Roots Of A Polynomial Function

A quadratic equation is given by:

$x^2 - (\text{sum of factors})x + \text{product of factors} = 0$

Given $\alpha$ and $\beta$ as the roots for $4x^2 + 8x + 1 = 0$, form an equation whose roots are $\alpha^2\beta$ and $\beta^2\alpha$.

Solution

Sum of roots: $\alpha^2\beta + \beta^2\alpha = \alpha\beta(\alpha + \beta)$

Product of roots: $(\alpha^2\beta)(\beta^2\alpha) = \alpha^3\beta^3 = (\alpha\beta)^3$

From the equation $4x^2 + 8x + 1 = 0$:

$$\alpha + \beta = -2, \quad \alpha\beta = \frac{1}{4}$$ $$\alpha^2\beta + \beta^2\alpha = \alpha\beta(\alpha + \beta) = \frac{1}{4}(-2) = -\frac{1}{2}$$ $$(\alpha^2\beta)(\beta^2\alpha) = (\alpha\beta)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$$

The required equation is:

$$x^2 - \left(-\frac{1}{2}\right)x + \frac{1}{64} = 0$$ $$\Rightarrow 64x^2 + 32x + 1 = 0$$

Example

The equation $3x^2 - 5x + 1 = 0$ has roots $\alpha$ and $\beta$.

a) Find the value of $\alpha^2 - \alpha\beta + \beta^2$

We use the identity:

$$\alpha^2 - \alpha\beta + \beta^2 = (\alpha + \beta)^2 - 3\alpha\beta$$

From the equation $3x^2 - 5x + 1 = 0$, we get:

$$\alpha + \beta = \frac{5}{3}, \quad \alpha\beta = \frac{1}{3}$$

So:

$$\alpha^2 - \alpha\beta + \beta^2 = \left(\frac{5}{3}\right)^2 - 3\left(\frac{1}{3}\right) = \frac{25}{9} - 1 = \frac{16}{9}$$

b) Find the value of $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$

We use the identity:

$$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}$$

And:

$$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$ $$= \left(\frac{5}{3}\right)^3 - 3\left(\frac{1}{3}\right)\left(\frac{5}{3}\right) = \frac{125}{27} - \frac{15}{9} = \frac{80}{27}$$

Now:

$$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{80}{27} \div \frac{1}{3} = \frac{80}{27} \times 3 = \frac{80}{9}$$

If $\alpha$, $\beta$, $\gamma$ are roots of a cubic equation, then:

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

Expanding:

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

Given a cubic equation $ax^3 + bx^2 + cx + d = 0$, dividing by $a$:

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

Equating coefficients:

1. $\alpha + \beta + \gamma = -\frac{b}{a}$; sum of roots
2. $\alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a}$; sum of products of roots taken two at a time
3. $\alpha\beta\gamma = -\frac{d}{a}$; product of roots

Example

The equation $3x^3 + 6x^2 - 4x + 7 = 0$ has roots $\alpha$, $\beta$, $\gamma$. Find the equation with roots $\frac{1}{\alpha}$, $\frac{1}{\beta}$, $\frac{1}{\gamma}$.

Solution

The required equation is:

$$x^3 - \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)x^2 + \left(\frac{1}{\alpha\beta} + \frac{1}{\alpha\gamma} + \frac{1}{\beta\gamma}\right)x - \frac{1}{\alpha\beta\gamma} = 0$$ $$x^3 - \left(\frac{\alpha\beta + \alpha\gamma + \beta\gamma}{\alpha\beta\gamma}\right)x^2 + \left(\frac{\alpha + \beta + \gamma}{\alpha\beta\gamma}\right)x - \frac{1}{\alpha\beta\gamma} = 0$$

From the equation $3x^3 + 6x^2 - 4x + 7 = 0$, divide through by 3:

$$x^3 + 2x^2 - \frac{4}{3}x + \frac{7}{3} = 0$$

Comparing with $x^3 + px^2 + qx + r = 0$, we get:

$$p = 2, \quad q = -\frac{4}{3}, \quad r = \frac{7}{3}$$

Thus:

• Sum of roots: $\alpha + \beta + \gamma = -p = -2$
• Sum of products of roots taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = q = -\frac{4}{3}$
• Product of roots: $\alpha\beta\gamma = -r = -\frac{7}{3}$

Substituting into the new equation:

$$x^3 - \left(\frac{-\frac{4}{3}}{-\frac{7}{3}}\right)x^2 + \left(\frac{-2}{-\frac{7}{3}}\right)x - \frac{1}{-\frac{7}{3}} = 0$$ $$x^3 - \frac{4}{7}x^2 + \frac{6}{7}x + \frac{3}{7} = 0$$ $$\Rightarrow 7x^3 - 4x^2 + 6x + 3 = 0$$

Example

If the roots of the equation $4x^3 + 7x^2 - 5x - 1 = 0$ are $\alpha$, $\beta$ and $\gamma$, find the equation whose roots are:

a) $\alpha + 1, \beta + 1, \gamma + 1$

b) $\alpha^2, \beta^2, \gamma^2$

Solution

$$4x^3 + 7x^2 - 5x - 1 = 0$$

Dividing by 4:

$$x^3 + \frac{7}{4}x^2 - \frac{5}{4}x - \frac{1}{4} = 0$$

From this:

(i) $\alpha + \beta + \gamma = -\frac{7}{4}$

(ii) $\alpha\beta + \alpha\gamma + \beta\gamma = -\frac{5}{4}$

(iii) $\alpha\beta\gamma = \frac{1}{4}$

(a) Roots: $\alpha + 1, \beta + 1, \gamma + 1$

Sum of new roots:

$$(\alpha + 1) + (\beta + 1) + (\gamma + 1) = (\alpha + \beta + \gamma) + 3 = -\frac{7}{4} + 3 = \frac{5}{4}$$

Sum of products of new roots taken two at a time:

$$(\alpha + 1)(\beta + 1) + (\alpha + 1)(\gamma + 1) + (\beta + 1)(\gamma + 1)$$ $$= \alpha\beta + \alpha + \beta + 1 + \alpha\gamma + \alpha + \gamma + 1 + \beta\gamma + \beta + \gamma + 1$$ $$= (\alpha\beta + \alpha\gamma + \beta\gamma) + 2(\alpha + \beta + \gamma) + 3$$ $$= -\frac{5}{4} + 2\left(-\frac{7}{4}\right) + 3 = -\frac{5}{4} - \frac{7}{2} + 3 = -\frac{7}{4}$$

Product of new roots:

$$(\alpha + 1)(\beta + 1)(\gamma + 1)$$ $$= \alpha\beta\gamma + (\alpha\beta + \alpha\gamma + \beta\gamma) + (\alpha + \beta + \gamma) + 1$$ $$= \frac{1}{4} + \left(-\frac{5}{4}\right) + \left(-\frac{7}{4}\right) + 1 = -\frac{7}{4}$$

The required equation is:

$$x^3 - \left(\frac{5}{4}\right)x^2 - \frac{7}{4}x - \frac{7}{4} = 0$$ $$\Rightarrow 4x^3 - 5x^2 - 7x - 7 = 0$$

(b) Roots: $\alpha^2, \beta^2, \gamma^2$

For this, we need to find:

• Sum: $\alpha^2 + \beta^2 + \gamma^2$
• Sum of products: $\alpha^2\beta^2 + \alpha^2\gamma^2 + \beta^2\gamma^2$
• Product: $\alpha^2\beta^2\gamma^2$

We know:

$$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma)$$ $$= \left(-\frac{7}{4}\right)^2 - 2\left(-\frac{5}{4}\right) = \frac{49}{16} + \frac{10}{4} = \frac{49}{16} + \frac{40}{16} = \frac{89}{16}$$ $$\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$$

For $\alpha^2\beta^2 + \alpha^2\gamma^2 + \beta^2\gamma^2$:

$$\alpha^2\beta^2 + \alpha^2\gamma^2 + \beta^2\gamma^2 = (\alpha\beta + \alpha\gamma + \beta\gamma)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)$$ $$= \left(-\frac{5}{4}\right)^2 - 2\left(\frac{1}{4}\right)\left(-\frac{7}{4}\right) = \frac{25}{16} + \frac{14}{16} = \frac{39}{16}$$

The required equation is:

$$x^3 - \frac{89}{16}x^2 + \frac{39}{16}x - \frac{1}{16} = 0$$ $$\Rightarrow 16x^3 - 89x^2 + 39x - 1 = 0$$