## 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 $a+bi\text{, }b\ne 0$, is a zero of $f\left(x\right)$. Then, by the Factor Theorem, $x-\left(a+bi\right)$ is a factor of $f\left(x\right)$. For f to have real coefficients, $x-\left(a-bi\right)$ must also be a factor of $f\left(x\right)$. This is true because any factor other than $x-\left(a-bi\right)$, when multiplied by $x-\left(a+bi\right)$, 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 $f\left(x\right)$. 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 $\left(x-c\right)$, 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 $\left(c\text{, }f(c)\right)$ 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 $\left(c,f\left(c\right)\right)$ 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 $f\left(-2\right)=100$.

### Solution

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

$\begin{cases}f\left(x\right)=a\left(x+3\right)\left(x - 2\right)\left(x-i\right)\left(x+i\right)\\ f\left(x\right)=a\left({x}^{2}+x - 6\right)\left({x}^{2}+1\right)\\ f\left(x\right)=a\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\right)\end{cases}$

We need to find a to ensure $f\left(-2\right)=100$. Substitute $x=-2$ and $f\left(2\right)=100$
into $f\left(x\right)$.

$\begin{cases}100=a\left({\left(-2\right)}^{4}+{\left(-2\right)}^{3}-5{\left(-2\right)}^{2}+\left(-2\right)-6\right)\hfill \\ 100=a\left(-20\right)\hfill \\ -5=a\hfill \end{cases}$

So the polynomial function is

$f\left(x\right)=-5\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\right)$

or

$f\left(x\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30$

### 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.

### 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 $f\left(1\right)=10$.

Solution