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Read polynomial zeros and multiplicities from factored form

Set each factor to zero

A factor $(x - r)^{m}$ gives a zero at $x = r$ with multiplicity $m$ . A nonzero constant factor changes neither the zeros nor their multiplicities.

Problem

For $f (x) = (x - 5) (x - 13) (x + 12)^{2} (x + 9)^{3}$ , list the zeros of multiplicity one, two and three.

Answer

Multiplicity 1:

5, 13

. Multiplicity 2:

- 12

. Multiplicity 3:

- 9

The exponents on the factors are the multiplicities.

Step-by-step solution

1. Find each zero

Set each linear factor to zero: $x - 5 = 0$ gives 5, $x - 13 = 0$ gives 13, $x + 12 = 0$ gives -12, and $x + 9 = 0$ gives -9.

2. Read the exponents

The first two factors each occur once, so 5 and 13 have multiplicity one. The factor $(x + 12)^{2}$ makes -12 a double zero; $(x + 9)^{3}$ makes -9 a triple zero. No expansion is needed.

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