Section 4.7: Complex Zeros; Fundamental Theorem of Algebra

Learning Outcomes

  • Use the Conjugate Pairs Theorem.
  • Find a polynomial function with specified zeros.
  • Find the complex zeros of a polynomial function.

In a previous section we found the real solutions of a quadratic equation. That is, we found the real zeros of a polynomial function of degree 2. Then in a previous section we found the complex solutions of a quadratic equation. That is, we found the complex zeros of a polynomial function of degree 2. In the last section we found the real zeros of a polynomial function of degree 3 or higher. In this section we will find the complex zeros of polynomial functions of degree 3 or higher.

Use the Fundamental Theorem of Algebra

Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.

Suppose f is a polynomial function of degree four, and f(x)=0f(x)=0. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it c1c1. By the Factor Theorem, we can write f(x)f(x) as a product of xc1xc1 and a polynomial quotient. Since xc1xc1 is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it c2c2. So we can write the polynomial quotient as a product of xc2xc2 and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of f(x)f(x).

A General Note: The Fundamental Theorem of Algebra states that, if f(x) is a polynomial of degree n > 0, then f(x) has at least one complex zero.

We can use this theorem to argue that, if f(x)f(x) is a polynomial of degree n>0n>0, and a is a non-zero real number, then f(x)f(x) has exactly n linear factors

f(x)=a(xc1)(xc2)(xcn)f(x)=a(xc1)(xc2)(xcn)

where c1,c2,,cnc1,c2,,cn are complex numbers. Therefore, f(x)f(x) has n roots if we allow for multiplicities.

Q & A

Does every polynomial have at least one imaginary zero?

No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.

Example 1: Finding the Zeros of a Polynomial Function with Complex Zeros

Find the zeros of f(x)=3x3+9x2+x+3f(x)=3x3+9x2+x+3.

Try It

Find the zeros of f(x)=2x3+5x211x+4f(x)=2x3+5x211x+4.

Try It

Use the Linear Factorization Theorem to find polynomials with given zeros

A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree n will have n zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into n factors. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form (x – c), where c is a complex number.

Let f be a polynomial function with real coefficients, and suppose a+bib0a+bib0, is a zero of f(x)f(x). Then, by the Factor Theorem, x(a+bi)x(a+bi) is a factor of f(x)f(x). For f to have real coefficients, x(abi)x(abi) must also be a factor of f(x)f(x). This is true because any factor other than x(abi)x(abi), when multiplied by x(a+bi)x(a+bi), will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function f with real coefficients has a complex zero a+bia+bi, then the complex conjugate abiabi must also be a zero of f(x)f(x). This is called the Complex Conjugate Theorem.

A General Note: Complex Conjugate Theorem

According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form (xc)(xc), where c is a complex number.

If the polynomial function f has real coefficients and a complex zero in the form a+bia+bi, then the complex conjugate of the zero, abiabi, is also a zero.

Example 2: Using the Conjugate Pairs Theorem

A polynomial function f of degree 5 whose coefficients are real numbers has the zeros 1, 5i5i, and 1+i1+i. Find the remaining zeros.

How To: Given the zeros of a polynomial function ff and a point (cf(c))(cf(c)) on the graph of ff, use the Linear Factorization Theorem to find the polynomial function.

  1. Use the zeros to construct the linear factors of the polynomial.
  2. Multiply the linear factors to expand the polynomial.
  3. Substitute (c,f(c))(c,f(c)) into the function to determine the leading coefficient.
  4. Simplify.

Example 3: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros

Find a fourth degree polynomial with real coefficients that has zeros of –3, 2, i, such that f(2)=100f(2)=100.

Q & A

If 2 + 3i were given as a zero of a polynomial with real coefficients, would 2 – 3i also need to be a zero?

Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.

Try It

Find a third degree polynomial with real coefficients that has zeros of 5 and –2i such that f(1)=10f(1)=10.

Try It

Example 4: Use the given complex zero to find the remaining Zeros

The polynomial 3x4+5x3+25x2+45x183x4+5x3+25x2+45x18 has a zero of 3i3i. Use this zero to find the other zeros.

Try It

The polynomial 2x45x3+11x220x+122x45x3+11x220x+12 has a zero of 2i2i. Use this zero to find the other zeros.

Key Concepts

  • According to the Fundamental Theorem, every polynomial function has at least one complex zero.
  • Every polynomial function with degree greater than 0 has at least one complex zero.

Glossary

Fundamental Theorem of Algebra
a polynomial function with degree greater than 0 has at least one complex zero

 

Section 4.7 Homework Exercises

In the exercises 1-6, information is given about a polynomial function f whose coefficients are real numbers.
Find the remaining zeros of f.

1. Degree 3; zeros: 3, 4i4i

2. Degree 3; zeros: 4, 3+i3+i

3. Degree 4; zeros: ii, 4i4i

4. Degree 4; zeros: ii, 2+i2+i

5. Degree 5; zeros: 1, ii, 5i5i

6. Degree: 5; zeros: 0, 1, 2, ii

In exercises 7-12, find a polynomial function f with real coefficients (with a leading coefficient of a) having the given degree and zeros.

7. Degree 4; zeros: 4 (multiplicity 2); 3+i3+i

8. Degree 4; zeros:  ii1+2i1+2i

9. Degree 5; zeros:  2;ii; 1+i1+i

10. Degree 6; zeros:  ii, 4i4i; 2+i2+i

11. Degree 4; zeros: 3 (multiplicity 2); ii

12. Degree 5; zeros: 1 (multiplicity 3); 1+i1+i

In problems 13-18, use the given zero to find all remaining zeros of each polynomial function.

13. f(x)=x35x2+9x45f(x)=x35x2+9x45; zero: 3i3i

14. f(x)=x3+3x2+25x+75f(x)=x3+3x2+25x+75; zero: 5i5i

15. f(x)=4x4+7x3+62x2+112x32f(x)=4x4+7x3+62x2+112x32; zero: 4i4i

16. h(x)=3x4+5x3+25x2+45x18h(x)=3x4+5x3+25x2+45x18; zero: 3i3i

17. h(x)=x47x3+23x215x522h(x)=x47x3+23x215x522; zero: 25i25i

18. f(x)=x47x3+14x238x60f(x)=x47x3+14x238x60; zero: 1+3i1+3i

For the following exercises, find all complex zeros (real and non-real). Write f in factored form.

19. f(x)=x31f(x)=x31

20. f(x)=x41f(x)=x41

21. f(x)=x38x2+25x26f(x)=x38x2+25x26

22. f(x)=x3+13x2+57x+85f(x)=x3+13x2+57x+85

23. f(x)=x4+5x2+4f(x)=x4+5x2+4

24. f(x)=x4+13x2+36f(x)=x4+13x2+36