### Learning Outcomes

- Use the Linear Factorization Theorem to find polynomials with given zeros.
- Use Descartes rule of signs to determine the maximum number of possible real zeros of a polynomial function
- Solve real-world applications of polynomial equations.

A vital implication of the **Fundamental Theorem of Algebra **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 each factor will be of the form (*x – c*) where *c* is a complex number.

### recall multiplying complex conjugates

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. We’ll use the same idea in the paragraph below.

Ex. [latex]\left(a+bi\right)\left(a-bi\right)\Rightarrow a^2-abi+abi-b^2i^2 \Rightarrow a^2-b^2(-1) \Rightarrow a^2+b^2[/latex].

That is, multiplying complex conjugates eliminates the imaginary part.

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 of 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 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.

### Example: 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].

### 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

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].

## Descartes’ Rule of Signs

There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order,** Descartes’ Rule of Signs** tells us of a relationship between the number of sign changes in [latex]f\left(x\right)[/latex] and the number of positive real zeros.

There is a similar relationship between the number of sign changes in [latex]f\left(-x\right)[/latex] and the number of negative real zeros.

### A General Note: Descartes’ Rule of Signs

According to **Descartes’ Rule of Signs**, if we let [latex]f\left(x\right)={a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+…+{a}_{1}x+{a}_{0}[/latex] be a polynomial function with real coefficients:

- The number of positive real zeros is either equal to the number of sign changes of [latex]f\left(x\right)[/latex] or is less than the number of sign changes by an even integer.
- The number of negative real zeros is either equal to the number of sign changes of [latex]f\left(-x\right)[/latex] or is less than the number of sign changes by an even integer.

### Example: Using Descartes’ Rule of Signs

Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for [latex]f\left(x\right)=-{x}^{4}-3{x}^{3}+6{x}^{2}-4x - 12[/latex].

### Try It

Use Descartes’ Rule of Signs to determine the maximum possible number of positive and negative real zeros for [latex]f\left(x\right)=2{x}^{4}-10{x}^{3}+11{x}^{2}-15x+12[/latex]. Use a graph to verify the number of positive and negative real zeros for the function.

## Solving Real-world Applications of Polynomial Equations

We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section.

### Example: Solving Polynomial Equations

A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

### Try It

A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be?