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Complex Factoring and Sphere Volume: Precalculus Math Lesson

Last updated: October 22, 2026 · Written by the Math Made Easy math teachers

Precalculus Math Lessons – Mathbot, the Math Made Easy robot coach, with free printable lessons for Precalculus

Quick answer: Because i² = −1, a sum of squares factors over the complex numbers: a² + b² = (a + bi)(a − bi). So x² + 4 = (x + 2i)(x − 2i). The volume of a sphere, (4/3)πr³, follows from Cavalieri’s principle by comparing a hemisphere with a cylinder minus a cone.

Standards: N-CN.C.8, G-GMD.A.2 · Level: Grade 12 Precalculus

How do the algebra identities extend to complex numbers?

Since i² = −1, a sum of squares is no longer prime. The difference-of-squares pattern a² − b² = (a + b)(a − b) becomes a² + b² = (a + bi)(a − bi). Check: (a + bi)(a − bi) = a² − b²i² = a² + b².

Sum of squares over the complex numbers

a² + b² = (a + bi)(a − bi). In particular, x² + k = (x + √k·i)(x − √k·i) for any k > 0.

Worked example

x² + 4 = (x + 2i)(x − 2i). Setting each factor to zero gives the solutions x = ±2i, which sit on the imaginary axis of the complex plane.

-4-3-2-11234-4-3-2-11234ReIm2i−2i
The zeros of x² + 4 are 2i and −2i.

How do you factor polynomials completely over the complex numbers?

Method

  1. Factor out any common factor and apply real factoring first, such as a difference of squares.
  2. Split every sum of squares with a² + b² = (a + bi)(a − bi).
  3. For a quadratic with a negative discriminant, use the quadratic formula; the zeros come in conjugate pairs.
  4. Multiply the factors back to check.

For example, x⁴ − 16 = (x² − 4)(x² + 4) = (x − 2)(x + 2)(x + 2i)(x − 2i), so it has four zeros: ±2 and ±2i.

Common mistake

Writing x² + 9 = (x + 3)². Expanding gives x² + 6x + 9, which has a middle term. The correct factors have opposite signs: (x + 3i)(x − 3i).

Why is the volume of a sphere (4/3)πr³?

Cavalieri’s principle says two solids with the same height have equal volumes if every horizontal slice has equal area. Compare a hemisphere of radius r with a cylinder of radius r and height r that has an upside-down cone removed. At height h, the hemisphere’s slice is a disk of radius √(r² − h²), with area π(r² − h²). The cylinder-minus-cone slice is a ring with outer radius r and inner radius h, with area πr² − πh² = π(r² − h²). The areas match at every height.

hr√(r² − h²)hringHemisphere, radius rCylinder minus coner
At every height h both slices have the same area, π(r² − h²).

So the hemisphere’s volume equals the cylinder minus the cone: πr³ − 13πr³ = 23πr³. A full sphere doubles this: V = 43πr³.

-4-3-2-11234-4-3-2-11234ReIm2−22i−2i
The four zeros of x⁴ − 16 form a square around the origin.
Mathbot’s tip

When a sum of squares appears, look for its square roots, then give the factors opposite signs and add an i to the second root.

Key points to remember

  • i² = −1, and powers of i repeat in a cycle of 4: i, −1, −i, 1.
  • a² + b² = (a + bi)(a − bi); for example x² + 25 = (x + 5i)(x − 5i).
  • Complete factoring over the complex numbers splits every sum of squares, so degree n gives n zeros.
  • Cavalieri’s principle: equal heights and equal slice areas at every level mean equal volumes.
  • Hemisphere = cylinder − cone = (2/3)πr³, so a sphere has volume (4/3)πr³.

Complex Factoring and Sphere Volume: frequently asked questions

Why can x² + 4 be factored over the complex numbers?

Because (x + 2i)(x − 2i) = x² − 4i² = x² + 4. The imaginary unit supplies the negative sign that a sum of squares lacks over the real numbers.

How do I evaluate a large power of i?

The powers repeat every four steps: i, −1, −i, 1. Divide the exponent by 4 and use the remainder. For example, i²⁷ has remainder 3, so i²⁷ = i³ = −i.

What does Cavalieri’s principle say?

If two solids have the same height and the horizontal slices at every height have equal areas, the solids have equal volumes. It lets you find a hard volume from an easy one.

Why do the hemisphere and the cylinder-minus-cone have the same slices?

At height h, the hemisphere’s slice has radius √(r² − h²) by the Pythagorean Theorem, giving area π(r² − h²). The ring has area πr² − πh², which is the same expression.

How many zeros does a polynomial have over the complex numbers?

A polynomial of degree n has exactly n zeros when repeated zeros are counted. For x⁴ − 16 they are 2, −2, 2i and −2i.

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