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Mathematics · Ch 3 — Theory of Equations

Formation of Polynomial Equations with given Roots

3.3.2.3

Formation of Polynomial Equations with given Roots

Building an equation from given roots, directly. One way to write a degree-nn equation with roots α1,…,αn\alpha_1,\ldots,\alpha_n is to just multiply out the factors: (x−α1)(x−α2)⋯(x−αn)=0(x-\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)=0. But using Vieta's relations (§3.3.2.2) in reverse is far more efficient — it lets us write down the equation's coefficients directly from the sums/products of the roots, without expanding a product of nn linear factors, and even lets us read off one particular coefficient without finding the rest.

A cubic with roots α,β,γ\alpha,\beta,\gamma is

x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ=0.x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma = 0.

A degree-nn equation with roots α1,…,αn\alpha_1,\ldots,\alpha_n is

xn−(∑α1)xn−1+(∑α1α2)xn−2−⋯+(−1)n(∑α1α2⋯αn)=0.x^n - \Big(\textstyle\sum\alpha_1\Big)x^{n-1} + \Big(\textstyle\sum\alpha_1\alpha_2\Big)x^{n-2} - \cdots + (-1)^n\Big(\textstyle\sum\alpha_1\alpha_2\cdots\alpha_n\Big) = 0.

Worked illustration. A cubic with roots 1,−2,31,-2,3: sum =2=2, pairwise-sum =(1)(−2)+(−2)(3)+(3)(1)=−2−6+3=−5= (1)(-2)+(-2)(3)+(3)(1)=-2-6+3=-5, product =−6=-6. So x3−2x2−5x−(−6)=0x^3-2x^2-5x-(-6)=0, i.e. x3−2x2−5x+6=0x^3-2x^2-5x+6=0 — matching what direct expansion of (x−1)(x+2)(x−3)(x-1)(x+2)(x-3) gives.

Building an equation whose roots are a function of a known equation's roots (Transformation of Equations). This same reverse-Vieta idea is the workhorse for constructing an equation whose roots are related to those of a given equation, without ever solving that given equation — only its Vieta sums are needed.

Squares of the roots (Example 3.6 pattern). If α,β,γ\alpha,\beta,\gamma are the roots of x3+ax2+bx+c=0x^3+ax^2+bx+c=0, so ∑α=−a\sum\alpha=-a, ∑αβ=b\sum\alpha\beta=b, αβγ=−c\alpha\beta\gamma=-c, then the equation with roots α2,β2,γ2\alpha^2,\beta^2,\gamma^2 has

∑α2=(∑α)2−2∑αβ=a2−2b,∑α2β2=(∑αβ)2−2(αβγ)(∑α)=b2−2ac,α2β2γ2=(αβγ)2=c2,\textstyle\sum\alpha^2 = (\sum\alpha)^2-2\sum\alpha\beta = a^2-2b, \qquad \sum\alpha^2\beta^2=(\sum\alpha\beta)^2-2(\alpha\beta\gamma)(\sum\alpha)=b^2-2ac, \qquad \alpha^2\beta^2\gamma^2=(\alpha\beta\gamma)^2=c^2,

giving x3−(a2−2b)x2+(b2−2ac)x−c2=0x^3-(a^2-2b)x^2+(b^2-2ac)x-c^2=0.

Sum of squares of the roots (Example 3.4 pattern). For a quartic ax4+bx3+cx2+dx+e=0ax^4+bx^3+cx^2+dx+e=0 with roots α,β,γ,δ\alpha,\beta,\gamma,\delta: using (∑α)2=∑α2+2∑αβ(\sum\alpha)^2=\sum\alpha^2+2\sum\alpha\beta,

α2+β2+γ2+δ2=( ⁣−ba)2−2⋅ca=b2−2aca2.\alpha^2+\beta^2+\gamma^2+\delta^2 = \Big(\!-\frac ba\Big)^2 - 2\cdot\frac ca = \frac{b^2-2ac}{a^2}.

Roots in a given ratio (Example 3.5 pattern). If the roots of x3+ax2+bx+c=0x^3+ax^2+bx+c=0 are in the ratio p:q:rp:q:r, write them as pλ,qλ,rλp\lambda,q\lambda,r\lambda for an unknown scale factor λ\lambda. Vieta's relations give three equations in λ\lambda (and p,q,r,a,b,cp,q,r,a,b,c); eliminating λ\lambda between the sum-relation and the product-relation produces a pure condition on a,b,c,p,q,ra,b,c,p,q,r — e.g. pqr a3=c (p+q+r)3pqr\,a^3 = c\,(p+q+r)^3. …

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