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 + 3 i 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 –2*i* such that [latex]f\left(1\right)=10[/latex].

## Analysis of the Solution

We found that both

iand –iwere zeros, but only one of these zeros needed to be given. Ifiis a zero of a polynomial with real coefficients, then–imust also be a zero of the polynomial because–iis the complex conjugate ofi.