Unit 2, lesson 28

The Fundamental Theorem of Algebra

Why every nonconstant polynomial has a complex zero, why a polynomial of degree n has exactly n zeros counted with multiplicity, and how real polynomials split into real linear and quadratic factors.

  • Theory part 55
  • Practice part 56
  • Complex Numbers and Complete Factorization

The real numbers are not enough to solve x2+1=0x^2 + 1 = 0, and the complex numbers come from adding a solution of that one equation. The surprise is that this one addition is enough for every polynomial equation of degree at least 11. No matter how high the degree, and even when the coefficients are themselves complex, such an equation has a solution among the complex numbers, so nothing larger is ever needed.

Key terms

Nonconstant complex polynomials

Definition A complex polynomial is an expression

P(z)=anzn+an−1zn−1+⋯+a1z+a0,\begin{aligned} P(z) &= a_n z^n + a_{n-1} z^{n-1} \\[0.4em] &\quad + \cdots + a_1 z + a_0, \end{aligned}

where nn is a nonnegative integer and the coefficients a0,…,ana_0, \ldots, a_n are complex numbers. If an≠0a_n \ne 0, the degree is nn and ana_n is the leading coefficient. The polynomial is nonconstant when its degree is at least 11. A zero of PP is a complex number cc with P(c)=0P(c) = 0. Real numbers are complex numbers, so every polynomial with real coefficients is also a complex polynomial.

In plain words The same kind of polynomial as before, except that the coefficients and the variable may be complex. Nonconstant means that the variable really appears.

Example The polynomial P(z)=(1+i)z3−2iz+5P(z) = (1 + i)z^3 - 2iz + 5 has degree 33, leading coefficient 1+i1 + i and constant term 55. The constant polynomial P(z)=7P(z) = 7 has degree 00 and no zeros at all, which is why the theorem below asks for a nonconstant polynomial.

Existence of a complex zero

Definition The Fundamental Theorem of Algebra states that every nonconstant polynomial with complex coefficients has at least one zero in C\mathbb{C}. The theorem is about existence. It does not give a formula for the zero, and for general polynomials of degree 55 and higher no formula built from the coefficients with arithmetic and roots exists. Its proofs use ideas from beyond algebra, such as continuity.

In plain words Any polynomial equation P(z)=0P(z) = 0 with PP nonconstant has a complex solution. Over the real numbers this fails, since x2+1x^2 + 1 has no real zero.

Example The polynomial z2−iz^2 - i has complex coefficients, and the theorem promises a zero. One is

c=22(1+i),c2=12(1+2i+i2)=12⋅2i=i.\begin{aligned} c &= \frac{\sqrt{2}}{2}(1 + i), \\[0.8em] c^2 &= \frac{1}{2}(1 + 2i + i^2) \\[0.8em] &= \frac{1}{2} \cdot 2i = i. \end{aligned}

The other zero is −c-c.

Complex factorization

Definition Every polynomial PP of degree n≥1n \ge 1 with complex coefficients can be written as

P(z)=an(z−c1)(z−c2)⋯(z−cn),\begin{aligned} &P(z) \\[0.4em] &= a_n (z - c_1)(z - c_2) \cdots (z - c_n), \end{aligned}

where ana_n is the leading coefficient and c1,…,cnc_1, \ldots, c_n are complex numbers, not necessarily different. The list c1,…,cnc_1, \ldots, c_n is unique apart from its order, and it consists exactly of the zeros of PP. The proof repeats one step. The theorem gives a zero c1c_1, the Factor Theorem gives P(z)=(z−c1)Q(z)P(z) = (z - c_1)Q(z) with QQ of degree n−1n - 1, and the same argument applies to QQ until a constant is left.

In plain words Over the complex numbers, every polynomial breaks all the way down into linear factors, one for each unit of degree.

Example Over the reals, the polynomial z4−16z^4 - 16 splits into (z−2)(z+2)(z2+4)(z - 2)(z + 2)(z^2 + 4), and the last factor has no real zeros. Over C\mathbb{C} it splits completely,

z4−16=(z2−4)(z2+4)=(z−2)(z+2)(z−2i)(z+2i).\begin{aligned} &z^4 - 16 \\[0.4em] &= (z^2 - 4)(z^2 + 4) \\[0.4em] &= (z - 2)(z + 2)(z - 2i)(z + 2i). \end{aligned}

Counting zeros with multiplicity

Definition Let PP be a nonzero polynomial and cc a complex number. The multiplicity of cc is the largest integer m≥0m \ge 0 such that (z−c)m(z - c)^m divides P(z)P(z). The number cc is a zero exactly when m≥1m \ge 1. For a zero this is the multiplicity from the lesson on zeros and multiplicity, the exponent mm with P(z)=(z−c)mQ(z)P(z) = (z - c)^m Q(z) and Q(c)≠0Q(c) \ne 0. In the complex factorization, the multiplicity of cc is the number of factors equal to z−cz - c. It follows that a polynomial of degree n≥1n \ge 1 has exactly nn complex zeros when each zero is counted as many times as its multiplicity, and therefore at most nn different zeros.

In plain words The degree tells you how many zeros there are, as long as you count over the complex numbers and count a repeated zero once for each time its factor appears.

Example The polynomial

P(x)=(x−2)3(x2+1)=x5−6x4+13x3−14x2+12x−8\begin{aligned} P(x) &= (x - 2)^3 (x^2 + 1) \\[0.4em] &= x^5 - 6x^4 + 13x^3 \\[0.4em] &\quad - 14x^2 + 12x - 8 \end{aligned}

has degree 55. Since x2+1=(x−i)(x+i)x^2 + 1 = (x - i)(x + i), its zeros are

ZeroMultiplicity
2233
ii11
−i-i11

The multiplicities add up to 3+1+1=53 + 1 + 1 = 5, the degree, while there are only three different zeros and just one of them is real.

Real coefficients and conjugate pairs

Definition Let PP have real coefficients. If a complex number cc is a zero of PP, then its conjugate c‾\overline{c} is also a zero, with the same multiplicity. The reason for the first part is that conjugation respects sums and products and leaves real coefficients unchanged, so

P(c‾)=P(c)‾=0‾=0.P(\overline{c}) = \overline{P(c)} = \overline{0} = 0.

So the nonreal zeros of a real polynomial come in conjugate pairs. As a consequence, a polynomial with real coefficients and odd degree has at least one real zero. The rule needs real coefficients. The polynomial z−iz - i has the zero ii but not −i-i.

In plain words For a polynomial with real coefficients, the nonreal zeros are mirror images of each other across the real axis. Knowing one of them gives you its partner for free.

Example The number 1+2i1 + 2i is a zero of P(x)=x3−3x2+7x−5P(x) = x^3 - 3x^2 + 7x - 5, which has real coefficients. So 1−2i1 - 2i is a zero too. The sum of the coefficients is 1−3+7−5=01 - 3 + 7 - 5 = 0, so 11 is a zero as well. A cubic has exactly three zeros counted with multiplicity, so the zeros are 11, 1+2i1 + 2i and 1−2i1 - 2i. As a check, the pair gives the factor x2−2x+5x^2 - 2x + 5, and multiplying out confirms

(x−1)(x2−2x+5)=x3−3x2+7x−5.\begin{aligned} &(x - 1)(x^2 - 2x + 5) \\[0.4em] &\quad = x^3 - 3x^2 + 7x - 5. \end{aligned}

Real irreducible quadratic factors

Definition A quadratic ax2+bx+cax^2 + bx + c with real coefficients and a≠0a \ne 0 is irreducible over the reals when it is not a product of two linear factors with real coefficients. This happens exactly when it has no real zeros, that is when b2−4ac<0b^2 - 4ac < 0. For a complex number ww, multiplying the linear factors of the pair ww and w‾\overline{w} always gives a quadratic with real coefficients,

(x−w)(x−w‾)=x2−2Re⁡(w) x+∣w∣2,\begin{aligned} &(x - w)(x - \overline{w}) \\[0.4em] &\quad = x^2 - 2\operatorname{Re}(w)\,x + \lvert w \rvert^2, \end{aligned}

and it is irreducible when ww is not real. Grouping each conjugate pair this way shows that every polynomial with real coefficients and degree at least 11 is its leading coefficient times a product of real linear factors x−rx - r and real irreducible quadratic factors x2+px+qx^2 + px + q with p2−4q<0p^2 - 4q < 0.

In plain words If you want to stay with real numbers, you cannot always split a polynomial into linear pieces. You can always split it into linear pieces and quadratics with no real zeros, and that is as far as real factoring goes.

Example The polynomial x4+4x^4 + 4 has no real zeros, yet it does factor over the reals. Its complex zeros are 1±i1 \pm i and −1±i-1 \pm i. For instance (1+i)2=2i(1 + i)^2 = 2i, so (1+i)4=(2i)2=−4(1 + i)^4 = (2i)^2 = -4. The real coefficients make the conjugate 1−i1 - i a zero too, and since only even powers of xx appear, the negatives −1−i-1 - i and −1+i-1 + i are zeros as well. A polynomial of degree 44 has at most four different zeros, so these are all of them. Pairing conjugates gives

(x−(1+i))(x−(1−i))=x2−2x+2,(x−(−1+i))(x−(−1−i))=x2+2x+2.\begin{aligned} &(x - (1 + i))(x - (1 - i)) \\[0.4em] &\quad = x^2 - 2x + 2, \\[0.8em] &(x - (-1 + i))(x - (-1 - i)) \\[0.4em] &\quad = x^2 + 2x + 2. \end{aligned}

Their product is the real factorization

x4+4=(x2−2x+2)(x2+2x+2).\begin{aligned} &x^4 + 4 \\[0.4em] &\quad = (x^2 - 2x + 2)(x^2 + 2x + 2). \end{aligned}

Both quadratics have discriminant 4−8=−4<04 - 8 = -4 < 0, so neither splits further over the reals. This kind of real factorization is the starting point for integrating rational functions by partial fractions.

Check yourself

  1. List the zeros of P(x)=x3(x2+9)2P(x) = x^3 (x^2 + 9)^2 with their multiplicities, and check that the count matches the degree.
Show the answer

Since x2+9=(x−3i)(x+3i)x^2 + 9 = (x - 3i)(x + 3i), the zero 00 has multiplicity 33 and the zeros 3i3i and −3i-3i each have multiplicity 22. The total is 3+2+2=73 + 2 + 2 = 7, the degree of PP.

  1. The number 2+i2 + i is a zero of P(x)=x3−5x2+9x−5P(x) = x^3 - 5x^2 + 9x - 5. Find all zeros and factor P(x)P(x) over the reals.
Show the answer

The coefficients are real, so 2−i2 - i is also a zero. These two give the real factor

(x−(2+i))(x−(2−i))=x2−4x+5.\begin{aligned} &(x - (2 + i))(x - (2 - i)) \\[0.4em] &\quad = x^2 - 4x + 5. \end{aligned}

Dividing gives P(x)=(x−1)(x2−4x+5)P(x) = (x - 1)(x^2 - 4x + 5), so the zeros are 11, 2+i2 + i and 2−i2 - i. The quadratic is irreducible over the reals, since its discriminant is 16−20=−416 - 20 = -4.

  1. Find a polynomial with real coefficients, leading coefficient 11 and the least possible degree that has the zeros 22 and 3−i3 - i.
Show the answer

Real coefficients force 3+i3 + i to be a zero as well, so the degree is at least 33. The pair gives x2−6x+10x^2 - 6x + 10, and

(x−2)(x2−6x+10)=x3−8x2+22x−20.\begin{aligned} &(x - 2)(x^2 - 6x + 10) \\[0.4em] &\quad = x^3 - 8x^2 + 22x - 20. \end{aligned}

How this lesson connects

Where this leads in Calculus