Zeroes and coefficients of a cubic

The story we just told for quadratics extends gracefully to cubics. If a cubic $p(x) = a x^3 + b x^2 + c x + d$ ($a \ne 0$) has zeroes $\alpha, \beta, \gamma$ (counted with multiplicity if necessary), then there are three classical identities relating the zeroes to the coefficients.

Idea

Suppose $\alpha, \beta, \gamma$ are zeroes of $a x^3 + b x^2 + c x + d$. Then $p(x) = a (x - \alpha)(x - \beta)(x - \gamma).$ Expand the right side (use the identity $(x-\alpha)(x-\beta)(x-\gamma) = x^3 - (\alpha+\beta+\gamma) x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha) x - \alpha\beta\gamma$): $p(x) = a x^3 - a(\alpha+\beta+\gamma) x^2 + a(\alpha\beta+\beta\gamma+\gamma\alpha) x - a\alpha\beta\gamma.$ Comparing with $a x^3 + b x^2 + c x + d$: $\alpha + \beta + \gamma = -\frac{b}{a},$ $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a},$ $\alpha\beta\gamma = -\frac{d}{a}.$

These are the Vieta identities for the cubic.

Theorem and uses

Theorem. Let $a x^3 + b x^2 + c x + d$ have zeroes $\alpha, \beta, \gamma$. Then $\alpha + \beta + \gamma = -\frac{b}{a},$ $\alpha \beta + \beta\gamma + \gamma \alpha = \frac{c}{a},$ $\alpha\beta\gamma = -\frac{d}{a}.$

A few useful consequences.

(i) Reconstruction. Given three zeroes $\alpha, \beta, \gamma$, the cubic with leading coefficient $1$ is $x^3 - (\alpha+\beta+\gamma) x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha) x - \alpha\beta\gamma.$

(ii) Symmetric expressions in $\alpha, \beta, \gamma$. Many useful quantities can be expressed in terms of the three elementary symmetric polynomials $e_1 = \alpha+\beta+\gamma$, $e_2 = \alpha\beta + \beta\gamma + \gamma\alpha$, $e_3 = \alpha\beta\gamma$. For example, $\alpha^2 + \beta^2 + \gamma^2 = e_1^2 - 2 e_2,$ $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{e_2}{e_3}.$

(iii) Finding one zero given two. If two zeroes are known and you want the third, the easiest move is usually to use $\alpha + \beta + \gamma = -b/a$.

(iv) Verifying given zeroes. Plug zeroes into the polynomial or check that the sum and product match $-b/a$ and $-d/a$.

For board exams expect questions where the cubic has one obvious integer zero (you spot it by trying $\pm 1, \pm 2$, etc.), you divide out the linear factor, and you finish with a quadratic. We will see this in the worked examples.

Worked examples

Example 1. Verify that $1, -1, 3$ are zeroes of $p(x) = x^3 - 3 x^2 - x + 3$, and check Vieta.

$p(1) = 1 - 3 - 1 + 3 = 0$, $p(-1) = -1 - 3 + 1 + 3 = 0$, $p(3) = 27 - 27 - 3 + 3 = 0$. ✓

Sum $= 1 + (-1) + 3 = 3 = -(-3)/1$. ✓ Sum of products $= -1 + (-3) + 3 = -1 = (-1)/1$. ✓ Product $= -3 = -3/1$. ✓

Example 2. Find a cubic with zeroes $2, 3, -1$.

$e_1 = 4$, $e_2 = 6 - 3 - 2 = 1$, $e_3 = -6$. Cubic: $x^3 - 4 x^2 + x + 6$.

Example 3. If two of the zeroes of $x^3 - 4 x^2 - 7 x + 10$ are $5$ and $-2$, find the third.

By Vieta, $\alpha + \beta + \gamma = -(-4)/1 = 4$. With $\alpha = 5, \beta = -2$: $\gamma = 4 - 5 + 2 = 1$.

Example 4. The zeroes of $x^3 - 3 x^2 + x + 1$ are $a - b$, $a$, $a + b$ (an AP). Find $a$.

Sum of zeroes $= 3a = 3$, so $a = 1$.

Example 5. The sum of two zeroes of $x^3 + a x^2 + b x + c$ is zero. Show that $c = ab$.

If $\alpha + \beta = 0$, then $\gamma = -a$ (from sum $\alpha + \beta + \gamma = -a$). Also $\alpha + \beta = 0$ means $\alpha = -\beta$, so $\alpha \beta = -\beta^2$. Now, $\alpha \beta \gamma = -c \implies -\beta^2 \cdot (-a) = -c \implies a \beta^2 = -c \implies \beta^2 = -c/a.$ And $\alpha\beta + \beta\gamma + \gamma\alpha = b$ becomes $-\beta^2 + \gamma(\alpha + \beta) = -\beta^2 = b$, so $\beta^2 = -b$. Comparing, $-c/a = -b$, hence $c = ab$. $\blacksquare$

Try it yourself

1. Verify that $-1, 2, 3$ are zeroes of $x^3 - 4 x^2 + x + 6$, and check Vieta.
2. Form the cubic with zeroes $1, -3, 4$.
3. Two zeroes of $x^3 - 6 x^2 + 11 x - 6$ are $1$ and $2$. Find the third.
4. If the zeroes of $x^3 + p x^2 + q x + r$ are in AP, show that $2 p^3 = 9 pq - 27 r$.
5. The zeroes of $x^3 - 3 x^2 + r x + s$ are $a, b, c$ with $a + b = 4$. Find $c$.
6. Find a cubic whose zeroes are $\sqrt{2}, -\sqrt{2}, 5$.
7. If $\alpha, \beta, \gamma$ are zeroes of $x^3 - 6 x^2 + 11 x - 6$, find $\alpha^2 + \beta^2 + \gamma^2$.
8. If $\alpha, \beta, \gamma$ are zeroes of $x^3 + a x^2 + b x + c$, find $\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}$.
9. Construct a cubic with zeroes $-1, -1, 2$.
10. If $\alpha, \beta, \gamma$ are zeroes of $2 x^3 + x^2 - 5 x + 2$, find $\alpha\beta\gamma$ and $\alpha + \beta + \gamma$.

Pitfalls / Insight

• Sign alternation. $b/a$ comes with a minus, $c/a$ comes positive, $d/a$ comes with a minus. Easy to mix up.
• "Two zeroes are known" problems , use the sum first, then the product to check.
• Symmetric expressions in three variables expand using $e_1, e_2, e_3$. Don't try to find each zero individually.

Insight. The pattern from quadratic to cubic is clean: degree-$n$ polynomial $\Rightarrow$ $n$ symmetric coefficient-zero relations, with signs alternating. This generalises to all degrees and is one of the most beautiful symmetries in algebra.