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Toolkit 30

Vieta's Formula

For a quadratic: \text{For a quadratic: }
ax2+bx+c=0(roots: r1,r2)ax^2+bx+c=0\qquad(\text{roots: }r_1,r_2)
r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1r_2=\frac{c}{a}
For a cubic: \text{For a cubic: }
ax3+bx2+cx+d=0(roots: r1,r2,r3)ax^3+bx^2+cx+d=0\qquad(\text{roots: }r_1,r_2,r_3)
r1+r2+r3=−ba,r1r2+r1r3+r2r3=ca,r1r2r3=−dar_1+r_2+r_3=-\frac{b}{a},\qquad r_1r_2+r_1r_3+r_2r_3=\frac{c}{a},\qquad r_1r_2r_3=-\frac{d}{a}
For a general polynomial: \text{For a general polynomial: }
anxn+an−1xn−1+⋯+a1x+a0=0(roots: r1,…,rn)a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0\qquad(\text{roots: }r_1,\ldots,r_n)
r1+r2+⋯+rn=−an−1anr_1+r_2+\cdots+r_n=-\frac{a_{n-1}}{a_n}
∑1≤i1<i2≤nri1ri2=an−2an\sum_{1\le i_1<i_2\le n}r_{i_1}r_{i_2}=\frac{a_{n-2}}{a_n}
∑1≤i1<i2<i3≤nri1ri2ri3=−an−3an\sum_{1\le i_1<i_2<i_3\le n}r_{i_1}r_{i_2}r_{i_3}=-\frac{a_{n-3}}{a_n}
⋯\cdots
∑1≤i1<⋯<in−1≤nri1⋯rin−1=(−1)n−1a1an\sum_{1\le i_1<\cdots<i_{n-1}\le n}r_{i_1}\cdots r_{i_{n-1}}=(-1)^{n-1}\frac{a_1}{a_n}
r1r2⋯rn=(−1)na0anr_1r_2\cdots r_n=(-1)^n\frac{a_0}{a_n}

Proof

For a quadratic polynomial with roots r1r_1 and r2r_2,

ax2+bx+c=a(x−r1)(x−r2).ax^2+bx+c = a(x-r_1)(x-r_2).

Expanding the right-hand side,

a(x−r1)(x−r2)=a(x2−(r1+r2)x+r1r2)=ax2−a(r1+r2)x+ar1r2.\begin{aligned} a(x-r_1)(x-r_2) &= a\left(x^2-(r_1+r_2)x+r_1r_2\right) \\ &= ax^2 - a(r_1+r_2)x + ar_1r_2. \end{aligned}

Comparing corresponding coefficients gives

b=−a(r1+r2)b = -a(r_1+r_2)

and

c=ar1r2.c = ar_1r_2.

Therefore,

r1+r2=−ba,r1r2=ca.r_1+r_2 = -\frac{b}{a}, \qquad r_1r_2 = \frac{c}{a}.

For a cubic polynomial with roots r1,r2,r3r_1, r_2, r_3,

ax3+bx2+cx+d=a(x−r1)(x−r2)(x−r3).ax^3+bx^2+cx+d = a(x-r_1)(x-r_2)(x-r_3).

Expanding,

a(x−r1)(x−r2)(x−r3)=a(x3−(r1+r2+r3)x2+(r1r2+r1r3+r2r3)x−r1r2r3).\begin{aligned} a(x-r_1)(x-r_2)(x-r_3) &= a\Bigl(x^3 - (r_1+r_2+r_3)x^2 \\ &\qquad + (r_1r_2+r_1r_3+r_2r_3)x - r_1r_2r_3\Bigr). \end{aligned}

Comparing coefficients gives

r1+r2+r3=−ba,r_1+r_2+r_3 = -\frac{b}{a},
r1r2+r1r3+r2r3=ca,r_1r_2+r_1r_3+r_2r_3 = \frac{c}{a},

and

r1r2r3=−da.r_1r_2r_3 = -\frac{d}{a}.

More generally, suppose

anxn+an−1xn−1+⋯+a1x+a0=an(x−r1)(x−r2)⋯(x−rn).a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).

We expand the right-hand side and compare corresponding coefficients term by term.

Coefficient of xnx^n: choosing xx from every factor gives

an=an.a_n = a_n.

Coefficient of xn−1x^{n-1}: pick the constant term −ri-r_i from exactly one factor and xx from all others. Summing over the choice of factor gives

an−1=−an(r1+r2+⋯+rn),a_{n-1} = -a_n(r_1+r_2+\cdots+r_n),

so

r1+r2+⋯+rn=−an−1an.r_1+r_2+\cdots+r_n = -\frac{a_{n-1}}{a_n}.

Coefficient of xn−2x^{n-2}: pick a constant term −ri-r_i from exactly two factors and xx from all remaining factors. Summing over the pairs {i,j}\{i,j\} gives

an−2=an(r1r2+r1r3+⋯+rn−1rn),a_{n-2} = a_n\bigl(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n\bigr),

so

r1r2+r1r3+⋯+rn−1rn=an−2an.r_1r_2+r_1r_3+\cdots+r_{n-1}r_n = \frac{a_{n-2}}{a_n}.

Coefficient of xn−3x^{n-3}: pick a constant term −ri-r_i from exactly three factors and xx from all remaining factors. Summing over the triples gives

an−3=−an(r1r2r3+r1r2r4+⋯+rn−2rn−1rn),a_{n-3} = -a_n\bigl(r_1r_2r_3+r_1r_2r_4+\cdots+r_{n-2}r_{n-1}r_n\bigr),

so

r1r2r3+r1r2r4+⋯+rn−2rn−1rn=−an−3an.r_1r_2r_3+r_1r_2r_4+\cdots+r_{n-2}r_{n-1}r_n = -\frac{a_{n-3}}{a_n}.
⋮\vdots

Continuing in the same way, the coefficient of xx (the case j=n−1j=n-1) comes from picking the constant term from all but one factor:

r1r2⋯rn−1+⋯+r2r3⋯rn=(−1)n−1a1an,r_1r_2\cdots r_{n-1}+\cdots+r_2r_3\cdots r_n = (-1)^{n-1}\frac{a_1}{a_n},

and the constant term comes from picking the constant term from every factor:

r1r2⋯rn=(−1)na0an.□r_1r_2\cdots r_n = (-1)^n\frac{a_0}{a_n}. \quad\square
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