A calculus of the absurd

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.

\begin{equation} x^2-(\alpha + \beta )x + \alpha \beta \label {expanded alpha beta} \end{equation}

We can then compare the coefficients of Equation 4.120 with those of \(ax^2 + bx + c\).

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

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

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

\begin{equation} \alpha \beta = \frac {c}{a} \end{equation}

Something similar is also the case for higher-degree polynomials ³¹³¹ TODO: add a proof for this.

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

\begin{align*} \alpha ^2\beta ^2 & = (\alpha \beta )^2 \\ & = 12^2 \\ & = 144 \end{align*}

\begin{align*} \alpha ^2 + \beta ^2 & = (\alpha + \beta )^2 - 2(\alpha \beta ) \\ & = 8^2 - 2(12) \\ & = 40 \end{align*}

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. ³²³² 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 \(P(x)\).³³³³ Because that way \(\sqrt {x} \mapsto x\), which also means that \(x \mapsto x^2\)

Therefore, our new quadratic should be

\begin{align*} P(\sqrt {x}) & = (\sqrt {x})^2 - 8\sqrt {x} + 12 \\ & = x - 8\sqrt {x} + 12 \end{align*}

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

³⁴³⁴ 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.

\begin{align*} & P(\sqrt {x}) = 0 \\ & x - 8\sqrt {x} + 12 = 0 \\ & x + 12 = 8\sqrt {x} \\ & (x+12)^2 = 64x \\ & x^2 + 24x + 144 = 64x \\ & x^2 - 40x + 144 = 0 \end{align*}

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

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

\begin{equation} X = (2x+1)^3 \end{equation}

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

\begin{align} & (2x + 1)^3 = X \\ & 2x + 1 = X^{\frac {1}{3}} \\ & 2x = X^{\frac {1}{3}} - 1 \\ & x = \frac {X^{\frac {1}{3}} - 1} {2} \end{align}

Using this we can then substitute into the original polynomial

\begin{align} & 8x^3 + 12x^2 + 2x - 3 = 0 \\ & \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 \end{align}

This can then be simplified, a lot.

\begin{align} & 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 \\ & \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 \\ & \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 \\ & \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 \\ & \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 \\ & \implies X - 2X^{\frac {1}{3}} - 2 = 0 \end{align}

Here, we pull the familiar trick³⁶³⁶ Where familiar means "did it once in the previous example". and rearrange

\begin{equation} X - 2 = 2X^{\frac {1}{3}} \end{equation}

From here, we cube both sides and march onwards

\begin{align} & (X-2)^3 = \left (2X^{\frac {1}{3}}\right )^3 \\ & \implies X^3 + 3\left (X^2\right )(-2) + (3)(X)\left ((-2)^2\right ) + (-2)^3 = 8X \\ & \implies X^3 - 6X^2 + 12X - 8 = 8X \\ & \implies X^3 - 6X^2 + 4X -9 = 0 \end{align}

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

\begin{equation} -\frac {-6}{1} = 6 \end{equation}