3.2: Factors and Zeros

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Oct 6, 2021
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- 3.1: Polynomial Division
- 3.3: Polynomial Equations

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1. Review of the Factor Theorem

Recall from last time, if $P(x)$ is a polynomial and $P(r) = 0$ , then the remainder produced when $P(x)$ is divided by $x - r$ is 0.

We can conclude that $r$ is a root of $P(x)$ if and only if the $x - r$ divides $P(x)$ .

Example 1

Given that 3 is a root of

$x^3- x^2 - 8x + 6$

Find the other two roots.

Solution:

We use synthetic division to divide by $x - 3$ . The result is

$x^2+2x - 2$

Next we use the quadratic formula to find

$x = -1 +\sqrt{3}\; \text{ or } \; x = -1 -\sqrt{3}.$

2. The Conjugate Root Theorem

Theorem: The Conjugate Root Theorem

1. Suppose that $P(x)$ has a root of the form

$a + b\sqrt{c}$

where $c$ is not a perfect square, then

$a - b\sqrt{c}$

is also a root.

2. Suppose that $P(x)$ has a root of the form

$a + bi$

then

$a - bi$

is also a root.

3. The Fundamental Theorem of Algebra

Theorem: The Fundamental Theorem of Algebra

Every polynomial has a root in the complex numbers, moreover if the polynomial has degree $n$ then the polynomial can be written as a product of $n$ linear factors. We define the multiplicity of a root $r$ to be the number of factors the polynomial has of the form $x - r$ .

Example 2

Suppose that a polynomial passes through the point $(0,3)$ and has the roots as follows:

$i$ of multiplicity 2.
$2 -\sqrt{3}$ of multiplicity 3 and
4 of multiplicity 2.

Find the polynomial.

Solution:

The polynomial has the form

$P(x) = a(x - i)^2(x + i)^2(x - (2 +\sqrt{3}))^3(x - (2 -\sqrt{3}))^3(x - 4)^2.$

To find $a$ we have that

$3 = P(0) = a(-1)(-1)(1)(16)$

hence

$a = \dfrac{3}{16}$

so that

$P(x) = \dfrac{3}{16}(x - i)^2(x + i)^2(x - (2 +\sqrt{3}))^3(x - (2 -\sqrt{3}))^3(x - 4)^2.$

3. The Rational Root Theorem

Theorem: The Rational Root Theorem

Let

$P(x) = a_nx^n + a_{n-1}x^{n-1}+ ... + a_1x + a_0$

and let

$\dfrac{p}{q}$

be a root of $P(x)$ ,

then $p$ is a positive or negative factor of $a_0$ and $q$ is a positive factor of $a_n$ .

Example 3

Factor

$P(x) = 6x^3 + x^2 - 15x + 4$

Solution

Note that the factors of 4 are 1,-1, 2,-2,4,-4, and

the positive factors of 6 are 1,2,3,6.

Hence the possibilities for rational roots are

$1,\;-1,\; 2,\;-2,\;4,\;-4,\;\dfrac{1}{2},\;-\dfrac{1}{2},\;\dfrac{1}{3},\;-\dfrac{1}{3},\;\dfrac{2}{3},\;-\dfrac{2}{3},\;\dfrac{4}{3},\;-\dfrac{4}{3}.$

$alt$

Using the graph we see that the roots are near $\frac{1}{3}$ , $\frac{1}{2}$ , and $\frac{4}{3}$ . We try synthetic division until we find that $\frac{4}{3}$ is a root of the equation and that

$6x^2 + 9x - 3$

is the quotient of $P(x)$ and $x - \frac{4}{3}$ .

We have

$6x^2 + 9x - 3 = 3(2x^2 + 3x - 1).$

Next we use the quadratic formula to find that the other two roots are

$\dfrac{(-3 + \sqrt{17})}{4} \; \text{ and } \dfrac{ (-3-\sqrt{17})}{4}.$

We can say that the factors are

$x - \dfrac{3}{4}, x - \left[\dfrac{(-3 + \sqrt{17})}{4}\right], \; \text{and } x - \left[\dfrac{(-3 - \sqrt{17})}{4}\right].$

Descartes' Rule of Signs

Theorem: Descartes' Rule of Signs

Let $P(x)$ be a polynomial with real coefficients:

The number of positive real roots either is equal to the number of variations in the sign of $P(x)$ or is less then that by an even integer.
The number of negative roots is either equal to the number of variations in the sign of $P(-x)$ or is less than that by an even integer. Alternatively, to compute the number of negative roots, you change the sign of all the odd terms of $P(x)$ and count the number of signs.

Example 4

Consider the function

$f(x) = x^3 + 8x + 5.$

Since all the signs are positive, there are no sign changes, hence there are no positive roots.

$f(-x) = -x^3 - 8x + 5$

We see one sign change hence there is exactly one negative real root. Therefore there must be two imaginary roots. Using synthetic division, we have that neither -5 nor -1 are roots, hence there are no rational roots. Since there is only one negative root, it cannot be of the form

$a + b\sqrt{c}.$

We can conclude that there is one irrational root that is not a square root, and two other roots $a + bi$ and $a - bi$ .

Steps for finding roots:

Use Descartes' rule of signs to determine positive and negative real roots.
Use the $\frac{p}{q}$ theorem (Rational Root Theorem) in coordination with Descartes' Rule of signs to find a possible roots.
Plug in 1 and -1 to see if one of these two possibilities is a root. If so go to step 5.
If not use synthetic division to test the other possibilities for roots until you find the first root. If there are no rational roots, give up or use a calculator.
Find the quotient of dividing $f(x)$ by $x - r$ .
If the quotient is a quadratic, use the quadratic formula to find the last two roots. Otherwise go back to Step 2 applied to the quotient.
Reread the question and answer it.

Exercises

Find the rational zeros of each polynomial function
Factor each over the set of integers
1. $x^3 + 7x^2 + 16x + 12$
2. $2x^3 - 9x^2 + 14x - 10$
3. $4x^4 - 28x^3 + 67x^2 - 63x + 18$
4. $x^5 + x^4 - 6x^3 + 8x - 16$

Answer: For a worked out example, click here

Larry Green (Lake Tahoe Community College)

Integrated by Justin Marshall.

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1. Review of the Factor Theorem

2. The Conjugate Root Theorem

3. The Fundamental Theorem of Algebra

3. The Rational Root Theorem

Descartes' Rule of Signs

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