roots & zeros of polynomials how the roots, solutions, zeros, x-intercepts and factors of a...

Roots & Zeros of Polynomials

How the roots, solutions, zeros, x-

intercepts and factors of a polynomial

function are related.

2.5 Zeros of Polynomial Functions

Polynomials

A Polynomial Expression can be a monomial or a sum of monomials. The Polynomial Expressions that we are discussing today are in terms of one variable.

In a Polynomial Equation, two polynomials are set equal to each other.

Factoring Polynomials

Terms are Factors of a Polynomial if, when they are multiplied, they equal that polynomial:

2 2 15 ( 3)( 5)x x x x (x - 3) and (x + 5) are

Factors of the polynomial 2

2 15x x

Since Factors are a Product...

…and the only way a product can equal zero is if one or more of the factors are zero…

…then the only way the polynomial can equal zero is if one or more of the factors are zero.

Solving a Polynomial Equation

The only way that x2 +2x - 15 can = 0 is if x = -5 or x = 3

Rearrange the terms to have zero on one side: 2 22 15 2 15 0x x x x

Factor: ( 5)( 3) 0x x

Set each factor equal to zero and solve: ( 5) 0 and ( 3) 0

5 3

x x

x x

Solutions/Roots a PolynomialSetting the Factors of a Polynomial Expression equal to zero gives the Solutions to the Equation when the polynomial expression equals zero. Another name for the Solutions of a Polynomial is the Roots of a Polynomial !

Zeros of a Polynomial Function

A Polynomial Function is usually written in function notation or in terms of x and y.

f (x) x2 2x 15 or y x2 2x 15

The Zeros of a Polynomial Function are the solutions to the equation you get when you set the polynomial equal to zero.

Zeros of a Polynomial Function

The Zeros of a Polynomial Function ARE the Solutions to the Polynomial Equation when the polynomial equals zero.

Graph of a Polynomial Function

Here is the graph of our polynomial function:

The Zeros of the Polynomial are the values of x when the polynomial equals zero. In other words, the Zeros are the x-values where y equals zero.

y x2 2x 15

y x2 2x 15

x-Intercepts of a Polynomial

The points where y = 0 are called the x-intercepts of the graph.

The x-intercepts for our graph are the points...

and(-5, 0) (3, 0)

x-Intercepts of a Polynomial

When the Factors of a Polynomial Expression are set equal to zero, we get the Solutions or Roots of the Polynomial Equation.

The Solutions/Roots of the Polynomial Equation are the x-coordinates for the x-Intercepts of the Polynomial Graph!

Factors, Roots, Zeros

y x2 2x 15For our Polynomial Function:

The Factors are: (x + 5) & (x - 3)

The Roots/Solutions are: x = -5 and 3

The Zeros are at: (-5, 0) and (3, 0)

Roots & Zeros of Polynomials II

Finding the Roots/Zeros of Polynomials:

• The Fundamental Theorem of

Algebra

• Descartes’ Rule of Signs

• The Complex Conjugate Theorem

Fundamental Theorem Of Algebra

Every Polynomial Equation with a degree higher than zero has at least one root in the set of Complex Numbers.

A Polynomial Equation of the form P(x) = 0 of degree ‘n’ with complex coefficients has exactly ‘n’ Roots in the set of Complex Numbers.

COROLLARY:

Real/Imaginary RootsIf a polynomial has ‘n’ complex roots will its graph have ‘n’ x-intercepts?

In this example, the degree n = 3, and if we factor the polynomial, the roots are x = -2, 0, 2. We can also see from the graph that there are 3 x-intercepts.

y x3 4x

Real/Imaginary RootsJust because a polynomial has ‘n’ complex roots doesn’t mean that they are all Real!

y x3 2x2 x 4In this example, however, the degree is still n = 3, but there is only one Real x-intercept or root at x = -1, the other 2 roots must have imaginary components.

Descartes’ Rule of Signs

Arrange the terms of the polynomial P(x) in descending degree:

• The number of times the coefficients of the terms of P(x) change sign = the number of Positive Real

Roots (or less by any even number)

• The number of times the coefficients of the terms of P(-x) change sign = the number of Negative

Real Roots (or less by any even number) In the examples that follow, use Descartes’ Rule of Signs to predict the number of + and - Real Roots!

Find Roots/Zeros of a Polynomial

We can find the Roots or Zeros of a polynomial by setting the polynomial equal to 0 and factoring.

Some are easier to factor than others!

f (x) x3 4x

x(x2 4)

x(x 2)(x 2)

The roots are: 0, -2, 2


If we cannot factor the polynomial, but know one of the roots, we can divide that factor into the polynomial. The resulting polynomial has a lower degree and might be easier to factor or solve with the quadratic formula.

We can solve the resulting polynomial to get the other 2 roots:

f (x) x3 5x2 2x 10

one root is x 5 x 5 x3 5x2 2x 10

x3 5x2

2x 10

2x 10

0

x2 2

(x - 5) is a factor

2, 2x

Complex Conjugates Theorem

Roots/Zeros that are not Real are Complex with an Imaginary component. Complex roots with Imaginary components always exist in Conjugate Pairs.

If a + bi (b ≠ 0) is a zero of a polynomial function, then its Conjugate, a - bi, is also a zero of the function.


If the known root is imaginary, we can use the Complex Conjugates Theorem.

Ex: Find all the roots of f (x) x3 5x2 7x 51

If one root is 4 - i.

Because of the Complex Conjugate Theorem, we know that another root must be 4 + i.

Can the third root also be imaginary?

Consider… Descartes: # of Pos. Real Roots = 2 or 0

Descartes: # of Neg. Real Roots = 1

Example (con’t)Ex: Find all the roots of f (x) x3 5x2 7x 51


If one root is 4 - i, then one factor is [x - (4 - i)], and

Another root is 4 + i, & another factor is [x - (4 + i)].

Multiply these factors:

2

2 2

2

2

4 4 4 4 4 4

4 4 16

8 16 ( 1)

8 17

x i x i x x i x i i i

x x xi x xi i

x x

x x

Example (con’t)Ex: Find all the roots of f (x) x3 5x2 7x 51


x2 8x 17

If the product of the two non-real factors is x2 8x 17

then the third factor (that gives us the neg. real root) is the quotient of P(x) divided by :

x2 8x 17 x3 5x2 7x 51

x3 5x2 7x 51

0

x 3

The third root is x = -3

Finding Roots/Zeros of Polynomials

We use the Fundamental Thm. Of Algebra, Descartes’ Rule of Signs and the Complex Conjugate Thm. to predict the nature of the roots of a polynomial.

We use skills such as factoring, polynomial division and the quadratic formula to find the zeros/roots of polynomials.

In future lessons you will learn other rules and theorems to predict the values of roots so you can solve higher degree polynomials!

The Rational Zero Theorem

The Rational Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will appear

somewhere in the list.


If f (x) anxn an-1x

n-1 … a1x a0 has integer coefficients and

(where is reduced) is a rational zero, then p is a factor of the

constant term a0 and q is a factor of the leading coefficient an.


If f (x) anxn an-1x

n-1 … a1x a0 has integer coefficients and

(where is reduced) is a rational zero, then p is a factor of the

constant term a0 and q is a factor of the leading coefficient an.

pq

pq

EXAMPLE: Using the Rational Zero Theorem

List all possible rational zeros of f (x) 15x3 14x2 3x – 2.

Solution The constant term is –2 and the leading coefficient is 15.

1 2 1 2 1 25 53 3 15 15

Factors of the constant term, 2Possible rational zerosFactors of the leading coefficient, 15

1, 21, 3, 5, 15

1, 2, , , , , ,

Divide 1

and 2 by 1.

Divide 1

and 2 by 3.

Divide 1

and 2 by 5.

Divide 1

and 2 by 15.

There are 16 possible rational zeros. The actual solution set to f (x) 15x3 14x2 3x – 2 = 0 is {-1, 1/3, 2/5}, which contains 3 of the 16 possible solutions.

EXAMPLE: Solving a Polynomial Equation

Solve: x4 6x2 8x + 24 0.

Solution Because we are given an equation, we will use the word "roots," rather than "zeros," in the solution process. We begin by listing all possible rational roots.

Factors of the constant term, 24Possible rational zerosFactors of the leading coefficient, 1

1, 2 3, 4, 6, 8, 12, 241

1, 2 3, 4, 6, 8, 12, 24


Solve: x4 6x2 8x + 24 0.

Solution The graph of f (x) x4 6x2 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation.

x-intercept: 2

The zero remainder indicates that 2 is a root of x4 6x2 8x + 24 0.

2 1 0 6 8 24 2 4 4 241 2 2 12 0


Solve: x4 6x2 8x + 24 0.

Solution Now we can rewrite the given equation in factored form.

(x – 2)(x3 2x2 2x 12) 0 This is the result obtained from the synthetic division.

x – 2 0 or x3 2x2 2x 12 Set each factor equal to zero.

x4 6x2 8x + 24 0 This is the given equation.

Now we must continue by factoring x3 + 2x2 - 2x - 12 = 0


Solve: x4 6x2 8x + 24 0.

Solution Because the graph turns around at 2, this means that 2 is a root of even multiplicity. Thus, 2 must also be a root of x3 2x2 2x 12 = 0.

x-intercept

: 2

2 1 2 2 12 2 8 12 1 4 6 0

These are the coefficients of x3 2x2 2x 12 = 0.

The zero remainder indicates that 2 is a root of x3 2x2 2x 12 = 0.


Solve: x4 6x2 8x + 24 0.

Solution Now we can solve the original equation as follows.

(x – 2)(x3 2x2 2x 12) 0 This was obtained from the first synthetic division.

x4 6x2 8x + 24 0 This is the given equation.

(x – 2)(x – 2)(x2 4x 6) 0 This was obtained from the second synthetic division.

x – 2 0 or x – 2 0 or x2 4x 6 Set each factor equal to zero.

x 2 x 2 x2 4x 6 Solve.

EXAMPLE: Solving a Polynomial EquationSolve: x4 6x2 8x + 24 0.

Solution We can use the quadratic formula to solve x2 4x 6

Let a 1, b 4, and c 6.

24 4 4 1 62 1

We use the quadratic formula because x2 4x 6 cannot be factored.

2 42

b b acxa

Simplify.2 2i

Multiply and subtract under the radical. 4 82

4 2 22

i 8 4(2)( 1) 2 2i

The solution set of the original equation is {2, 2 i 2 i }.2,i 2i

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