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.
Ask a Question
Have questions about this article or topic? AskRelated Questions
See More QuestionsPopular Articles
 Kindergarten Sight Words List
 First Grade Sight Words List
 10 Fun Activities for Children with Autism
 Grammar Lesson: Complete and Simple Predicates
 Definitions of Social Studies
 Child Development Theories
 Signs Your Child Might Have Asperger's Syndrome
 How to Practice Preschool Letter and Name Writing
 Social Cognitive Theory
 Theories of Learning