Polynomial roots: multiplicity and complex solutions

Optional final course topic. Use degree reduction, factor signs and complex numbers.

A graph may touch the axis and turn back or cross it. Sometimes it never intersects even though complex roots exist. A root corresponds to a linear factor; multiplicity counts repetitions of that factor. This explains the different graph behavior.

From remainder to multiplicity

By A014’s remainder theorem:

$P(x)=(x-r)Q(x)+P(r),\qquad P(r)=0\iff(x-r)\mid P(x).$

The vertical bar means divides without remainder. A root r has multiplicity m if integer m≥1 and:

$P(x)={(x-r)}^mG(x),\qquad G(r)\ne0.$

G(r)≠0 means no further copy of that factor can be extracted. For example:

$P(x)={(x-1)}^2(x+2)=x^3-3x+2.$

The roots are −2, simple, and 1, double. There are two distinct roots but 1+2=3 counting multiplicity, matching the degree. Near a real r, continuous G keeps its sign because G(r)≠0. The power (x−r)^m changes sign for odd m and retains it for even m. Thus the graph crosses at odd multiplicity and touches without changing sign at even multiplicity. OpenStax

Simple and double roots

P(x)=(x−1)²(x+2)

No real zeros

Q(x)=x²+1; Q(x)≥1

The first graph crosses at −2 (multiplicity 1) and touches at 1 (multiplicity 2). The second, x²+1, never crosses: complex roots ±i do not lie on the real axis. Dimensionless uniform scales; only graph windows shown.

The second graph x²+1 has no real zeros, but over complex numbers:

$x^2+1=(x-i)(x+i).$

Roots i and −i are not points on the real x-axis. Drawing the real graph more accurately cannot reveal these complex roots on it.

Conjugate pairs require real coefficients

If all coefficients are real, root a+bi comes with a−bi. Conjugation preserves sums and products and fixes real coefficients, so P(a+bi)=0 implies P(a−bi)=0. Their multiplicities also agree. This need not hold for complex coefficients: z−i has only root i.

What the fundamental theorem guarantees

Every nonconstant degree-n polynomial with complex coefficients, n≥1, has exactly n complex roots counting multiplicity. It factors as:

$P(z)=a_n\prod_{j=1}^{n}(z-r_j),\qquad a_n\ne0,\quad n\ge1.$

The rⱼ may repeat and may be real. The theorem promises neither n distinct nor n real roots and does not apply to nonzero constants. It guarantees existence rather than easy calculation. Precise statement, NIST DLMF §1.11(ii).

Cubic equations: an outline

A real-coefficient cubic has at least one real root: far left and right its signs differ, and a continuous graph between opposite signs must pass through zero. Once a root is found, division extracts its linear factor and leaves a quadratic.

A substitution removes the quadratic term of a general cubic; p,q are the resulting real coefficients:

$ax^3+bx^2+cx+d=0,\quad a\ne0,\qquad x=y-\frac b{3a}\ \Rightarrow\ y^3+py+q=0.$

For Δ≥0 one step of Cardano’s method is below; cube roots here are real, with the radicand’s sign:

$\Delta={(q/2)}^2+{(p/3)}^3,\qquad y=\sqrt[3]{-q/2+\sqrt\Delta}+\sqrt[3]{-q/2-\sqrt\Delta}\quad(\Delta\ge0).$

For p=−3,q=−2, Δ=0 and the formula gives y=2. Division finds the rest:

$y^3-3y-2=(y-2){(y+1)}^2.$

Thus 2 is simple and −1 double. For Δ<0 there are three distinct real roots, yet the radical route uses complex intermediate numbers; cube-root branches must be coordinated. This is a direction for further study, without independently choosing arbitrary roots. Cubic reduction and formulas, NIST DLMF §1.11(iii). NIST uses discriminant D=−108Δ: check the notation’s sign.

At the course’s end, always name the number domain, admissible values and how roots are counted. Then formula, graph and answer describe the same problem.

Two additional checks

Expand a(x−r₁)(x−r₂)(x−r₃) and compare coefficients to obtain Vieta’s formulas, repeating roots by multiplicity:

$r_1+r_2+r_3=-b/a,\quad r_1r_2+r_1r_3+r_2r_3=c/a,\quad r_1r_2r_3=-d/a.$

For 2,−1,−1 the sum is 0, pair-product sum −3 and product 2, recovering y³−3y−2. This checks coefficients without solving again.

An additional method after derivative rules: for a nonzero polynomial, a real r has multiplicity m when P and derivatives of orders 1,…,m−1 vanish there, while derivative m does not. Repeated differentiation removes one x−r factor at a time. For P=x³−3x+2 at r=1: P=0, P′=0, P″=6, hence multiplicity 2. This method is unnecessary for the factor-sign explanation above.