Using the Fundamental Theorem of Algebra and the Linear Factorization Theorem

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 $f(x)=0$. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it $c_{1}$. By the Factor Theorem, we can write $f(x)$ as a product of $x-c_{1}$ and a polynomial quotient. Since $x-c_{1}$ 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 $c_{2}$. So we can write the polynomial quotient as a product of $x-c_{2}$ 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 $f(x)$.

THE FUNDAMENTAL THEOREM OF ALGEBRA

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 $f(x)$ is a polynomial of degree $n > 0$, and a is a non-zero real number, then $f(x)$ has exactly $n$ linear factors

$f(x)=a\left(x-c_{1}\right)\left(x-c_{2}\right) \ldots\left(x-c_{n}\right)$

where $c_{1}, c_{2}, \ldots, c_{n}$ are complex numbers. Therefore, $f(x)$ has $n$ roots if we allow for multiplicities.

Q&A

Does every polynomial have at least one imaginary zero?

No. Real numbers are a subset of complex numbers, but not the other way around. A complex number is not necessarily imaginary. Real numbers are also complex numbers.

EXAMPLE 6

Finding the Zeros of a Polynomial Function with Complex Zeros

Find the zeros of $f(x)=3 x^{3}+9 x^{2}+x+3$.

Solution

The Rational Zero Theorem tells us that if $\frac{p}{q}$ is a zero of $f(x)$, then $p$ is a factor of $3$ and $q$ is a factor of $3$.

$\begin{aligned} \frac{p}{q} &=\frac{\text { factor of constant term }}{\text { factor of leading coefficient }} \\ &=\frac{\text { factor of } 3}{\text { factor of } 3} \end{aligned}$

The factors of $3$ are $\pm 1$ and $\pm 3$. The possible values for pq, and therefore the possible rational zeros for the function, are $\pm 3$, $\pm 1$, and $\pm \frac{1}{3}$. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of $0$. Let's begin with $–3$.

Dividing by $(x+3)$ gives a remainder of $0$, so $–3$ is a zero of the function. The polynomial can be written as

$(x+3)\left(3 x^{2}+1\right)$

We can then set the quadratic equal to $0$ and solve to find the other zeros of the function.

$\begin{aligned} 3 x^{2}+1 &=0 \\ x^{2} &=-\frac{1}{3} \\ x &=\pm \sqrt{-\frac{1}{3}}=\pm \frac{i \sqrt{3}}{3} \end{aligned}$

The zeros of $f(x)$ are $–3$ and $\pm \frac{i \sqrt{3}}{3}$.

Analysis

Look at the graph of the function $f$ in Figure 2. Notice that, at $x=−3$, the graph crosses the $x$-axis, indicating an odd multiplicity ($1$) for the zero $x=–3$. Also note the presence of the two turning points. This means that, since there is a 3rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of $x=−3$ is $1$ and there are two complex solutions, which is what we found, or the multiplicity at $x=−3$ is three. Either way, our result is correct.

Figure 2

TRY IT #4

Find the zeros of $f(x)=2 x^{3}+5 x^{2}-11 x+4$.

Source: Rice University, https://openstax.org/books/college-algebra/pages/5-5-zeros-of-polynomial-functions
This work is licensed under a Creative Commons Attribution 4.0 License.

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 $a+bi$, $b≠0$, is a zero of $f(x)$. Then, by the Factor Theorem, $x−(a+bi)$ is a factor of $f(x)$. For $f$ to have real coefficients, $x−(a−bi)$ must also be a factor of $f(x)$. This is true because any factor other than $x−(a−bi)$, when multiplied by $x−(a+bi)$, 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 $a+bi$, then the complex conjugate $a−bi$ must also be a zero of $f(x)$. This is called the Complex Conjugate Theorem.

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 $(x−c)$, where c is a complex number.

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

How To

Given the zeros of a polynomial function $f$ and a point $(c, f(c))$ on the graph of $f$, 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 $(c,f(c))$ into the function to determine the leading coefficient.
4. 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 $f(−2)=100$.

Solution

Because $x=i$ is a zero, by the Complex Conjugate Theorem $x=–i$ is also a zero. The polynomial must have factors of $(x+3)$, $(x−2)$, $(x−i)$, and $(x+i)$. 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.

$f(x)=a(x+3)(x−2)(x−i)(x+i)$

$f(x)=a(x^2+x−6)(x^2+1)$

$f(x)=a(x^4+x^3−5x^2+x−6$

We need to find a to ensure $f(–2)=100$. Substitute $x=–2$  and $f(2)=100$ into $f(x)$.

$100=a((−2)^4+(−2)^3−5(−2)^2+(−2)−6)$

$100=a(−20)$

$−5=a$

So the polynomial function is

$f(x)=−5(x^4+x^3−5x^2+x−6)$

or

$f(x)=−5x^4−5x^3+25x^2−5x+30$

Analysis

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 $i$.

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 $f(1)=10$.