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 [latex]a+bi\text{, }b\ne 0[/latex], is a zero of [latex]f\left(x\right)[/latex]. Then, by the Factor Theorem, [latex]x-\left(a+bi\right)[/latex] is a factor of [latex]f\left(x\right)[/latex]. For f to have real coefficients, [latex]x-\left(a-bi\right)[/latex] must also be a factor of [latex]f\left(x\right)[/latex]. This is true because any factor other than [latex]x-\left(a-bi\right)[/latex], when multiplied by [latex]x-\left(a+bi\right)[/latex], 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 [latex]a+bi[/latex], then the complex conjugate [latex]a-bi[/latex] must also be a zero of [latex]f\left(x\right)[/latex]. 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 [latex]\left(x-c\right)[/latex], where c is a complex number.
If the polynomial function f has real coefficients and a complex zero in the form [latex]a+bi[/latex], then the complex conjugate of the zero, [latex]a-bi[/latex], is also a zero.
How To: Given the zeros of a polynomial function [latex]f[/latex] and a point [latex]\left(c\text{, }f(c)\right)[/latex] on the graph of [latex]f[/latex], use the Linear Factorization Theorem to find the polynomial function.
- Use the zeros to construct the linear factors of the polynomial.
- Multiply the linear factors to expand the polynomial.
- Substitute [latex]\left(c,f\left(c\right)\right)[/latex] into the function to determine the leading coefficient.
- Simplify.
Example 7: 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 [latex]f\left(-2\right)=100[/latex].
Solution
Because [latex]x=i[/latex] is a zero, by the Complex Conjugate Theorem [latex]x=-i[/latex] is also a zero. The polynomial must have factors of [latex]\left(x+3\right),\left(x - 2\right),\left(x-i\right)[/latex], and [latex]\left(x+i\right)[/latex]. Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors.
We need to find a to ensure [latex]f\left(-2\right)=100[/latex]. Substitute [latex]x=-2[/latex] and [latex]f\left(2\right)=100[/latex]
into [latex]f\left(x\right)[/latex].
So the polynomial function is
or
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 5
Find a third degree polynomial with real coefficients that has zeros of 5 and –2i such that [latex]f\left(1\right)=10[/latex].
Analysis of the Solution
We found that both i and –i were zeros, but only one of these zeros needed to be given. If i is a zero of a polynomial with real coefficients, then –i must also be a zero of the polynomial because –i is the complex conjugate of i.