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Roots of Polynomials

Tree roots, square roots, root beer. There are many things in life that deal with roots of some kind, but here we'll be talking about roots of polynomials.

Polynomials have many uses, from modelling the curves of a rollercoaster to predicting growth patterns in economics. The roots of a polynomial are solutions where the value of the polynomial is equal to zero, and in this article we'll be covering different methods of calculating these roots using summations and recurrence relations.

Equations used in roots of polynomials

This section covers the general equations for each of the three types of polynomials discussed in this article.

The quadratic equation takes the general form $$ax^{2} + bx + c = 0$$.

It is a polynomial where the highest power of $$x$$ is 2.

It is often further simplified to

$$x^{2} + \frac{b}{a}x + \frac{c}{a} = 0$$,

known as the monic form of the polynomial, because the leading coefficient is equal to 1.

This equation can also be written as

$$(x - x_{1})(x - x_{2}) = 0$$,

where $$x_{1}$$ and $$x_{2}$$ are the two roots of the equation.

Cubic equations

A cubic equation takes the general form $$ax^{3} + bx^{2} + cx + d = 0$$.

It is a polynomial in which the highest power of $$x$$ is 3.

As with the quadratic equation, is often simplified to

$$x^{3} + \frac{b}{a}x^{2} + \frac{c}{a}x + \frac{d}{a} = 0$$.

It is also written as

$$(x - x_{1})(x - x_{2})(x - x_{3}) = 0$$,

where $$x_{1}, \; x_{2} \text{ and } x_{3}$$ represent the roots of the equation.

Quartic equations

Quartic equations take the general form $$ax^{4} + bx^{3} + cx^{2} + dx + e = 0$$.

They are polynomials in which the highest power of $$x$$ is 4.

Likewise, they can be simplified to

$$x^{4} + \frac{b}{a}x^{3} + \frac{c}{a}x^{2} + \frac{d}{a}x + \frac{e}{a} = 0$$.

They are also written as

$$(x - x_{1})(x - x_{2})(x - x_{3})(x - x_{4}) = 0$$,

where $$x_{1}, \; x_{2}, \; x_{3} \text{ and } x_{4}$$ represent the roots of equation.

Roots of polynomials using alpha, beta, gamma and delta

Quadratic equations using alpha and beta

Take the quadratic equation, $$ax^{2} + bx + c = 0$$. Divide the equation by the coefficient of $$x^{2}, a$$ to rewrite the equation as $$x^{2} + \frac{b}{a}x + \frac{c}{a} = 0$$.

Now let us take $$\alpha$$ and $$\beta$$ to be the roots of our equation. We can thus also write the equation as

$$(x - \alpha)(x - \beta) = 0$$.

If we multiply this out, get

$$x^{2} - \alpha x - \beta x + \alpha\beta = 0$$.

Simplify the equation further to get

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

We now compare this with our other equation. What do you notice?

$x^{2} + \frac{b}{a}x + \frac{c}{a} = 0\qquad \text{and} \qquadx^{2} - (\alpha + \beta)x + \alpha\beta = 0$

From this comparison, we can see that:

• the sum of the roots, $$\alpha + \beta$$, is equal to $$-\frac{b}{a}$$;

• the product of the roots, $$\alpha\beta$$, is equal to $$\frac{c}{a}$$.

We can use these properties to calculate many other results, but a new notation needs to be defined first.

• $$\alpha + \beta$$ can be written as $$\Sigma{\alpha}$$;

• $$\alpha\beta$$ can be written as $$\Sigma{\alpha\beta}$$.

This form is called summation notation.

How would we determine $$\alpha^{2} + \beta^{2}$$, using what we have listed above?

First, let us square $$\alpha + \beta$$ to get the two squared terms. We know that $$(\alpha + \beta)^{2} = \alpha^{2} + \beta^{2} + 2\alpha\beta$$ so to get the final answer of $$\alpha^{2} + \beta^{2}$$, we simply subtract $$2\alpha\beta$$.

In summation notation, this calculation will look like this:

$\alpha^{2} + \beta^{2} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta},$

where $$\alpha^{2} + \beta^{2}$$ is denoted as $$\Sigma{\alpha^{2}}$$.

Note that $$\Sigma{\alpha^{2}}$$ and $$(\Sigma{\alpha})^{2}$$ mean different things. $$(\Sigma{\alpha})^{2} = (\alpha + \beta)^{2}$$ whereas $$\Sigma{\alpha^{2}} = \alpha^{2} + \beta^{2}$$.

We can determine $$\frac{1}{\alpha} + \frac{1}{\beta}$$, denoted as $$\Sigma{\frac{1}{\alpha}}$$, using similar methods.

Combine the fractions to get $$\frac{\alpha + \beta}{\alpha\beta}$$ which we can rewrite in summation notation to get $$\frac{\Sigma{\alpha}}{\Sigma{\alpha\beta}}$$.

Recurrence relations

Another way of working with roots of polynomials is to make use of recurrence relations. The relation used for each type of polynomial is summarized in the table below:

 Type of polynomial Quadratic equations $S_{n} = \alpha^{n} + \beta^{n}$ Cubic equations $S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n}$ Quartic equations $S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} + \delta^{n}$

Let's look at an example right away.

Consider the quadratic equation $$x^{2} + 6x + 8 = 0$$.

We know that $$\alpha + \beta = -6$$. We can also view this as $$S_{1}$$. Using $$S_{1}$$, we can calculate $$\alpha^{2} + \beta^{2}$$ with a different method.

Since $$\alpha$$ and $$\beta$$ are roots of the equation, we can say that $$\alpha^{2} + 6\alpha + 8 = 0 \; \text{and} \; \beta^{2} + 6\beta + 8 = 0$$. We add these two together to get $$(\alpha^{2} + \beta^{2}) + 6({\alpha + \beta}) + 16 = 0$$, which can also be written as $$S_{2} + 6S_{1} + 16 = 0$$.

Remember that $$S_{n} = \alpha^{n} + \beta^{n}$$, so $$S_{1} = \alpha + \beta$$ and $$S_{2} = \alpha^{2} + \beta^{2}$$.

By substituting $$S_{1}$$ back into the equation, we can calculate that $$\alpha^{2} + \beta^{2} = S_{2} = 20$$.

This can also be found by using the summation method mentioned above, $$\alpha^{2} + \beta^{2} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}$$.

Let's look at another example.

Assuming that $$3x^{2} + 4x + 12 = 0$$ has roots $$\alpha$$ and $$\beta$$, find the value of $$S_{3}$$.

Solution

First, substitute $$\alpha \; \text{and}\; \beta$$ into the equation to obtain the following:

\begin{align}3\alpha^{2} + 4\alpha + 12 & = 0 \\\newline 3\beta^{2} + 4\beta + 12 & = 0 \end{align}

Add the two equations together to get:$3(\alpha^{2} + \beta^{2}) + 4(\alpha + \beta) + 24 = 0$

Which is the same as:$3S_{2} + 4S_{1} + 24 = 0$

We know that $$S_{1} = \alpha + \beta = -\frac{b}{a}$$

$\therefore \qquad S_{1} = -\frac{4}{3}$

Substitute $$S_{1}$$ back into the equation to calculate that $$S_{2} = -\frac{56}{9}$$.

Using $$S_{n} = \alpha^{n} + \beta^{n}$$, we know that $$S_{3}$$ must thus equal $$\alpha^{3} +\beta^{3}$$. Returning to the original equation, $$3x^{2} + 4x + 12 = 0$$, we can see that there are no cubes present, so we multiply the whole equation by $$x$$ to get $$3x^{3} + 4x^{2} + 12x = 0$$.

As we did earlier, substitute $$\alpha \; \text{and} \; \beta$$ in the equation to get:

\begin{align}3\alpha^{3} + 4\alpha^{2} + 12\alpha & = 0 \\\newline 3\beta^{3} + 4\beta^{2} + 12\beta & = 0 \end{align}

Add the two equations together and apply the recurrence relation, $$S_{n} = \alpha^{n} + \beta^{n}$$ to result in:

$3S_{3} + 4S_{2} + 12S_{1} = 0$

Since we already know the values of $$S_{1} \; \text{and} \; S_{2}$$, we just substitute them back into the equation to calculate the value of $$S_{3}$$.

\begin{align}3S_{3} + 4\left(-\frac{56}{9} \right) + 12\left(-\frac{4}{3} \right) & = 0 \\\newline S_{3} & = \frac{368}{27}\end{align}

Alternative method

Alternatively, this can also be solved using summation notation.

We know that $$S_{3} = \alpha^{3} + \beta^{3}$$ and that $$\alpha^{3} + \beta^{3}\ = (\alpha + \beta)^{3} - 3\alpha\beta(\alpha + \beta)$$

In summation form this gives us:

$\Sigma{\alpha^{3}} = (\Sigma{\alpha})^{3} - 3\Sigma{\alpha\beta}\Sigma{\alpha}$

The sum of the roots, $$\Sigma{\alpha}$$, is $$-\frac{4}{3}$$ and the product of the roots, $$\Sigma{\alpha\beta}$$, is $$4$$.

Substitute these values into the summation form above to get:

\begin{align}\Sigma{\alpha^{3}} & = \left(-\frac{4}{3} \right)^{3} - 3(4)\left(-\frac{4}{3} \right) = \frac{368}{27}\end{align}

Recurrence relations can also be used to find values such $$S_{-1}$$. This can be done similarly to how we found $$S_{3}$$ in the previous example, but there is a multitude of methods one can use to calculate it.

The following example explains two of these different methods in detail:

You are given the quadratic equation $$3x^{2} + 5x - 8 = 0$$ with roots $$\alpha$$ and $$\beta$$. Calculate the value of $$S_{-1}$$.

Solution - Method 1

First, calculate the values of $$S_{1}$$.

\begin{align}S_{1} & = -\frac{b}{a} \\ & = -\frac{5}{3} \end{align}

Next, divide by $$x$$ to obtain the following:

$3x + 5 - \frac{8}{x} = 0$

Substitute the roots into the equation and apply your recurrence relation to get:

$3S_{1} + 10 - 8S_{-1} = 0$

You have already calculated the value of $$S_{1}$$, so substitute that value back into the equation and simplify to calculate $$S_{-1}$$.

\begin{align} 3 \left(-\frac{5}{3} \right) + 10 - 8S_{-1} & = 0 \\-8S_{-1} & = -5 \\ S_{-1} & = \frac{5}{8} \end{align}

Method 2

$$S_{-1}$$ is equivalent to $$\frac{1}{\alpha} + \frac{1}{\beta}$$.

Remember that $$S_{n} = \alpha^{n} + \beta^{n}$$ for quadratic equations. It therefore follows that $$S_{-1} = \alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta}$$.

It was established in the previous section that

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\Sigma{\alpha}}{\Sigma{\alpha}{\beta}}$

You know that $$\Sigma{\alpha} = -\frac{b}{a} \text{ and } \Sigma{\alpha\beta} = \frac{c}{a}$$. Substitute these into the equations to get the following:

$S_{-1} = -\frac{b}{c}$

From there, all you need to do is substitute the values for $$b$$ and $$c$$ in and simplify.

We thus get that $$S_{-1} = \frac{5}{8}$$.

Cubic equations using alpha, beta and gamma

Cubic equations are very similar to quadratic equations, except the roots will now be represented by $$\alpha,\;\beta \; \text{and} \; \gamma$$.

We start off with $$ax^{3} + bx^{2} + cx + d = 0$$. As with quadratic equations, we divide by $$a$$ to get $$x^{3} + \frac{b}{a}x^{2} + \frac{c}{a}x + \frac{d}{a} = 0$$.

We've established that we will be using $$\alpha, \; \beta \; \text{and} \; \gamma$$ to represent the roots of the cubic equation. We thus get

$$(x - \alpha)(x - \beta)(x - \gamma) = 0$$,

which, when expanded, gives us:

$x^{3} - (\alpha + \beta + \gamma)x^{2} + (\alpha\beta + \alpha\gamma + \beta\gamma)x -\alpha\beta\gamma = 0.$

When we compare this to $$x^{3} + \frac{b}{a}x^{2} + \frac{c}{a}x + \frac{d}{a} = 0$$, we can see that

• $$\alpha + \beta + \gamma = -\frac{b}{a}$$, which is denoted by $$\Sigma{\alpha}$$ or $$S_{1}$$;

• $$\alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}$$ which we denote with $$\Sigma{\alpha\beta}$$;

• $$\alpha\beta\gamma = -\frac{d}{a}$$, which we write as $$\Sigma{\alpha\beta\gamma}$$.

We know from quadratics that $$\Sigma{\alpha^{2}} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}$$. This also holds true for cubic equations where

$$(\Sigma{\alpha})^{2} = (\alpha + \beta + \gamma)^{2} = \alpha^{2} + \beta^{2} + \gamma^{2} + 2\alpha\beta + 2\alpha\gamma + 2\beta\gamma$$,

and $$\Sigma{\alpha^{2}}$$ therefore equals $$(\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}$$.

We can now use $$S_{n} = \alpha^{n} + \beta{n} + \gamma^{n}$$ to represent the roots of the cubic equations. For example, $$S_{3}$$ will be used to represent $$\alpha^{3} + \beta^{3} + \gamma^{3}$$, e.t.c.

All the results that have been derived from quadratic equations hold true for cubic equations, but the calculations get more complicated as we work with higher powers. It is easier to use recurrence relations rather than summation forms to find $$S_{n}$$ from powers of 3 and higher.

Refer to the following example that shows this:

Find $$S_{3}$$ for the equation $$6x^{3} - 15x^{2} - 17x + 6 = 0$$.

Solution - Using Summation Form

We know that $$S_{3} = (\Sigma{\alpha})^{3} - 3\Sigma{\alpha\beta}\Sigma{\alpha} + 3\Sigma{\alpha\beta\gamma}$$.

First, let us calculate the values of $$\Sigma{\alpha}, \; \Sigma{\alpha\beta} \; \text{and} \; \Sigma{\alpha\beta\gamma}$$:

\begin{align}\Sigma{\alpha} & = -\frac{b}{a} \\& = -\frac{-15}{6} \\& = \frac{5}{2}\end{align}

\begin{align}\Sigma{\alpha\beta} & = \frac{c}{a} \\& = -\frac{17}{6}\end{align}

\begin{align}\Sigma{\alpha\beta\gamma} & = -\frac{d}{a} \\& = -\frac{6}{6} \\& = -1\end{align}

Now that we have these values, we can substitute them back into the summation form to calculate $$S_{3}$$.

\begin{align}S_{3} & = (\Sigma{\alpha})^{3} - 3\Sigma{\alpha\beta}\Sigma{\alpha} + 3\Sigma{\alpha\beta\gamma} \\& = \left(\frac{5}{2} \right)^{3} - 3\left(-\frac{17}{6} \right) \left(\frac{5}{2} + 3(-1) \right) \\& = \frac{271}{8}\end{align}

$\therefore \qquad S_{3} = \frac{169}{8}$

As you can see, this method does work, but it's easy to make mistakes, and it gets messy quickly. There is a more efficient way of solving this problem using recurrence relations.

Alternative Solution - Using Recurrence Relations

$$\alpha, \; \beta \text{ and } \gamma$$ all satisfy our cubic equation, so we know that:

\begin{align}6\alpha^{3} - 15\alpha^{2} - 17\alpha + 6 & = 0 \\6\beta^{3} - 15\beta^{2} - 17\beta + 6 & = 0 \\6\gamma^{3} - 15\gamma^{2} - 17\gamma + 6 & = 0\end{align}

When we add these three equations, we get:

$6(\alpha^{3} + \beta^{3} + \gamma^{3}) - 15(\alpha^{2} + \beta^{2} + \gamma^{2}) - 17(\alpha + \beta + \gamma) + 18 = 0$

Which is the same as:

$6S_{3} - 15S_{2} - 17S_{1} + 18 = 0$

We know that:

$S_{1} = -\frac{b}{a} = \frac{5}{2}$

and,

\begin{align}S_{2} & = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta} \\& = \left(\frac{5}{2}\right)^{2} - 2\left(-\frac{17}{6}\right) \\& = \frac{143}{12} \\\end{align}

\begin{align}\therefore \qquad 6S_{3} - 15S_{2} - 17S_{1} + 18 & = 0 \qquad \text{Substitute } S_{1} \text{ and } S_{2} \text{ in} \\6S_{3} - 15 \left( \frac{143}{12}\right) - 17 \left(\frac{5}{2} \right) + 18 & = 0 \\S_{3} & = \frac{271}{8}\end{align}

Quartic equations using alpha, beta, gamma and delta

For quartic equations, we will be using $$\alpha, \; \beta, \; \gamma \text{ and } \delta$$ to represent the roots of the equation.

As with quadratic and cubic equations, the sum of the roots, $$\alpha + \beta + \gamma + \delta$$, is denoted by $$\Sigma{\alpha}$$ and $$S_{1}$$ and is equivalent to $$-\frac{b}{a}$$.

It is further given that, for quartic equations in the form $$ax^{4} + bx^{3} + cx^{2} + dx + e = 0$$.

When working with quartic equations, make sure to use the recurrence formula wherever possible, especially when working with the sum of the cubes, $$(\alpha^{3} + \beta^{3} + \gamma^{3} + \delta^{3})$$.

The best method for determining the sum of the cubes of a general quartic equation is to calculate $$S_{1}$$ first and then to determine $$S_{2}$$ and $$S_{-1}$$. Once we have determined those values, we can then use $$aS_{4} + bS_{3} + cS_{2} + dS_{1} + 4e = 0$$ and divide by $$x$$ to calculate $$S_{3}$$.

For quartic equations in the form $$ax^{4} + bx^{3} + cx^{2} + dx + e = 0$$, $\begin{gather*}\Sigma{\alpha} = \alpha + \beta + \gamma + \delta = -\frac{b}{a} \\\Sigma{\alpha\beta} = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = \frac{c}{a} \\\Sigma{\alpha\beta\gamma} = \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -\frac{d}{a} \end{gather*}$

and,

$\Sigma{\alpha\beta\gamma\delta} = \alpha\beta\gamma\delta = \frac{e}{a}$

These formulas come from expanding $$(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = 0$$ and comparing it to $$x^{4} + \frac{b}{a}x^{3} + \frac{c}{a}x^{2} + \frac{d}{a}x + e = 0$$.

\begin{align} (x + \alpha)(x + \beta)(x + \gamma)(x + \delta) & = 0 \\ x^{4} - (\alpha + \beta + \delta + \gamma)x^{3} + (\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma +\beta\delta + \gamma\delta)x^{2} - (\alpha\beta\gamma + \alpha\beta\delta + \alpha\delta\gamma\beta\gamma\delta)x + \alpha\beta\gamma\delta & = 0 \\x^{4} + \Sigma{\alpha}x^{3} + \Sigma{\alpha\beta}x^{2} + \Sigma{\alpha\beta\gamma}x +\Sigma{\alpha\beta\gamma\delta} & = 0 \end{align}

$$\therefore \qquad$$ when we compare the two equations, we can conclude the following:

$\begin{gather*} \Sigma{\alpha} = -\frac{b}{a} \\ \hspace{1cm} \\ \Sigma{\alpha\beta} = \frac{c}{a} \\ \hspace{1cm} \\ \Sigma{\alpha\beta\gamma} = -\frac{d}{a} \\ \hspace{1cm} \\ \Sigma{\alpha\beta\gamma\delta} = \frac{e}{a} \end{gather*}$The full workings of the expansion of $$(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = 0$$ are not shown here as it is not necessary for you to know them. It is only necessary for you to know how to apply the formulas that result from the expansion.

For recurrence relations involving quartic equations, use the following formulas:

$\begin{gather*} S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} + \delta^{n} \\ \hspace{1cm} \\S_{-1} = \Sigma{\frac{1}{\alpha}} = \frac{\Sigma{\alpha\beta\gamma}}{\Sigma{\alpha\beta\gamma\delta}} \\\end{gather*}$

As with quadratic and cubic equations, $$S_{2} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}$$.

You are given the quartic equation $$3x^{4} - x^{3} + 2x + 7 = 0$$ with roots $$\alpha, \; \beta, \; \gamma \text{ and } \delta$$. Calculate the values of $$S_{3}$$ and $$S_{4}$$.

Solution

First, calculate $$S_{1}$$:

$S_{1} = -\frac{b}{a} = \frac{1}{3}$

Next, calculate the value of $$S_{2}$$:

\begin{align}S_{2} & = \left( \Sigma{\alpha} \right)^{2} – 2\Sigma{\alpha\beta} \\& = \left ( \frac{1}{3} \right )^{2} – 2\left( \frac{0}{2} \right) \\& = \frac{1}{9}\end{align}

Divide the original equation by $$x$$ to get $$3x^{3} - x^{2} + 2 + \frac{7}{x} = 0$$ and use this to calculate the value of $$S_{-1}$$.

You know from quadratic equations that $$S_{-1} = \Sigma{\frac{1}{\alpha}}$$. It therefore

Substitutions

Suppose that you are given a quadratic equation with roots $$\alpha$$ and $$\beta$$. You are asked to find another quadratic that has roots $$3\alpha$$ and $$3\beta$$. How would you go about solving this?

The following example explains the two methods you could use in detail.

$$6x^{2} - x - 12 = 0$$ has roots $$\alpha$$ and $$\beta$$. Find the quadratic equation with roots $$3\alpha$$ and $$3\beta$$.

Solution - Method 1

Start by considering the quadratic $$(y - 3\alpha)(y - 3\beta)\ = 0$$. This equation has roots $$3\alpha$$ and $$3\beta$$, which is what you want.

Expand it to get $$y^{2} - 3(\alpha + \beta)y + 9\alpha\beta = 0$$.

Now compare this to the original equation. From the original, we know that $$\alpha + \beta = \frac{1}{6}$$ and that $$\alpha\beta = -2$$.

Input these values into the new equation to get the final result of $$y^{2} - \frac{1}{2}y - 18 = 0$$, which can also be multiplied by 2 to give $$2y^{2} - y - 36 = 0$$.

Method 2

The second method of determining the new equation works by relating the roots of the different equations.

We know that the roots of the new equation ($$y$$) are three times that of the original.

The relation is thus shown as:

$y =3x$

Rearrange the equation so that $$x$$ is the subject of the formula.

$x = \frac{y}{3}$

Next, substitute for $$x$$ in the original equation and simplify

\begin{align}6\left(\frac{y}{3} \right)^{2} - \frac{y}{3} - 12 & = 0 \\\frac{6}{9}y^{2} - \frac{y}{3} - 12 & = 0 \\2y^{2} - y - 36 & = 0\end{align}

A more complicated use of substitution involves reciprocal functions. The following example details how one would go about using reciprocal functions in substitution:

Consider the given cubic equation, $$x^{3} - 2x^{2} - 6x + 4 = 0$$ with roots $$\alpha, \; \beta \text{ and } \gamma$$. Find the cubic equation with roots $$\frac{1}{\alpha}, \; \frac{1}{\beta}, \; \frac{1}{\gamma}$$.

Solution

First, relate the roots to each other and make $$x$$ the subject of the formula.

\begin{align} y & = \frac{1}{x} \\x & = \frac{1}{y}\end{align}

Next, substitute for $$x$$ in the original equation and simplify.

\begin{align}\left(\frac{1}{y} \right)^{3} - 2\left(\frac{1}{y} \right)^{2} - 6\left(\frac{1}{y} \right) + 4 & = 0 \\\hspace{1cm} \\\frac{1}{y^{3}} - \frac{2}{y^{2}} - \frac{6}{y} + 4 & = 0 \\\hspace{1cm} \\ 4y^{3} - 6y^{2} - 2y + 1 & = 0 \qquad \text{Multiply by } y^{3}\end{align}

The cubic equation with roots $$\alpha, \; \beta \text{ and } \gamma$$ is thus $$4y^{3} - 6y^{2} - 2y + 1 = 0$$.

There are two methods one can use when substituting with powers of roots.

You are given the quadratic equation $$2x^{2} - 4x + 7 = 0$$ with roots $$\alpha \text{ and } \beta$$. Find the quadratic equation with roots $$\alpha^{2}$$ and $$\beta^{2}$$.

Solution - Method 1

We can state that $$y = x^{2}$$ and therefore $$x = \sqrt{y}$$.

Substituting this value into the equation results in the following:

$\begin{split}2((y^{\frac{1}{2}})^{2} - 4y^{\frac{1}{2}} + 7 = 0 \\2y - 4y^{\frac{1}{2}} + 7 = 0\end{split}$

We can then rearrange the equation and square both sides. This will result in the quadratic equation that is the solution to this question.

\begin{align}2y + 7 & = 4y^{\frac{1}{2}} \\4y^{2} + 28y + 49 & = 16y \\4y^{2} + 12y + 49 & = 0 \end{align}

Method 2

In this method, we reverse the order we followed to solve the problem in the first method.

First, rearrange the equation. We want to perform an operation so that we get even powers for every term of $$x$$. This is done by squaring both sides.

\begin{align}-4x & = -2x^{2} - 7 \\16x^{2} & = 4x^{4} + 28x^{2} + 49 \qquad \text{Square both sides} \end{align}

Now substitute $$y = x^{2}$$ in, like we did in the first method.

\begin{align}16y & = 4y^{2} + 28y + 49 \\4y^{2} + 12y + 49 & = 0 \end{align}

As you can see, this method results in the same answer as the first method.

For both methods, you must ensure that you are using the correct the powers of $$x$$.

Another use of substitution methods is determining the values of recurrence relations, such as $$S_{6}$$.

Suppose you were asked to determine the value of $$S_{6}$$ for the cubic equation $$x^{3} + 3x^{2} - 1 = 0$$. Not only is the process for doing this long, it is also easy to make mistakes.

Now, imagine that there was another cubic equation for which $$y= x^{3}$$. For the $$x$$ cubic, $$S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n}$$.

Since the roots of the $$y$$ cubic are equal to the roots of the $$x$$ cubic cubed, the $$y$$ cubic would have $$S_{n} = \alpha^{3n} + \beta^{3n} + \gamma^{3n}$$. The power for each root has been tripled, so we only need to find $$S_{2}$$ of the $$y$$ cubic to determine the value of $$S_{6}$$ for the $$x$$ cubic.

The following example demonstrates how one would do this:

Consider the cubic equation $$2x^{3} - 5x + 1 = 0$$ with roots $$\alpha, \; \beta \text{ and } \gamma$$. Determine the value of $$S_{6}$$ using the substitution of $$y = x^{3}$$.

Solution

Label the cubic and rearrange it as follows:

$$$\tag{1}2x^{3} + 1 = 5x$$$

Cube both sides of the cubic and then simplify to get:

\begin{align}8x^{9} + 6x^{6} + 6x^{3} + 1 & = 125x^{3} \\8x^{9} + 6x^{6} - 119x^{3} + 1 & = 0\end{align}

Next, substitute $$y$$ in for $$x^{3}$$ and label this new equation.

$$$\tag{2}8y^{3} + 6y^{2} - 119y + 1 = 0$$$

Determine $$S_{1}$$ for equation 2.

\begin{align}S_{1} & = -\frac{b}{a} \\\hspace{1cm} \\ & = -\frac{6}{8} \\\hspace{1cm} \\ & = -\frac{3}{4}\end{align}

Finally, determine $$S_{2}$$ using the $$S_{1}$$ value you just calculated, and state the value of $$S_{6}$$ for Equation 1.

\begin{align} S_{2} & = (S_{1})^{2} - \frac{c}{a} \\\hspace{1cm} \\ & = \left(-\frac{3}{4} \right)^{2} - \left(-\frac{119}{8} \right) \\\hspace{1cm} \\& = \frac{247}{16} \\ \end{align}

$\begin{split}\therefore \qquad S_{2} \text{ for Equation } 2 = \frac{247}{16} \\\hspace{1cm} \\ \therefore \qquad S_{6} \text{ for Equation 1 } = \frac{247}{16}\end{split}$

Roots of polynomial formulas

This section summarises the formulas derived in previous sections.

For quadratic equations in the form $$ax^{2} + bx + c = 0$$:

$\begin{gather*} \Sigma{\alpha} = \alpha + \beta = -\frac{b}{a} \\ \hspace{1cm} \\\Sigma{\alpha^{2}} = \alpha^{2} + \beta^{2} \\ \hspace{1cm} \\\Sigma{\alpha\beta} = \alpha\beta = \frac{c}{a} \\ \hspace{1cm} \\S_{n} = \alpha^{n} + \beta^{n} \\\hspace{1cm} \\ S_{-1} = \Sigma{\frac{1}{\alpha}} = \frac{\Sigma{\alpha}}{\Sigma{\alpha\beta}} \\\hspace{1cm} \\ \Sigma{\frac{1}{\alpha}} = \frac{1}{\alpha} + \frac{1}{\beta} \end{gather*}$

Formulas for cubic equations

For cubic equations in the form $$ax^{3} + bx^{2} + cx + d = 0$$:

$\begin{gather*} \Sigma{\alpha} = \alpha + \beta + \gamma = -\frac{b}{a} \\ \hspace{1cm} \\\Sigma{\alpha\beta} = \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a} \\ \hspace{1cm} \\\Sigma{\alpha\beta\gamma} = \alpha\beta\gamma = -\frac{d}{a} \\ \hspace{1cm} \\S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} \\\hspace{1cm} \\ S_{-1} = \Sigma{\frac{1}{\alpha}} = \frac{\Sigma{\alpha\beta}}{\Sigma{\alpha\beta\gamma}} \\ \hspace{1cm} \\\Sigma{\frac{1}{\alpha}} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\end{gather*}$

Formulas for quartic equations

For quartic equations in the form $$ax^{4} + bx^{3} + cx^{2} + dx + e = 0$$:

$\begin{gather*}\Sigma{\alpha} = \alpha + \beta + \gamma + \delta = -\frac{b}{a} \\\hspace{1cm} \\\Sigma{\alpha\beta} = \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta= \frac{c}{a} \\\hspace{1cm} \\\Sigma{\alpha\beta\gamma} = \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -\frac{d}{a} \\ \hspace{1cm} \\\Sigma{\alpha\beta\gamma\delta} = \alpha\beta\gamma\delta = \frac{e}{a} \\ \hspace{1cm} \\S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} + \delta^{n} \\ \hspace{1cm} \\ S_{-1} = \Sigma{\frac{1}{\alpha}} = \frac{\Sigma{\alpha\beta\gamma}}{\Sigma{\alpha\beta\gamma\delta}} \\ \hspace{1cm} \\\Sigma{\frac{1}{\alpha}} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} \\\end{gather*}$

Recurrence notation

The formulas you need to know for recurrence notation are as follows:

$\begin{gather*}S_{1} = \Sigma{\alpha} \\ \hspace{1cm} \\ S_{2} = \Sigma{\alpha^{2}} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta} \\ \hspace{1cm} \\ S_{-1} = \Sigma{\frac{1}{\alpha}}\end{gather*}$

Roots of polynomial calculation

In this section, we will work through a multitude of practice questions involving roots of polynomials.

You are given the cubic equation $$3x^{3} + mx^{2} + nx + p = 0$$ with $$S_{1} = -5, \; S_{2} = \frac{79}{3} \text{ and } S_{-1} = \frac{1}{2}$$. Find the values of the constants $$m, \; n \text{ and } p$$.

Solution

Calculate the value of $$m$$. You already know that $$S_{1} = \alpha + \beta = -\frac{b}{a}$$, so make use of this relationship to find $$m$$.

\begin{align}S_{1} & = -\frac{b}{a} \qquad \quad \text{ Substitute in the values you know.} \\ -5 & = -\frac{m}{3} \\ \therefore \qquad m & = 15\end{align}

Similarly, when calculating $$n$$, you know that $$S_{2} = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta}$$ and also that $$\Sigma{\alpha} = S_{1} \text { and } \Sigma{\alpha\beta} = \frac{c}{a}$$. Substitute the values you know in and simplify to solve for $$n$$.

\begin{align} S_{2} & = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha}{\beta} \\ \frac{79}{3} & = (-5)^{2} - 2\frac{n}{3} \\ \frac{4}{3} & = -2\frac{n}{3} \\ n & = -2 \end{align}

Finally, to calculate $$p$$, you use $$S_{-1} = \frac{\Sigma{\alpha\beta}}{\Sigma{\alpha\beta\gamma}}$$ (there are other ways of doing this, but this method is the easiest).

\begin{align}S_{-1} & = \frac{\Sigma{\alpha}{\beta}}{\Sigma{\alpha\beta\gamma}} \\ \frac{1}{2} & = -\frac{c}{d} \\ \frac{1}{2} & = -\frac{n}{p} \\ \frac{1}{2} & = -\frac{-2}{p} \\ p & = 4\end{align}

Let's see another example.

You are given the quartic equation $$x^{4} + 3x^{3} - 8x^{2} + 5 = 0$$ with roots $$\alpha, \beta, \; \gamma \text{ and } \delta$$. Show that $$S_{4} = -22S_{3} + 1$$.

Solution

First, you must calculate the values of $$S_{1}, \; S_{2}$$ and $$S_{-1}$$: \begin{align}S_{1} & = \alpha + \beta + \gamma + \delta \\ & = -\frac{b}{a} \\ & = -3 \end{align}

\begin{align} S_{2} & = (\Sigma{\alpha})^{2} - 2\Sigma{\alpha\beta} \\ & = (-3)^{2} - 2(8) \\ & = -7 \end{align}

\begin{align} S_{-1} & = \frac{\Sigma{\alpha\beta\gamma}}{\Sigma{\alpha\beta\gamma\delta}} \\ & = -\frac{d}{e} \\ & = -\frac{0}{5} \\ & = 0\end{align}

Next, you need to calculate the value of $$S_{3}$$. This is done by dividing the original equation by $$x$$ so that it looks like this $$x^{3} + 3x^{2} - 8x + \frac{5}{x} = 0$$ and then substituting each of the four roots, $$\alpha, \; \beta, \; \gamma \text{ and } \delta$$ into the equation and adding the four resulting equations together to get the following:

$(\alpha^{3} + \beta^{3} + \gamma^{3} + \delta^{3}) + 3(\alpha^{2} + \beta^{2} + \gamma^{2} + \delta^{2} ) - 8(\alpha + \beta + \gamma + \delta) + 5(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}) = 0$

Apply the recurrence relation $$S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n} + \delta^{n}$$ and your equation should now look like this:

$S_{3} + 3S_{2} - 8S_{1} + 5S_{-1} = 0$

Substitute $$S_{1}, \; S_{2} \text{ and } \S_{-1}$$ into the equation and solve for $$S_{3}$$:

\begin{align} S_{3} + 3(-7) - 8(-3) + 5(0) & = 0 \\ S_{3} & = -3 \\ \end{align}

To solve for $$S_{4}$$, we repeat a similar process to solving for $$S_{3}$$, except for dividing by $$x$$. Substitute each of the roots, $$\alpha, \; \beta, \; \gamma \text{ and } \delta$$ into the original equation and add the four resulting equations to get the following:

$(\alpha^{4} + \beta^{4} + \gamma^{4} + \delta^{4}) + 3(\alpha^{3} + \beta^{3} + \gamma^{3} + \delta^{3}) - 8(\alpha^{2} + \beta^{2} + \gamma^{2} + \delta^{2}) + 20 = 0$

Apply the recurrence relation and you get:

$S_{4} + 3S_{3} - 8S_{2} + 20 = 0$Finally, substitute the values for $$S_{3}$$ and $$S_{2}$$ in and solve for $$S_{4}$$:

\begin{align}S_{4} + 3(-3) - 8(-7) + 20 & = 0 \\S_{4} & = 67\end{align}

$$\therefore \qquad -22S_{3} + 1 = -22(-3) + 1 = 67 = S_{4}$$

This final example covers a more complicated problem involving substitution.

The cubic equation $$x^{3} + 4x^{2} - 7x - 1 = 0$$ has roots $$\alpha, \beta$$ and $$\gamma$$. Find the cubic equations with roots $$\frac{2\alpha - 3}{\alpha}, \; \frac{2\beta - 3}{\beta} \text{ and } \frac{2\gamma - 3}{\gamma}$$.

Solution

We know that the roots of this new cubic equation are:

$y = \frac{2x - 3}{x}$

Make $$x$$ the subject of the formula:

\begin{align} xy & = 2x - 3 \\ xy - 2x & = -3 \\ x(y - 2) & = -3 \\ x & = -\frac{3}{y-2} \end{align}

Substitute this value for $$x$$ back into the original equation and simplify to obtain the new cubic.

\begin{align} \left( -\frac{3}{y - 2} \right)^{3} + 4\left(-\frac{3}{y-2}\right)^{2} - 7\left(-\frac{3}{y-2}\right) + 1 & = 0 \\ \hspace{1cm} \\ -\frac{27}{(y-2)^{3}} + \frac{36}{(y-2)^{2}} + \frac{21}{y-2} + 1 & = 0 \\ \hspace{1cm} \\ -27 + 36(y-2) + 21(y-2)^{2} + (y-2)^{3} & = 0 \\ \hspace{1cm} \\-27 + 36y - 72 + 21y^{2} - 84y + 84 + y^{3} - 6y^{2} + 12y - 8 & = 0 \\ \hspace{1cm} \\y^{3} + 15y^{2} - 36y - 23 & = 0\end{align}

The cubic equation with roots $$\frac{2\alpha - 3}{\alpha}, \; \frac{2\beta - 3}{\beta} \text{ and } \frac{2\gamma - 3}{\gamma}$$ is thus:

$y^{3} + 15y^{2} - 36y - 23 = 0$

Roots of Polynomials - Key takeaways

• We use $$\alpha$$ and $$\beta$$ to represent the roots of a quadratic equation, $$\alpha, \; \beta \text{ and } \gamma$$ to represent the roots of a cubic equation, and $$\alpha, \; \beta, \; \gamma \text{ and } \delta$$ to represent the roots of a quartic equation.
• The sum of the roots of a polynomial equation can be calculated using $$\Sigma{\alpha} = -\frac{b}{a}$$.
• It is better to use recurrence relations when working with powers of 3 and higher.
• Substitution can be used to find a polynomial by relating the roots of the polynomial you want to find with those of a known polynomial, followed by substituting that relation into the known polynomial and simplifying.
• The sum of the inverse roots of a polynomial, $$\Sigma{1}{\alpha}$$, is always equal to the negative of the coefficient of the linear term divided by the constant term.

Flashcards in Roots of Polynomials 4

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How to find complex roots of a polynomial?

Considering a, b, c, d, e the coefficients, you analyse the sign of the polynomial's discriminant:

• for quadratic equations, if Δ = b2 – 4ac < 0, then the roots are complex;
• for cubic equations, if Δ = b2c2 – 4ac3 – 4b3d – 27a2d2 + 18abcd < 0, then there's one real root and two complex conjugate roots;
• for quartic equations, if Δ = 256a3e3 – 192a2bde2 – 128a2c2e2 + 144a2cd2e – 27a2d4 + 144ab2ce2 – 6ab2d2e – 80abc2de + 18abcd3 + 16ac4e – 4ac3d2 – 27b4e2 + 18b3cde – 4b3d3 – 4b2c3e + b2c2d2 > 0, there are two real and two complex roots, and if Δ < 0, all roots are real or complex.

How to find real roots of polynomial equations?

Considering a, b, c, d, e the coefficients, you analyse the sign of the polynomial's discriminant:

• for quadratic equations, if Δ = b2 – 4ac > 0 or Δ = 0, then the roots are real;
• for cubic equations, if Δ = b2c2 – 4ac3 – 4b3d – 27a2d2 + 18abcd > 0 or Δ= 0, then all the roots are real;
• for quartic equations, if Δ = 256a3e3 – 192a2bde2 – 128a2c2e2 + 144a2cd2e – 27a2d4 + 144ab2ce2 – 6ab2d2e – 80abc2de + 18abcd3 + 16ac4e – 4ac3d2 – 27b4e2 + 18b3cde – 4b3d3 – 4b2c3e + b2c2d2 > 0, there are two real and two complex roots, and if Δ < 0, all roots are real or complex.

How to find integer roots of polynomials?

To find integer roots of polynomials, you'll need to follow the same steps for finding the real roots of (2nd, 3rd and 4th degree) polynomials and then analyse which of the real roots are integer.

How to find roots of a cubic polynomial?

To find the roots of a cubic polynomial, you can:

• use factorisation, reducing it to three linear equations or one linear and one quadratic equation;
• use the division method;
• use the graphical method.

How to find rational roots of polynomial equations?

To find rational roots of polynomials, you'll need to follow the same steps for finding the real roots of (2nd, 3rd and 4th degree) polynomials and then analyse which of the real roots are rational.

Test your knowledge with multiple choice flashcards

The products of the roots of any polynomial is always equal to the constant term.

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