Further Maths/Lessons/Roots of Polynomials/Roots of Cubic & Quartic Equations
Core Pure · Roots of Polynomials · Year 1

Roots of Cubic & Quartic Equations

The quadratic's sum-and-product pattern extends naturally to cubics and quartics — every symmetric combination of roots is readable from the coefficients alone.

17 min read 6 worked examples

Roots of a cubic equation

Theorem 11.1 — Vieta's formulae for a cubic

If $\alpha,\beta,\gamma$ are the roots of $ax^3+bx^2+cx+d=0$ ($a\neq0$), then $$\alpha+\beta+\gamma=-\frac{b}{a},$$ $$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a},$$ $$\alpha\beta\gamma=-\frac{d}{a}$$

ProofVieta's formulae for a cubic
1
$ax^3+bx^2+cx+d\equiv a(x-\alpha)(x-\beta)(x-\gamma)$
Factor form using the three roots.
2
Expanding: $a(x-\alpha)(x-\beta)(x-\gamma)=a\big[x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma\big]$
Systematic expansion of three linear factors, collecting by descending power of $x$.
3
Comparing coefficients of $x^2$: $b=-a(\alpha+\beta+\gamma) \Rightarrow \alpha+\beta+\gamma=-\dfrac{b}{a}$.
Coefficient matching.
4
Comparing coefficients of $x$: $c=a(\alpha\beta+\beta\gamma+\gamma\alpha) \Rightarrow \alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{c}{a}$.
Coefficient matching.
5
Comparing constant terms: $d=-a\alpha\beta\gamma \Rightarrow \alpha\beta\gamma=-\dfrac{d}{a}$.
Coefficient matching.
∎
Example 1 · Finding all three symmetric sums
The roots of $x^3-6x^2+11x-6=0$ are $\alpha,\beta,\gamma$. Find $\alpha+\beta+\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$, and $\alpha\beta\gamma$.

$\alpha+\beta+\gamma=6$, $\alpha\beta+\beta\gamma+\gamma\alpha=11$, $\alpha\beta\gamma=6$.

Example 2 · Evaluating α² + β² + γ²
Using the same equation as Example 1, find $\alpha^2+\beta^2+\gamma^2$.

$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=6^2-2(11)=36-22=14$.

Reminder — complex roots come in pairs

A cubic with real coefficients has at least one real root, and its other two roots are either both real or a complex conjugate pair. The sums $\alpha+\beta+\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$ and $\alpha\beta\gamma$ are still guaranteed to be real in every case, because whenever a conjugate pair $p+qi$ and $p-qi$ appears, their sum $2p$ and product $p^2+q^2$ are both real.

Example 3 · Working backwards from the roots to the equation
A cubic equation $ax^3+bx^2+cx+d=0$ has roots $\alpha=3-i$, $\beta=3+i$ and $\gamma=-2$. Find integer values for $a$, $b$, $c$ and $d$.

$\alpha+\beta+\gamma=(3-i)+(3+i)+(-2)=4$, so $4=-\dfrac ba$.

$\alpha\beta+\beta\gamma+\gamma\alpha=(3-i)(3+i)+(-2)(3+i)+(-2)(3-i)$. Since $(3-i)(3+i)=9+1=10$ and $(-2)(3+i)+(-2)(3-i)=-2\big[(3+i)+(3-i)\big]=-2(6)=-12$, the total is $10-12=-2$, giving $-2=\dfrac ca$.

$\alpha\beta\gamma=10\times(-2)=-20$, so $-20=-\dfrac da$.

Setting $a=1$: $b=-4$, $c=-2$, $d=20$, giving $x^3-4x^2-2x+20=0$.

Roots of a quartic equation

Theorem 11.2 — Vieta's formulae for a quartic

If $\alpha,\beta,\gamma,\delta$ are the roots of $ax^4+bx^3+cx^2+dx+e=0$, then, with the pattern of alternating signs continuing:

$$\sum\alpha=-\frac{b}{a}, \quad \sum\alpha\beta=\frac{c}{a}, \quad \sum\alpha\beta\gamma=-\frac{d}{a}, \quad \alpha\beta\gamma\delta=\frac{e}{a}$$ where $\sum\alpha\beta$ means the sum of all $\binom42=6$ pairwise products, and $\sum\alpha\beta\gamma$ means the sum of all $\binom43=4$ triple products.

Exam technique — remembering the sign pattern

The coefficients of $x^{n-1},x^{n-2},x^{n-3},\ldots$ correspond to $-\sum\alpha,\ +\sum\alpha\beta,\ -\sum\alpha\beta\gamma,\ldots$ — signs alternate starting with a minus, matching how many roots are being multiplied together (odd numbers of roots give a minus, even numbers give a plus), always relative to $a$, the leading coefficient.

Example 4 · A cubic with an unknown coefficient
The roots of $2x^3+px^2+qx+24=0$ are $2,3,k$. Find $p$, $q$, and $k$.

Product of roots: $2\times3\times k=-\dfrac{24}{2}=-12 \Rightarrow 6k=-12 \Rightarrow k=-2$.

Sum of roots: $2+3+(-2)=3=-\dfrac{p}{2} \Rightarrow p=-6$.

Sum of pairwise products: $2(3)+3(-2)+(-2)(2)=6-6-4=-4=\dfrac{q}{2} \Rightarrow q=-8$.

Example 5 · Quartic symmetric sum
The roots of $x^4-2x^3-7x^2+8x+12=0$ are $\alpha,\beta,\gamma,\delta$. Find $\alpha\beta\gamma\delta$ and $\alpha+\beta+\gamma+\delta$.

Product of roots: $\alpha\beta\gamma\delta=\dfrac{12}{1}=12$ (all four multiplied, even number, so $+$ sign, and $a=1$).

Sum of roots: $\alpha+\beta+\gamma+\delta=-\dfrac{-2}{1}=2$.

Common mistake

For a quartic, $\sum\alpha\beta$ has 6 terms (all $\binom42$ pairs), not 4 — students often only write the "adjacent" pairs. Systematically list every pair $\alpha\beta,\alpha\gamma,\alpha\delta,\beta\gamma,\beta\delta,\gamma\delta$ to avoid missing terms.

Example 6 · Working backwards for a quartic
A quartic equation $ax^4+bx^3+cx^2+dx+e=0$ has roots $\alpha=1$, $\beta=-1$, $\gamma=2+i$ and $\delta=2-i$. Find integer values for $a,b,c,d,e$.

$\sum\alpha=1+(-1)+(2+i)+(2-i)=4$, so $4=-\dfrac ba$.

$(2+i)(2-i)=4+1=5$ and $(1)(-1)=-1$; the remaining four cross pairs are $1(2+i)+1(2-i)+(-1)(2+i)+(-1)(2-i)=(2+i)+(2-i)-(2+i)-(2-i)=0$. So $\sum\alpha\beta=-1+5+0=4$, giving $4=\dfrac ca$.

$\sum\alpha\beta\gamma$: the four triple products are $1(-1)(2+i)+1(-1)(2-i)+1(2+i)(2-i)+(-1)(2+i)(2-i)=-(2+i)-(2-i)+5-5=-4$. So $-4=-\dfrac da \Rightarrow \dfrac da=4$.

$\alpha\beta\gamma\delta=(1)(-1)(5)=-5$, so $-5=\dfrac ea$.

Setting $a=1$: $b=-4$, $c=4$, $d=-4$, $e=-5$, giving $x^4-4x^3+4x^2-4x-5=0$.

Exam-style questions

Question 1 5 marks

The roots of $x^3+2x^2-5x-3=0$ are $\alpha,\beta,\gamma$. Find (a) $\alpha\beta+\beta\gamma+\gamma\alpha$, (b) $\alpha^2+\beta^2+\gamma^2$.

$\alpha+\beta+\gamma=-2$, $\alpha\beta+\beta\gamma+\gamma\alpha=-5$, $\alpha\beta\gamma=3$ (B1)

(a) $-5$ (read directly from Vieta's) (A1)

(b) $(-2)^2-2(-5)=4+10=14$ (M1, A1, A1)

Question 2 4 marks

The roots of $2x^3-5x^2+kx-3=0$ are $\alpha,\beta,\gamma$, and $\alpha\beta+\beta\gamma+\gamma\alpha=4$. Find the value of $k$.

$\alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{k}{2}$ (M1, A1)

$\dfrac{k}{2}=4 \Rightarrow k=8$ (M1, A1)

Question 3 8 marks

The roots of $x^4+ax^3+bx^2+8x-16=0$ are $1,-1,2,k$. Find $a$, $b$, and $k$.

Product of roots: $1\times(-1)\times2\times k=-16 \Rightarrow -2k=-16 \Rightarrow k=8$ (M1, A1)

Sum of roots: $1-1+2+8=10=-a \Rightarrow a=-10$ (M1, A1)

Sum of pairwise products: $(1)(-1)+(1)(2)+(1)(8)+(-1)(2)+(-1)(8)+(2)(8)$

$=-1+2+8-2-8+16=15=b \Rightarrow b=15$ (M1, A1, A1, A1)

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Roots of a Quadratic Equation