Linear Factorization Theorem and Complex Conjugate Theorem
A vital implication of the Fundamental Theorem of Algebra is that a polynomial function of degree [latex]n[/latex] will have [latex]n[/latex] zeros in the set of complex numbers if we allow for multiplicities. This means that we can factor the polynomial function into [latex]n[/latex] factors.
Linear Factorization Theorem
The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and each factor will be of the form [latex](x – c)[/latex] where [latex]c[/latex] is a complex number.
When you learned to divide complex numbers, you multiplied the top and bottom of the quotient of complex numbers deliberately by the conjugate of the denominator so that the imaginary part would eliminate from the denominator. That is, multiplying complex conjugates eliminates the imaginary part.
Let [latex]f[/latex] 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 [latex]f[/latex] 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 [latex]f[/latex] 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.
Complex Conjugate Theorem
If the polynomial function [latex]f[/latex] has real coefficients and a complex zero of 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.
Find a fourth degree polynomial with real coefficients that has zeros of [latex]–3[/latex], [latex]2[/latex], [latex]i[/latex], such that [latex]f\left(-2\right)=100[/latex].
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 [latex]4[/latex] polynomial and now have four zeros, we have all four factors. Let’s begin by multiplying these factors.
We need to find [latex]a[/latex] 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].
We found that both [latex]i[/latex] and [latex]–i[/latex] were zeros, but only one of these zeros needed to be given. If [latex]i[/latex] is a zero of a polynomial with real coefficients, then [latex]–i[/latex] must also be a zero of the polynomial because [latex]–i[/latex] is the complex conjugate of [latex]i[/latex].