Roots of Polynomials

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 \(S_{2} + 6S_{1} + 16 = 0\).

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

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:

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

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 \(\frac{1}{\alpha} + \frac{1}{\beta}\).

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 \(S_{-1} = \frac{5}{8}\).

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,

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

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

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

\(6x^{2} - x - 12 = 0\) 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

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

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 \(4y^{3} - 6y^{2} - 2y + 1 = 0\).

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 \(x\).

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:

\[\begin{equation}\tag{1}2x^{3} + 1 = 5x \end{equation}\]

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.

\[\begin{equation}\tag{2}8y^{3} + 6y^{2} - 119y + 1 = 0\end{equation}\]

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.

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

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

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).

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

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 \(S_{4}\):

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

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

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:

\[ y^{3} + 15y^{2} - 36y - 23 = 0 \]