The Fundamental Theorem of Algebra Help

• 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

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 x − x 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 ³ − x ² − 10 x −8)

  = ax ³ − ax ² − 10ax − 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.