Finding zeroes and constructing polynomials

In the previous two topics we used Vieta's identities to relate zeroes and coefficients. We now turn the wheel both ways:

1. Forward problem. Given a polynomial, find its zeroes.
2. Inverse problem. Given the zeroes (or some symmetric data about them), construct the polynomial.

Strategies for finding zeroes

Quadratic $ax^2 + bx + c$. Three reliable methods.

• Factorisation. If you can split the middle term , find two numbers whose product is $ac$ and whose sum is $b$ , you have the zeroes immediately.
• Completing the square. Rewrite as $a(x + b/2a)^2 + (c - b^2/4a) = 0$.
• Quadratic formula. $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which we cover thoroughly in Chapter 4. For now, treat it as a backup tool.

Cubic $ax^3 + bx^2 + cx + d$. The plan is to find one zero by inspection, divide it out, and then solve the leftover quadratic.

• Try small integer values: $x = \pm 1, \pm 2, \pm 3, \ldots$. If $d/a$ is an integer, the rational root candidates are factors of $d/a$.
• Once a zero $\alpha$ is found, perform polynomial division: $p(x) = (x - \alpha) \cdot q(x)$, where $q(x)$ is a quadratic.
• Solve $q(x) = 0$ for the remaining two zeroes.

Strategies for constructing polynomials

Quadratic from zeroes $\alpha, \beta$. The simplest answer is $p(x) = x^2 - (\alpha + \beta) x + \alpha \beta.$ Any non-zero scalar multiple is also a valid answer.

Cubic from zeroes $\alpha, \beta, \gamma$. $p(x) = x^3 - (\alpha + \beta + \gamma) x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha) x - \alpha\beta\gamma.$

Quadratic from sum $s$ and product $p$. Just write $x^2 - s x + p$.

Theorem: division algorithm (informal)

If $p(x)$ is divided by a non-zero polynomial $g(x)$, then we can find polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that $p(x) = g(x) \cdot q(x) + r(x),$ where $r(x) = 0$ or $\deg r(x) < \deg g(x)$. In particular, $(x - \alpha)$ divides $p(x)$ exactly when $\alpha$ is a zero of $p(x)$ , this is the factor theorem, the workhorse for splitting cubics.

A practical recipe for the cubic problem:

1. Find one rational zero $\alpha$ by trying factors of $d/a$.
2. Divide $p(x)$ by $(x - \alpha)$ , long division or synthetic division , to obtain a quadratic $q(x)$.
3. Solve $q(x) = 0$ to obtain the other two zeroes.

Worked examples

Example 1. Find the zeroes of $p(x) = x^2 + 7 x + 10$ by factorisation.

Split the middle term: we need two numbers whose product is $10$ and sum is $7$. They are $2$ and $5$. So $p(x) = x^2 + 2x + 5x + 10 = x(x + 2) + 5(x + 2) = (x + 2)(x + 5)$. Zeroes: $-2$ and $-5$.

Example 2. Find a quadratic whose zeroes are $\dfrac{1}{4}$ and $-1$.

Sum $= 1/4 - 1 = -3/4$. Product $= -1/4$. Quadratic: $x^2 + (3/4) x - 1/4$ or, clearing denominators, $4 x^2 + 3 x - 1$.

Example 3. Find all zeroes of $p(x) = x^3 - 4 x^2 + x + 6$.

Try $x = -1$: $-1 - 4 - 1 + 6 = 0$. ✓ So $(x + 1)$ is a factor. Divide: $\frac{x^3 - 4 x^2 + x + 6}{x + 1} = x^2 - 5 x + 6.$ Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$. Zeroes: $-1, 2, 3$.

Example 4. If the zeroes of $x^3 + p x^2 + q x + r$ are reciprocals of the zeroes of $x^3 + a x^2 + b x + c$, find $p, q, r$ in terms of $a, b, c$.

Let $\alpha, \beta, \gamma$ be zeroes of the second cubic. So $\alpha + \beta + \gamma = -a$, $\alpha\beta + \beta\gamma + \gamma\alpha = b$, $\alpha\beta\gamma = -c$.

Reciprocals $1/\alpha, 1/\beta, 1/\gamma$ have:

• sum $= (\beta\gamma + \gamma\alpha + \alpha\beta)/(\alpha\beta\gamma) = b/(-c) = -b/c$,
• sum of products in pairs $= (\alpha + \beta + \gamma)/(\alpha\beta\gamma) = (-a)/(-c) = a/c$,
• product $= 1/(\alpha\beta\gamma) = 1/(-c) = -1/c$.

So $p = -(-b/c) = b/c$, $q = a/c$, $r = -(-1/c) = 1/c$. The cubic is $x^3 + (b/c) x^2 + (a/c) x + 1/c$, or clearing denominators, $c x^3 + b x^2 + a x + 1$.

Example 5. Construct a cubic whose zeroes are $-2, -2, 5$.

Sum $= 1$. Sum of products $= 4 - 10 - 10 = -16$. Product $= 20$.

Cubic: $x^3 - x^2 - 16 x - 20$.

Try it yourself

1. Find the zeroes of $6 x^2 - x - 2$ by factorisation.
2. Find a quadratic whose zeroes are $\dfrac{2}{3}$ and $-\dfrac{1}{2}$.
3. Construct a cubic with zeroes $1, -2, 4$.
4. Find all zeroes of $x^3 - x^2 - 4 x + 4$. (Try $x = 1$ first.)
5. The sum and product of zeroes of a quadratic are $-1$ and $-6$ respectively. Write the quadratic.
6. Find all zeroes of $2 x^3 + 5 x^2 - 4 x - 12$. (Hint: try $x = 3/2$ or factors of $6$.)
7. If $\alpha, \beta$ are zeroes of $x^2 - 6 x + a$, find the quadratic with zeroes $\alpha + 2, \beta + 2$.
8. Find a quadratic whose zeroes are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.
9. Find the value of $k$ for which the polynomial $x^3 - 3 x^2 + (k+5) x - 6$ has $x = 1$ as a zero.
10. Find a cubic whose zeroes are $1, 1, -2$.

Pitfalls / Insight

• Rational root candidates for a polynomial with integer coefficients are limited to ratios of factors of the constant by factors of the leading coefficient.
• After finding one zero of a cubic, always divide , don't try to guess the other two unless you have a strong reason.
• "Construct a polynomial" problems are easier than they look: just remember $(x - \alpha)(x - \beta)\cdots$ and expand.

Insight. Forward (find zeroes) and inverse (construct polynomial) are dual problems. Mastering both turns this chapter into a small but reliable source of marks , and a launchpad for Chapter 4 on quadratic equations.