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# The Fundamental Theorem of Algebra Help (page 2)

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By — McGraw-Hill Professional
Updated on Oct 4, 2011

## Multiplicity of Zeros

A consequence of the Fundamental Theorem of Algebra is that a polynomial of degree n will have n zeros, though not necessarily n different zeros. For example, the polynomial f(x) = ( x − 2) 3 = ( x − 2)( x − 2)( x − 2) has x = 2 as a zero three times. The number of times an x -value is a zero is called its multiplicity. In the above example, x = 2 is a zero with multiplicity 3.

#### Example

• f(x) = x 4 ( x + 3) 2 ( x − 6)
• x = 0 is a zero with multiplicity 4 (We can think of x 4 as ( x − 0) 4 .)

x = −3 is a zero with multiplicity 2

x = 6 is a zero with multiplicity 1

Now, instead of finding the zeros for a given polynomial, we will find a polynomial with the given zeros. Because we will know the zeros, we will know the factors. Once we know the factors of a polynomial, we pretty much know the polynomial.

#### Examples

Find a polynomial with integer coefficients having the given degree and zeros.

• Degree 3 with zeros 1, 2, and 5 Because x = 1 is a zero, x − 1 is a factor. Because x = 2 is a zero, x − 2 is a factor. And because x = 5 is a zero, x − 5 is a factor. Such a polynomial will be of the form a ( x − 1)( x − 2)( x − 5), where a is some nonzero number. We will want to choose a so that the coefficients are integers.

a ( x − 1)( x − 2)( x − 5) = a(x − 1)[( x − 2)( x − 5)]

= a(x − 1)( x 2 − 7 x + 10)

= a(x 3 − 7 x 2 + 10 xx 2 + 7 x − 10)

= a(x 3 − 8 x 2 + 17 x − 10)

Because the coefficients are already integers, we can let a = 1. One polynomial of degree three having integer coefficients and 1, 2, and 5 as zeros is x 3 − 8 x 2 + 17 x − 10.

• Degree 4 with zeros −3, 2 − 5 i , with −3 a zero of multiplicity 2 Because −3 is a zero of multiplicity 2, ( x + 3) 2 = x 2 + 6 x + 9 is a factor. Because 2 − 5 i is a zero, 2 + 5 i is another zero. Another factor of the polynomial is

( x − (2 − 5 i )) ( x − (2 + 5 i )) = ( x − 2 + 5 i )( x − 2 − 5 i )

= x 2 − 2 x − 5 ix − 2 x + 4 + 10 i + 5 ix

− 10 i − 25 i 2

= x 2 − 4 x + 4 − 25(−1) = x 2 −4 x + 29.

The polynomial has the form a ( x 2 + 6 x + 9)( x 2 − 4 x + 29), where a is any real number that makes all coefficients integers.

a ( x 2 + 6 x + 9)( x 2 − 4 x + 29) = a ( x 4 − 4 x 3 + 29 x 2 + 6 x 3 − 24 x 2 + 174 x + 9 x 2 − 36 x + 261)

= a ( x 4 + 2 x 3 + 14 x 2 + 138 x + 261)

Because the coefficients are already integers, we can let a = 1. One polynomial that satisfies the given conditions is x 4 + 2 x 3 + 14 x 2 + 138 x + 261.

In the previous problems, there were infinitely many answers because a could be any integer. In the following problem, there will be exactly one polynomial that satisfies the given conditions. This means that a will likely be a number other than 1.

• Degree 3 with zeros − 1, −2, and 4, where the coefficient for x is − 20

a ( x + 1)( x + 2)( x − 4) = a( x + 1)[( x + 2)( x − 4)]

= a ( x + 1)( x 2 − 2 x − 8)

= a ( x 3 − 2 x 2 − 8 x + x 2 − 2 x − 8)

= a ( x 3x 2 − 10 x −8)

= ax 3ax 210ax − 8 a

Because we need the coefficient of x to be −20, we need − 10 ax = −20 x , so we need a = 2 (from −10 a = −20). The polynomial that satisfies the conditions is 2 x 3 − 2 x 2 − 20 x − 16.

Find practice problems and solutions for these concepts at  The Fundamental Theorem of Algebra Practice Problems.

More practice problems for this concept can be found at: Polynomial Functions Practice Test.

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