# Linear Factorization and Descartes Rule of Signs

### Learning Objectives

- Use linear factorization to find the equation of a polynomial function given it's zeros and a point on it's graph
- Use Descartes rule of signs to determine the maximum number of possible real zeros of a polynomial function
- Solve a polynomial function application involving volume

**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 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: 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].

Answer: 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.

[latex]\begin{array}{l}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{array}[/latex]

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].

[latex]\begin{array}{l}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{array}[/latex]

So the polynomial function is[latex]f\left(x\right)=-5\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\right)[/latex]

or[latex]f\left(x\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30[/latex]

#### 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 + 3**

*i*were given as a zero of a polynomial with real coefficients, would 2 – 3*i*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].

Answer: [latex]f\left(x\right)=-\frac{1}{2}{x}^{3}+\frac{5}{2}{x}^{2}-2x+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. For example, the polynomial function below has one sign change. This tells us that the function must have 1 positive real zero. 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. In this case, [latex]f\left(\mathrm{-x}\right)[/latex] has 3 sign changes. This tells us that [latex]f\left(x\right)[/latex] could have 3 or 1 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].Answer: Begin by determining the number of sign changes. There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine [latex]f\left(-x\right)[/latex] to determine the number of negative real roots.

[latex]\begin{array}{l}f\left(-x\right)=-{\left(-x\right)}^{4}-3{\left(-x\right)}^{3}+6{\left(-x\right)}^{2}-4\left(-x\right)-12\hfill \\ f\left(-x\right)=-{x}^{4}+3{x}^{3}+6{x}^{2}+4x - 12\hfill \end{array}[/latex]

Again, there are two sign changes, so there are either 2 or 0 negative real roots. There are four possibilities, as we can see below.Positive Real Zeros | Negative Real Zeros | Complex Zeros | Total Zeros |
---|---|---|---|

2 | 2 | 0 | 4 |

2 | 0 | 2 | 4 |

0 | 2 | 2 | 4 |

0 | 0 | 4 | 4 |

#### Analysis of the Solution

We can confirm the numbers of positive and negative real roots by examining a graph of the function. We can see from the graph that the function has 0 positive real roots and 2 negative real roots.### Try It

Use Descartes’ Rule of Signs to determine the maximum possible numbers 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 numbers of positive and negative real zeros for the function.Answer: There must be 4, 2, or 0 positive real roots and 0 negative real roots. The graph shows that there are 2 positive real zeros and 0 negative real zeros.

## Solve 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?Answer: Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\frac{1}{3}w[/latex]. Let’s write the volume of the cake in terms of width of the cake.

[latex]\begin{array}{l}V=\left(w+4\right)\left(w\right)\left(\frac{1}{3}w\right)\\ V=\frac{1}{3}{w}^{3}+\frac{4}{3}{w}^{2}\end{array}[/latex]

Substitute the given volume into this equation.[latex]\begin{array}{l}\text{ }351=\frac{1}{3}{w}^{3}+\frac{4}{3}{w}^{2}\hfill & \text{Substitute 351 for }V.\hfill \\ 1053={w}^{3}+4{w}^{2}\hfill & \text{Multiply both sides by 3}.\hfill \\ \text{ }0={w}^{3}+4{w}^{2}-1053 \hfill & \text{Subtract 1053 from both sides}.\hfill \end{array}[/latex]

Descartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\pm 3,\pm 9,\pm 13,\pm 27,\pm 39,\pm 81,\pm 117,\pm 351[/latex], and [latex]\pm 1053[/latex]. We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[/latex]. Since 1 is not a solution, we will check [latex]x=3[/latex]. Since 3 is not a solution either, we will test [latex]x=9[/latex]. Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.[latex]l=w+4=9+4=13\text{ and }h=\frac{1}{3}w=\frac{1}{3}\left(9\right)=3[/latex]

The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.### 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?Answer: 3 meters by 4 meters by 7 meters

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