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Week 02 · Every Polynomial Factors

Tuesday

Every polynomial has its zeros
// Planks in Pascal's rows
⏱ about 20 min

Tuesday: Every Polynomial Has Its Zeros

"Four quadratics on the chalkboard," Wren says. "Two have real zeros. One has a double zero. One looked like it had none."

"x² + 9 has zeros now," Comet says. "-3i and 3i. What do you notice about the counts?"

"Two, two, two and two," Wren says. "Every quadratic has exactly two zeros, if you count 3 twice for x² - 6x + 9."

Nova projects a cubic beside them. "Would you like a hint? Try the same count for degree 3."

"x³ - x is x(x - 1)(x + 1). Three zeros," Comet says. "Degree 3, three zeros. Is that always true?"

"That is the Fundamental Theorem of Algebra," Wren says. "Degree n means n complex zeros, counting repeats."

"Then nothing is unfactorable," Comet says. "Every polynomial splits all the way down into linear pieces."

The theorem

The Fundamental Theorem of Algebra: every polynomial of degree n with complex coefficients has exactly n complex zeros, counting multiplicity.

Multiplicity means a repeated zero is counted each time it repeats. The zero 3 in (x - 3)² counts twice.

So every polynomial of degree n factors into n linear factors (x - z₁)(x - z₂)...(x - zₙ), times a constant.

The theorem does not say the zeros are real. It says they exist, somewhere on the complex plane.

Way one: factor when you can

x² - 5x + 6 = (x - 2)(x - 3). Two real zeros, 2 and 3.

x² - 6x + 9 = (x - 3)². One zero, 3, with multiplicity 2. Still two zeros when counted.

x² + 9 = (x + 3i)(x - 3i). Two complex zeros, -3i and 3i, from Monday.

Way two: the quadratic formula always works

For x² - 4x + 13, factoring by eye is hard. The quadratic formula gives x = (4 ± √(16 - 52))/2 = (4 ± √(-36))/2.

√(-36) = 6i, so x = (4 ± 6i)/2 = 2 - 3i and 2 + 3i. Two zeros again, a conjugate pair.

The discriminant b² - 4ac decides the kind. Positive gives two real zeros, zero gives one double zero, negative gives a conjugate pair.

Whatever the sign, the count is two. That is the theorem for degree 2.

The chalkboard quadratics

QuadraticDiscriminantZerosCount with multiplicity
x² - 5x + 612 and 32
x² - 6x + 903 (twice)2
x² + 9-36-3i and 3i2
x² - 4x + 13-362 - 3i and 2 + 3i2
COUNT THE ZEROS
  • Read the question.
  • Tap your answer.
Counting multiplicity, how many complex zeros does x² + 4 have?
Nova writes x³ - 2x² + x on the board. Counting multiplicity, how many complex zeros does it have?
Counting multiplicity, how many complex zeros does x⁵ - 3x² + 1 have?
Counting multiplicity, how many complex zeros does (x² + 4)(x² - 1) have?
FIND THE ZEROS
  • Read the question.
  • Tap your answer.
Solve x² + 4 = 0. What are its two zeros?
What are the solutions of x² - 4x + 13 = 0?
What are the zeros of p(x) = x³ - x?
What is the multiplicity of the zero 3 in (x - 3)² (x + 2)?
READ THE DISCRIMINANT
  • Read the question.
  • Tap your answer.
x² - 4x + 13 = 0 has a discriminant of -36. How many solutions does it have, and of what kind?
x² - 5x + 6 = 0 has a discriminant of 1. How many solutions does it have, and of what kind?
A polynomial of degree 7 has how many complex zeros, counting multiplicity? Type the number.
In x² - 6x + 9 = (x - 3)², the zero 3 has what multiplicity? Type the number.
A negative discriminant gives two zeros of what kind? Type one word.
StatementTrue or false?
Every quadratic has exactly two complex zeros, counting multiplicity.?
A polynomial of degree 4 can have five zeros.?
The zeros of a quadratic with a negative discriminant are a conjugate pair.?
x² + 9 has no zeros.?
4 × 4 - 4 × 13 = -36?
WHY THIS EXERCISEThe Fundamental Theorem of Algebra turns factoring from a hope into a promise.

Excellent counting. Tomorrow is Boathouse Lab: the planks come off the hull and stack into Pascal's triangle.

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