Roots and Coefficients of a Cubic

Three brackets, multiplied out

For a quadratic, the two roots add to $-\frac{b}{a}$−b/a and multiply to $\frac{c}{a}$c/a. A cubic has three roots, and the same idea gives three relationships.

If the cubic $a{x}^3+b{x}^2+$ax³ + bx² + cx + d = 0 has roots $\alpha$α, $\beta$β and $\gamma$γ, it factors as $a\left(x-\alpha \right)\left(x-\beta \right)\left(x-\gamma \right)=0$a(x − α)(x − β)(x − γ) = 0. Multiply out two brackets first: $\left(x-\alpha \right)\left(x-\beta \right)=x^2-\left(\alpha +\beta \right)x+\alpha \beta$(x − α)(x − β) = x² − (α + β)x + αβ.

Now multiply by $x-\gamma$x − γ. Times x gives $x^3-\left(\alpha +\beta \right)x^2+\alpha \beta x$x³ − (α + β)x² + αβx. Times $-\gamma$−γ gives $-\gamma x^2+\left(\alpha \gamma +\beta \gamma \right)x-\alpha \beta \gamma$−γx² + (αγ + βγ)x − αβγ. Collect the like terms: $x^3-\left(\alpha +\beta +\gamma \right)x^2+\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)x-\alpha \beta \gamma$x³ − (α + β + γ)x² + (αβ + βγ + γα)x − αβγ. Finally multiply through by a.

So $a{x}^3+b{x}^2+$ax³ + bx² + cx $+d=a{x}^3-a\left(\alpha +\beta +\gamma \right)x^2+a\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)x-a\alpha \beta \gamma$+ d = ax³ − a(α + β + γ)x² + a(αβ + βγ + γα)x − aαβγ.

Match the coefficients

Two polynomials that are equal for every x have the same coefficients. Match them one power at a time.

The $x^2$x² terms: $b=-a\left(\alpha +\beta +\gamma \right)$b = −a(α + β + γ), so $\alpha +\beta +\gamma =-\frac{b}{a}$α + β + γ = −b/a. The x terms: $c=a\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)$c = a(αβ + βγ + γα), so $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$αβ + βγ + γα = c/a. The constants: $d=-a\alpha \beta \gamma$d = −aαβγ, so $\alpha \beta \gamma =-\frac{d}{a}$αβγ = −d/a.

The signs alternate: minus, plus, minus. The sum $\alpha \beta +\beta \gamma +\gamma \alpha$αβ + βγ + γα adds the products of the roots two at a time, and it is often written $\sum \alpha \beta$Σαβ, just as $\alpha +\beta +\gamma$α + β + γ is written $\sum \alpha$Σα.

Check on $x^3-6x^2+11x-6=\left(x-1\right)\left(x-2\right)\left(x-3\right)$x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3), whose roots are 1, 2 and 3. Here a = 1, b = −6, c = 11 and d = −6. The sum is $1+2+3=6=-\frac{b}{a}$1 + 2 + 3 = 6 = −b/a. The pairs give $1\times 2+2\times 3+3\times 1=11=\frac{c}{a}$1 × 2 + 2 × 3 + 3 × 1 = 11 = c/a. The product is $1\times 2\times 3=6=-\frac{d}{a}$1 × 2 × 3 = 6 = −d/a.

The gold curve is $y=x^3-6x^2+11x-6$y = x³ − 6x² + 11x − 6, and the plain curve is twice it, $y=2x^3-12x^2+22x-12$y = 2x³ − 12x² + 22x − 12. Both cross the x-axis at x = 1, 2 and 3. Doubling every coefficient leaves the roots where they are, and $-\frac{b}{a}$−b/a stays $\frac{12}{2}=6$12/2 = 6: that is why each sum is divided by a.

Read off without solving

For $2x^3-6x^2+4x-10=0$2x³ − 6x² + 4x − 10 = 0, a = 2, b = −6, c = 4 and d = −10. So $\alpha +\beta +\gamma =-\frac{\left(-6\right)}{2}=3$α + β + γ = −(−6)/2 = 3, $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{4}{2}=2$αβ + βγ + γα = 4/2 = 2, and $\alpha \beta \gamma =-\frac{\left(-10\right)}{2}=5$αβγ = −(−10)/2 = 5.

None of the roots has been found, and none needs to be. These three values hold whether the roots are real or not; this cubic has only one real root, close to 2.904, and the other two are complex, yet the three sums are still 3, 2 and 5.

Other symmetric expressions follow from the three. The sum of the reciprocals is $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{\left(\beta \gamma +\gamma \alpha +\alpha \beta \right)}{\left(\alpha \beta \gamma \right)}$1/α + 1/β + 1/γ = (βγ + γα + αβ)/(αβγ), over a common denominator. For this cubic that is $\frac{2}{5}$2/5.

A quartic, one row longer

A quartic $a{x}^4+b{x}^3+$ax⁴ + bx³ + cx² + dx + e = 0 has four roots $\alpha$α, $\beta$β, $\gamma$γ and $\delta$δ, and four sums. The roots one at a time: $\sum \alpha =-\frac{b}{a}$Σα = −b/a. Two at a time: $\sum \alpha \beta =\frac{c}{a}$Σαβ = c/a. Three at a time: $\sum \alpha \beta \gamma =-\frac{d}{a}$Σαβγ = −d/a. All four multiplied: $\alpha \beta \gamma \delta =\frac{e}{a}$αβγδ = e/a. The signs still alternate, starting with minus.

Check on $\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=x^4-10x^3+35x^2-50x+24$(x − 1)(x − 2)(x − 3)(x − 4) = x⁴ − 10x³ + 35x² − 50x + 24. The roots add to 1 + 2 + 3 + 4 = 10. The six pairs give 2 + 3 + 4 + 6 + 8 + 12 = 35. The four triples give 6 + 8 + 12 + 24 = 50. The product is 24. With a = 1, those are −b, c, −d and e, as the pattern says.

The sum of the squares

Square the sum of the roots: ${\left(\alpha +\beta +\gamma \right)}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2\alpha \beta +2\beta \gamma +2\gamma \alpha$(α + β + γ)² = α² + β² + γ² + 2αβ + 2βγ + 2γα. Each product of two different roots appears twice. So ${\left(\sum \alpha \right)}^2=\sum {\alpha }^2+2\sum \alpha \beta$(Σα)² = Σα² + 2Σαβ, and rearranging, $\sum {\alpha }^2={\left(\sum \alpha \right)}^2-2\sum \alpha \beta$Σα² = (Σα)² − 2Σαβ.

If $\sum \alpha =5$Σα = 5 and $\sum \alpha \beta =6$Σαβ = 6, then ${\alpha }^2+{\beta }^2+{\gamma }^2=25-12=13$α² + β² + γ² = 25 − 12 = 13. For the roots 1, 2 and 3: $\sum {\alpha }^2=6^2-2\times 11=36-22=14$Σα² = 6² − 2 × 11 = 36 − 22 = 14, and directly 1 + 4 + 9 = 14.

For $2x^3-6x^2+4x-10=0$2x³ − 6x² + 4x − 10 = 0, $\sum {\alpha }^2=3^2-2\times 2=5$Σα² = 3² − 2 × 2 = 5, again without finding a root.

Building a cubic from its sums

The relationships work backwards too. With a = 1, a cubic whose roots have sum S, pair sum P and product Q is $x^3-$x³ − Sx² + Px − Q = 0. Roots with S = 12, P = 47 and Q = 60 give $x^3-12x^2+47x-60=0$x³ − 12x² + 47x − 60 = 0.

The roots of that cubic are 3, 4 and 5: 3 + 4 + 5 = 12, 12 + 20 + 15 = 47, and 3 × 4 × 5 = 60.

The usual mistakes

Dropping the minus sign. For $x^3-6x^2+11x-6$x³ − 6x² + 11x − 6, $\alpha +\beta +\gamma =-\frac{b}{a}=-\left(-6\right)=6$α + β + γ = −b/a = −(−6) = 6, not −6.

Dropping the minus from the product. For $2x^3+4x^2-6x+8$2x³ + 4x² − 6x + 8, $\alpha \beta \gamma =-\frac{d}{a}=-\frac{8}{2}=-4$αβγ = −d/a = −8/2 = −4, not 4.

Mixing up the coefficients. c gives the pair sum, not the sum of the roots, so 11 is $\alpha \beta +\beta \gamma +\gamma \alpha$αβ + βγ + γα for $x^3-6x^2+11x-6$x³ − 6x² + 11x − 6, and $-3=\frac{c}{a}$−3 = c/a is not the product for $2x^3+4x^2-6x+8$2x³ + 4x² − 6x + 8.

Giving the pair sum a minus sign. $\sum \alpha \beta =\frac{c}{a}$Σαβ = c/a: the signs alternate starting with minus on b, so the c term arrives positive.

Taking ${\left(\sum \alpha \right)}^2$(Σα)² as the sum of the squares. It still holds $2\sum \alpha \beta$2Σαβ: with $\sum \alpha =5$Σα = 5 and $\sum \alpha \beta =6$Σαβ = 6, the sum of the squares is 25 − 12 = 13, not 25, and taking off one $\sum \alpha \beta$Σαβ instead of two gives 19.

A gift box

In the application below, the edges, the card and the volume of a box give the sum, the pair sum and the product of its three dimensions. Those are the three sums of a cubic, and the longest rod in the box comes from the sum of the squares before the cubic is solved.