# 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.

• 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 $$\alpha ^2 + \beta ^2$$.

\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
• Example 4.11.2 35: the cubic $$C$$, with roots $$\alpha$$, $$\beta$$, and $$\gamma$$ is given by

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

The integer function $$S_n$$ is defined as

\begin{equation} S_n = (2\alpha + 1)^n + (2\beta + 1)^n + (2\gamma + 1)^n \end{equation}

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

\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}