To review these concepts, go to The Fundamental Theorem of Algebra Help
The Fundamental Theorem of Algebra Practice Problems
Problems
Directions: For Problems 16 solve the equations and write complex solutions in the form a + bi, where a and b are real numbers.
 9 x ^{2} + 4 = 0
 2. 6 x ^{2} + 8 x + 9 = 0
 3. x ^{4} − 81 = 0
 4. x ^{3} + 13 x − 34 = 0
 5. x ^{4} − x ^{3} + 8 x ^{2} − 9 x − 9 = 0; x = −3 i is a solution.
 6. x ^{3} − 5 x ^{2} + 7 x + 13 = 0; x = 3 − 2 i is a solution.
Directions: For Problems 710 find a polynomial with integer coefficients having the given conditions.
 Degree 3 with zeros 0, −4, and 6
 Degree 4 with zeros −1 and 6 − 7 i , where x = − 1 has multiplicity 2.
 Degree 3, zeros 4, and ±1, with leading coefficient 3
 Degree 4 with zeros i and 4i , with constant term −16
 State each zero and its multiplicity for f(x) = x ^{2} ( x + 4)( x + 9) ^{6} ( x − 5) ^{3}
Solutions

9 x ^{2} + 4 = 0
9 x ^{2} = −4


x ^{4} − 81 = ( x ^{2} − 9)( x ^{2} + 9) = ( x − 3)( x + 3)( x ^{2} + 9)
x ^{2} + 9 = 0
x ^{2} = −9
The solutions are ±3, ±3 i .

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.
The quotient is x ^{2} + 2 x + 17. We will find the other solutions by solving x ^{2} + 2 x + 17 = 0.
The solutions are 2 and −1 ±4 i .

x = −3 i is a solution, so x = 3 i is a solution, also. One factor of x ^{4} − x ^{3} + 8 x ^{2} − 9 x − 9 is ( x − 3 i ) ( x + 3 i ) = x ^{2} + 9 = x ^{2} + 0 x + 9.
Solve x ^{2} − x − 1 = 0.
The solutions are ±3 i , .

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.
The solutions are 3 ± 2 i and −1.

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 .

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.

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 ^{2} − x + 4)
= ax ^{3} − 4 ax ^{2} − ax + 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.

The factors are x + i , x − i , x − 4 i , and x + 4i .
a(x + i) ( x − i )( 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.

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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