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The Fundamental Theorem of Algebra Practice Problems

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

To review these concepts, go to The Fundamental Theorem of Algebra Help

The Fundamental Theorem of Algebra Practice Problems

Problems

Directions: For Problems 1-6 solve the equations and write complex solutions in the form a + bi, where a and b are real numbers.

  1. 9 x 2 + 4 = 0
  2. 2. 6 x 2 + 8 x + 9 = 0
  3. 3. x 4 − 81 = 0
  4. 4. x 3 + 13 x − 34 = 0
  5. 5. x 4x 3 + 8 x 2 − 9 x − 9 = 0; x = −3 i is a solution.
  6. 6. x 3 − 5 x 2 + 7 x + 13 = 0; x = 3 − 2 i is a solution.

Directions: For Problems 7-10 find a polynomial with integer coefficients having the given conditions.

  1. Degree 3 with zeros 0, −4, and 6
  2. Degree 4 with zeros −1 and 6 − 7 i , where x = − 1 has multiplicity 2.
  3. Degree 3, zeros 4, and ±1, with leading coefficient 3
  4. Degree 4 with zeros i and 4i , with constant term −16
  5. State each zero and its multiplicity for f(x) = x 2 ( x + 4)( x + 9) 6 ( x − 5) 3

Solutions

  1. 9 x 2 + 4 = 0

    9 x 2 = −4

    Polynomial Functions Solutions

    Polynomial Functions Solutions

  2.  

     Polynomial Functions Solutions

  3. x 4 − 81 = ( x 2 − 9)( x 2 + 9) = ( x − 3)( x + 3)( x 2 + 9)

    x 2 + 9 = 0

    x 2 = −9

    Polynomial Functions Solutions

    The solutions are ±3, ±3 i .

  4. x = 2 is a solution, so x − 2 is a factor of x 3 + 13 x − 34. Using synthetic division, we can find the quotient, which will be another factor.

    Polynomial Functions Solutions

    The quotient is x 2 + 2 x + 17. We will find the other solutions by solving x 2 + 2 x + 17 = 0.

    Polynomial Functions Solutions

     

    Polynomial Functions Solutions

    The solutions are 2 and −1 ±4 i .

  5. x = −3 i is a solution, so x = 3 i is a solution, also. One factor of x 4x 3 + 8 x 2 − 9 x − 9 is ( x − 3 i ) ( x + 3 i ) = x 2 + 9 = x 2 + 0 x + 9.

    Polynomial Functions Solutions

    Solve x 2x − 1 = 0.

    Polynomial Functions Solutions

    The solutions are ±3 i , Polynomial Functions Solutions .

  6. x = 3 − 2 i is a solution, so x = 3 + 2i is also a solution. One factor of x 3 - 5 x 2 + 7 x + 13 is

    ( x −(3 − 2 i ))( x − (3 + 2 i )) = ( x − 3 + 2 i )( x − 3 − 2 i )

    = x 2 − 3 x − 2 ix − 3 x + 9 + 6 i + 2 ix − 6 i −4 i 2

    = x 2 − 6 x + 9 − 4(−1) = x 2 − 6 x + 13.

    Polynomial Functions Solutions

    The solutions are 3 ± 2 i and −1.

  7. One polynomial with integer coefficients, with degree 3 and zeros 0, −4 and 6 is

    x ( x + 4)( x − 6) = x ( x 2 − 2 x − 24) = x 3 − 2 x 2 − 24 x .

  8. One polynomial with integer coefficients, with degree 4 and zeros − 1, 6 − 7 i , where x = − 1 has multiplicity 2 is

    ( x +1) 2 ( x −(6−7 i ))( x −(6+7 i )) = ( x +1) 2 ( x −6+7 i )( x −6−7 i )

    = [( x +1)(x+1)][ x 2 −6 x −7 ix −6 x +36+42 i +7 ix −42 i −49 i 2 ]

    = ( x 2 +2 x +1)( x 2 −12 x +85)

    = x 4 −12 x 3 +85 x 2 +2 x 3 −24 x 2 +170 x + x 2 −12 x +85

    = x 4 −10 x 3 +62 x 2 +158 x +85.

  9. The factors are x − 4, x − 1, and x + 1.

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

    = a [ (x − 4) (x 2 − 1)] = a(x 3 − 4 x 2x + 4)

    = ax 3 − 4 ax 2ax + 4 a

    We want the leading coefficient to be 3, so a = 3. The polynomial that satisfies the conditions is 3 x 3 − 12 x 2 − 3 x + 12.

  10. The factors are x + i , xi , x − 4 i , and x + 4i .

    a(x + i) ( xi )( x −4 i )( x +4 i ) = a [( x +i)( x −i)][( x −4 i )( x +4 i )]

    = a(x 2 +1)( x 2 +16)= a ( x 4 +17 x 2 +16)

    = ax 4 +17 ax 2 +16 a

    We want 16 a = −16, so a = − 1. The polynomial that satisfies the conditions is − x 4 − 17 x 2 − 16.

  11. x = 0 is a zero with multiplicity 2.

    x = −4 is a zero with multiplicity 1.

    x = −9 is a zero with multiplicity 6.

    x = 5 is a zero with multiplicity 3.

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