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The fundamental theorem of algebra

Numbers · level 8, lesson 5 of 6

A question to start

The polynomial \(x^4 + 1\) is never zero for real \(x\): \(x^4\) is never negative, so \(x^4 + 1 \ge 1\). You might guess that it cannot be broken into smaller pieces with real numbers either. Yet

\[x^4 + 1 = \big(x^2 + \sqrt2\,x + 1\big)\big(x^2 - \sqrt2\,x + 1\big).\]

Multiply it out and check: the \(x^3\) terms cancel, the \(x^2\) terms give \(x^2 - 2x^2 + x^2 = 0\), and the \(x\) terms cancel. How would anyone find this? The answer is to step outside the real numbers, find the four complex roots of \(x^4 + 1\), and pair them up. Behind that method is one of the great theorems of mathematics: the complex numbers are enough. No polynomial equation ever needs a bigger family of numbers.

The idea

The theorem

A polynomial of degree \(n\) is an expression

\[p(z) = a_n z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0 \qquad (a_n \ne 0)\]

where the coefficients \(a_0, \ldots, a_n\) are numbers (here: complex numbers, which includes real ones). A root of \(p\) is a number \(r\) with \(p(r) = 0\).

The fundamental theorem of algebra. Every polynomial of degree \(n \ge 1\) with complex coefficients has at least one complex root.

That sounds modest. But with one more fact it becomes much stronger.

The factor theorem

Factor theorem. \(p(r) = 0\) exactly when \((z - r)\) is a factor of \(p(z)\), that is, \(p(z) = (z - r)q(z)\) for some polynomial \(q\) of degree \(n - 1\).

Why. If \(p(z) = (z - r)q(z)\) then clearly \(p(r) = 0\). For the other direction, look at

\[p(z) - p(r) = a_n(z^n - r^n) + a_{n-1}(z^{n-1} - r^{n-1}) + \cdots + a_1(z - r).\]

Each bracket \(z^k - r^k\) has \(z - r\) as a factor, because \(z^k - r^k = (z - r)(z^{k-1} + z^{k-2}r + \cdots + r^{k-1})\) (multiply out and everything cancels except the two end terms). So \(p(z) - p(r) = (z - r)q(z)\). If \(p(r) = 0\), this says \(p(z) = (z - r)q(z)\). ∎

\(n\) roots for degree \(n\)

Now combine the two. Take \(p\) of degree \(n \ge 1\). The theorem gives a root \(r_1\); the factor theorem gives \(p(z) = (z - r_1)q_1(z)\) with \(q_1\) of degree \(n - 1\). If \(n - 1 \ge 1\), the theorem applies to \(q_1\): it has a root \(r_2\), and \(q_1(z) = (z - r_2)q_2(z)\). Keep going. Each step lowers the degree by one, and after \(n\) steps only a constant is left, which must be \(a_n\). So:

Every polynomial of degree \(n \ge 1\) factors completely into \(n\) linear factors: \(\;p(z) = a_n(z - r_1)(z - r_2)\cdots(z - r_n)\).

The roots \(r_1, \ldots, r_n\) need not be different. If a factor \((z - r)\) appears \(m\) times, \(r\) is a root of multiplicity \(m\) (a double root for \(m = 2\), a triple root for \(m = 3\)). Then the statement is:

A polynomial of degree \(n\) has exactly \(n\) complex roots, counted with multiplicity.

It cannot have more: if \(r\) is not among \(r_1, \ldots, r_n\), then each \((r - r_k) \ne 0\), and a product of non-zero complex numbers is not zero, so \(p(r) \ne 0\).

Multiplicity on a graph. For a real polynomial, a real root of odd multiplicity is where the graph crosses the \(x\)-axis; a root of even multiplicity is where it touches and turns back.

−2: crosses 1: touches (double) y = (x − 1)²(x + 2)

What the theorem says, and what it does not

  • It says roots exist. It does not give a formula for them. There are formulas for degrees 2, 3 and 4 (the cubic one is in level 7). For degree 5 and higher there is no general formula using only \(+\), \(-\), \(\times\), \(\div\) and roots; Niels Henrik Abel proved this in 1824. The roots exist, but in general they can only be calculated numerically.
  • It is about complex coefficients and complex roots. Over \(\mathbb{R}\) it is false (\(x^2 + 1\) has no real root). So the theorem is really a statement about how good \(\mathbb{C}\) is: building one new number \(i\) was enough to solve every polynomial equation, even ones whose coefficients involve \(i\). Mathematicians say \(\mathbb{C}\) is algebraically closed. This is where the ladder of level 7, lesson 1 stops for polynomial equations.
  • It is hard to prove. Many mathematicians attempted it in the 1700s. Carl Friedrich Gauss's doctoral thesis of 1799 is often counted as the first serious proof, although by modern standards it has a gap; complete proofs came in the 1800s. Every known proof uses some idea from outside algebra, such as continuity or the "no holes" property of \(\mathbb{R}\), which is why it sits more naturally in analysis than its name suggests.

A picture of why it is true (not a proof)

Think of \(p(z) = z^n + \cdots + a_0\) as a machine that sends each point \(z\) of the plane to a point \(p(z)\). Let \(z\) walk once round a circle of radius \(R\) centred at 0, and watch the path traced by \(p(z)\).

  • A huge circle. When \(R\) is very large, the term \(z^n\) is far bigger than all the others, so \(p(z)\) behaves like \(z^n\). By De Moivre, as \(z\) goes round once, \(z^n\) goes round 0 \(n\) times. So the path of \(p(z)\) winds round 0 \(n\) times.
  • A tiny circle. When \(R\) is very small, \(p(z)\) stays close to \(p(0) = a_0\). If \(a_0 \ne 0\), the path is a small loop near \(a_0\) that does not go round 0 at all.

Now shrink the circle slowly from huge to tiny. The path changes smoothly, from winding \(n\) times round 0 to winding 0 times. A loop cannot "unwind" from round a point without passing through that point at some stage. At that moment some \(z\) on the circle has \(p(z) = 0\): a root. Making "winding" and "passing through" exact is the work of a real proof, which you will meet in university analysis.

Real polynomials: linear and quadratic factors

If the coefficients are real, level 7 showed the non-real roots come in conjugate pairs \(w\), \(\bar w\). Multiply each pair's factors together:

\[(z - w)(z - \bar w) = z^2 - (w + \bar w)z + w\bar w = z^2 - 2\operatorname{Re}(w)\,z + |w|^2,\]

a quadratic with real coefficients (and negative discriminant). So:

Every polynomial with real coefficients factors into real linear factors and real quadratic factors with no real roots.

That is exactly how the opening factorisation of \(x^4 + 1\) is found.

Sums and products of roots

Expanding \(a_n(z - r_1)\cdots(z - r_n)\) and comparing with \(p(z)\) gives the relations named after François Viète. The two simplest are

\[r_1 + r_2 + \cdots + r_n = -\frac{a_{n-1}}{a_n}, \qquad r_1 r_2 \cdots r_n = (-1)^n\frac{a_0}{a_n}.\]

For \(z^n - 1\) the coefficient of \(z^{n-1}\) is 0 (when \(n \ge 2\)), so the roots of unity add to 0: a third proof of the fact in lesson 3.

Worked examples

1. Factor \(x^4 + 1\).

  1. Roots: \(x^4 = -1 = e^{i\pi}\). By lesson 3 the roots have length 1 and angles \(\tfrac\pi4, \tfrac{3\pi}4, \tfrac{5\pi}4, \tfrac{7\pi}4\), that is \(\dfrac{\pm1 \pm i}{\sqrt2}\) (all four sign choices).
  2. Pair conjugates. \(w = e^{i\pi/4} = \tfrac{1 + i}{\sqrt2}\) and \(\bar w\): \(2\operatorname{Re}(w) = \sqrt2\) and \(|w|^2 = 1\), giving \(x^2 - \sqrt2\,x + 1\).
  3. \(w = e^{3i\pi/4} = \tfrac{-1 + i}{\sqrt2}\) and \(\bar w\): \(2\operatorname{Re}(w) = -\sqrt2\), giving \(x^2 + \sqrt2\,x + 1\).
  4. So \(x^4 + 1 = (x^2 - \sqrt2\,x + 1)(x^2 + \sqrt2\,x + 1)\) over \(\mathbb{R}\), and over \(\mathbb{C}\) it is the product of the four factors \(\left(x - \tfrac{\pm 1 \pm i}{\sqrt2}\right)\).
  5. A purely algebraic route, for comparison: \(x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2 = (x^2 + 1)^2 - (\sqrt2\,x)^2\), a difference of two squares. The complex roots told us that such a trick must exist.

2. Factor \(x^3 - 8\) over \(\mathbb{R}\) and over \(\mathbb{C}\).

  1. \(x = 2\) is a root, so \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\).
  2. \(x^2 + 2x + 4\) has \(D = 4 - 16 = -12 < 0\), so it is the real quadratic factor. Over \(\mathbb{R}\) we stop.
  3. Its roots are \(\dfrac{-2 \pm 2i\sqrt3}{2} = -1 \pm i\sqrt3\). Over \(\mathbb{C}\): \(x^3 - 8 = (x - 2)(x + 1 - i\sqrt3)(x + 1 + i\sqrt3)\). (These are 2, \(2\omega\), \(2\omega^2\).)

3. Multiplicity. Find the roots of \(p(x) = x^3 - 3x + 2\) with their multiplicities.

  1. Try small integers: \(p(1) = 1 - 3 + 2 = 0\). So \((x - 1)\) is a factor: \(p(x) = (x - 1)(x^2 + x - 2)\).
  2. \(x^2 + x - 2 = (x - 1)(x + 2)\). So \(p(x) = (x - 1)^2(x + 2)\).
  3. Roots: \(1\) with multiplicity 2, and \(-2\) with multiplicity 1. Count: \(2 + 1 = 3\), the degree. ✓ The graph above touches at 1 and crosses at \(-2\).

4. Build a polynomial. Find the real polynomial of lowest degree, with leading coefficient 1, that has a double root at 1 and a root at \(2i\).

  1. Real coefficients force \(-2i\) to be a root too. So the roots are \(1, 1, 2i, -2i\): degree 4.
  2. \((x - 2i)(x + 2i) = x^2 + 4\) and \((x - 1)^2 = x^2 - 2x + 1\).
  3. \(p(x) = (x^2 - 2x + 1)(x^2 + 4) = x^4 - 2x^3 + 5x^2 - 8x + 4\).

Try it

  1. How many complex roots, counted with multiplicity, does \(3z^7 - z + 5\) have? Must at least one be real? Why?
  2. Factor \(x^4 - 16\) completely over \(\mathbb{C}\), and over \(\mathbb{R}\).
  3. Find the roots and multiplicities of \((x^2 - 1)^3\).
  4. Factor \(x^4 + 4\) into two real quadratics. (Hint: find the roots of \(x^4 = -4\), or write it as \((x^2 + 2)^2 - (2x)^2\).)
  5. Given that \(2i\) is a root of \(x^4 - 2x^3 + 6x^2 - 8x + 8\), find all its roots.
  6. Factor \(x^6 - 1\) into real linear and quadratic factors.
  7. Without solving, find the sum and the product of the roots of \(2x^3 - 5x^2 + 4x - 7\).
  8. Explain why a real polynomial of degree 4 cannot have exactly three non-real roots (counted with multiplicity).
Answers
  1. Degree 7, so 7 roots counted with multiplicity. The coefficients are real, so the non-real roots pair off. Seven is odd, so at least one root is left unpaired, and it must be real.
  2. \(x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)\) over \(\mathbb{R}\). Over \(\mathbb{C}\), also split \(x^2 + 4 = (x - 2i)(x + 2i)\).
  3. \((x^2 - 1)^3 = (x - 1)^3(x + 1)^3\): root \(1\) with multiplicity 3 and root \(-1\) with multiplicity 3. Total 6, the degree.
  4. \((x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2)\). (The roots of \(x^4 = -4\) are \(\pm1 \pm i\); the pair \(1 \pm i\) gives \(x^2 - 2x + 2\), the pair \(-1 \pm i\) gives \(x^2 + 2x + 2\).)
  5. Real coefficients, so \(-2i\) is also a root, and \((x - 2i)(x + 2i) = x^2 + 4\) is a factor. Dividing: \(x^4 - 2x^3 + 6x^2 - 8x + 8 = (x^2 + 4)(x^2 - 2x + 2)\). The second factor has roots \(1 \pm i\). All roots: \(2i, -2i, 1 + i, 1 - i\).
  6. \(x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)\). The two quadratics have negative discriminants (\(1 - 4 = -3\)), so this is complete over \(\mathbb{R}\). Their roots are the four non-real sixth roots of unity.
  7. Sum \(= -\dfrac{-5}{2} = \dfrac52\). Product \(= (-1)^3\dfrac{-7}{2} = \dfrac72\).
  8. Non-real roots of a real polynomial come in conjugate pairs, each pair with matching multiplicities, so the number of non-real roots is even. Three is odd.

Common mistake

Thinking the theorem gives a way to find the roots. It only promises that they exist. For most polynomials of degree 5 or more, the roots can only be found approximately, by numerical methods such as Newton's method.

Forgetting multiplicity when counting. \((x - 1)^2(x + 2)\) has only two different roots, but three roots counted with multiplicity. "Degree \(n\) means \(n\) roots" is only true with that counting.

Where it leads