13+1)+(X131)3=0\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)
(X3X23+3X131)+(3X236X13+3)+(X131)3=0\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)
X+(3X23+3X23)+(+3X136X13+X13)+(1+313)=0\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)
X2X132=0\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

X2=2X13X-2=2X^{\frac{1}{3}} (4.139)

From here, we cube both sides and march onwards

(X2)3=(2X13)3\displaystyle(X-2)^{3}=\left(2X^{\frac{1}{3}}\right)^{3} (4.140)
X3+3(X2)(2)+(3)(X)((2)2)+(2)3=8X\displaystyle\implies X^{3}+3\left(X^{2}\right)(-2)+(3)(X)\left((-2)^{2}\right% )+(-2)^{3}=8X (4.141)
X36X2+12X8=8X\displaystyle\implies X^{3}-6X^{2}+12X-8=8X (4.142)
X36X2+4X9=0\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 ba-\frac{b}{a}, the value of S3S_{3} is

61=6-\frac{-6}{1}=6 (4.144)

4.11.2 Some further examples

Example 4.11.3

The equation x3+bx3+cx2+dx+ex^{3}+bx^{3}+cx^{2}+dx+e has roots pp, 2p2p, 3p3p and 4p4p. Show that 125e=310b4125e=\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 ee and bb is one way to think about this. A logical next question is whether we can find a simple way to relate ee and bb 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 ee in terms of bb by first expressing ee in terms of some third quantity, and then expressing this quantity in terms of bb. Considering the question gives us the answer; we can express ee in terms of pp, and bb in terms of pp (which also means we can express pp in terms of bb).

Using Vieta’s formulae we can express pp in terms of ee,

(p)(2p)(3p)(4p)=24p4=e(p)(2p)(3p)(4p)=24p^{4}=e (4.145)

We can also express pp in terms of bb,

p+2p+3p+4p=10p=bp+2p+3p+4p=10p=-b (4.146)

Therefore, p=b10p=-\frac{b}{10}. We can then just substitute this into Equation 4.145,

e\displaystyle e =24p4\displaystyle=24p^{4} (4.147)
=24(b10)4\displaystyle=24\left(-\frac{b}{10}\right)^{4} (4.148)
=2410,000b4\displaystyle=\frac{24}{10,000}b^{4} (4.149)
=31250b4\displaystyle=\frac{3}{1250}b^{4} (4.150)

Therefore, after multiplying by 125125, the end result is that

125e=310b4125e=\frac{3}{10}b^{4} (4.151)