Roots & Zeros of Polynomials III

Using the Rational Root Theorem to Predict the Rational Roots of a

Polynomial

Created by K. Chiodo, HCPS

Find the Roots of a PolynomialFor higher degree polynomials, finding the complex roots (real and imaginary) is easier if we know one of the roots.

Descartes’ Rule of Signs can help get you started. Complete the table below:

Polynomial# + Real

Roots# - RealRoots

# Imag.Roots

y x 4 2x2 3

y x3 7x2 17x 15

y 3x 4 x3 3x 2 x 1

The Rational Root Theorem

The Rational Root Theorem gives us a tool to predict the values of rational roots:

10 1 1If ( ) ... , where

the coefficients are all integers, and a rational

zero of ( ) in reduced form is , then

must be a factor of (the constant term) and

must be a

n nn n

n

P x a x a x a x a

pP x

q

p a

q

0factor of (the leading coefficient).a

List the Possible Rational Roots

For the polynomial: f (x) x3 3x2 5x 15

All possible values of: p: 1, 3, 5

q: 1

All possible rational roots of the form p/q:

p

q: 1, 3, 5

Narrow the List of Possible Roots

For the polynomial:

Descartes’ Rule: # + Real Roots = 3 or 1

# Real Roots = 0

# Imag. Roots = 2 or 0

All possible Rational Roots of the form p/q:

f (x) x3 3x2 5x 15

: 1, 3, 5, 15p

q

Find a Root That Works

For the polynomial:

Substitute each of our possible rational roots into f(x). If f(a) = 0, then a is a root of f(x). (Roots are the solutions to the polynomial set equal to zero!)

f (1) 1 3 5 15 12

f (3) 27 27 15 15 0 *

f (5) 125 75 25 15 60

f (x) x3 3x2 5x 15

Find the Other Roots

Now that we know one root is x = 3, do the other two roots have to be imaginary? What other category have we left out?

To find the other roots, divide the root we know into the original polynomial:

3 1 3 5 15

Find the Other Roots (con’t)

The degree of the resulting polynomial is 1 less than the original polynomial. When the resulting polynomial is a QUADRATIC, we can solve it by FACTORING or by using the QUADRATIC FORMULA!

3 1 3 5 15

+3 0 +15

1 0 5 0 x 2 5

Find the Other Roots (con’t)

x2 5 0

x2 5

x2 5

x i 5

This quadratic does not have real factors, but it can be solved easily by moving the 5 to the other side of the equation.

Find the Other Roots (con’t)

The roots of the polynomial equation:

are:

x 3, i 5, i 5

f (x) x3 3x2 5x 15

Another Example

• Descartes’ rule of signs: # Total Roots = 4

# + Real Roots = 1

# - Real Roots = 3 or 1

# Imag. Roots = 0 or 2

• The possible RATIONAL roots:

f (x) 3x4 2x3 4x2 7x 2

p

q 1,2,

1

3,2

3

Another Example

• Find a Rational Root that works:

f(2) = 0, so x = 2 is a root

• Synthetic Division with the root, x = 2:

2 3 2 4 7 2

+6 +8 +8 2

3 4 4 1 0

f (x) 3x4 2x3 4x2 7x 2

Another Example

• The reduced polynomial is:

3x 3 4x2 4x 1

• Since the degree > 2, we must do synthetic division again, go back to the list of possible roots and try them in the REDUCED polynomial. The same root might work again - so try it also!

f(-1/3) = 0, so x = - 1/3 is a root

Another Example

• Synthetic Division with the root, x = - 1/3 into the REDUCED POLYNOMIAL:

13 3 4 4 1

-1 -1 -1

3 3 3 0• The reduced polynomial is now a quadratic:

3x 2 3x 3

Another Example

• Solve the resulting quadratic using the quad. formula:

• The 4 roots of the polynomial are:

3x 2 3x 3 3 x2 x 1 0

x 1 1 4(1)(1)

2

1 i 3

2

2, 1

3,

1 i 3

2

More Practice

For each of the polynomials on the next page, find the roots of the polynomials.• Know what to expect ---- Make a table using

Descartes’ Rule of Signs.

• List the possible RATIONAL roots, p/q.

• Find one number from your p/q list that makes the polynomial = 0. You can check this either by evaluating f(#) or by synthetic division.

• Do synthetic division with a root - solve the resulting polynomial.

More Practice

Find the roots of the polynomials:

1. f (x) x3 3x2 3x 10

2. g(x) 2x3 x 2 16x 15

3. h(x) 2x4 2x3 9x2 2x 7

4. p(x) 4x4 8x3 x 2 2x