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6-5
Finding Real Roots of
Polynomial Equations
Warm Up
Factor completely.
1. 2y3 + 4y2 – 30y 2y(y – 3)(y + 5)
2. 3x4 – 6x2 – 24
3(x – 2)(x + 2)(x2 + 2)
Solve each equation.
3. x2 – 9 = 0
x = – 3, 3
4. x3 + 3x2 – 4x = 0
Holt McDougal Algebra 2
x = –4, 0, 1
6-5
Finding Real Roots of
Polynomial Equations
Objectives
Identify the multiplicity of roots.
Use the Rational Root Theorem and the
irrational Root Theorem to solve
polynomial equations.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
In Lesson 6-4, you used several methods for
factoring polynomials. As with some quadratic
equations, factoring a polynomial equation is
one way to find its real roots.
Recall the Zero Product Property from Lesson
5-3. You can find the roots, or solutions, of the
polynomial equation P(x) = 0 by setting each
factor equal to 0 and solving for x.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Example 1A: Using Factoring to Solve Polynomial
Equations
Solve the polynomial equation by factoring.
4x6 + 4x5 – 24x4 = 0
4x4(x2 + x – 6) = 0
Factor out the GCF, 4x4.
4x4(x + 3)(x – 2) = 0
Factor the quadratic.
4x4 = 0 or (x + 3) = 0 or (x – 2) = 0 Set each factor
equal to 0.
Solve for x.
x = 0, x = –3, x = 2
The roots are 0, –3, and 2.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Example 1A Continued
Check Use a graph. The
roots appear to be
located at x = 0, x = –3,
and x = 2. 
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Example 1B: Using Factoring to Solve Polynomial
Equations
Solve the polynomial equation by factoring.
x4 + 25 = 26x2
x4 – 26 x2 + 25 = 0
(x2 – 25)(x2 – 1) = 0
Set the equation equal to 0.
Factor the trinomial in
quadratic form.
(x – 5)(x + 5)(x – 1)(x + 1) Factor the difference of two
squares.
x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0
x = 5, x = –5, x = 1 or x = –1
The roots are 5, –5, 1, and –1.
Holt McDougal Algebra 2
Solve for x.
6-5
Finding Real Roots of
Polynomial Equations
Check It Out! Example 1a
Solve the polynomial equation by factoring.
2x6 – 10x5 – 12x4 = 0
2x4(x2 – 5x – 6) = 0
Factor out the GCF, 2x4.
2x4(x – 6)(x + 1) = 0
Factor the quadratic.
2x4 = 0 or (x – 6) = 0 or (x + 1) = 0 Set each factor
equal to 0.
x = 0, x = 6, x = –1
Solve for x.
The roots are 0, 6, and –1.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Check It Out! Example 1b
Solve the polynomial equation by factoring.
x3 – 2x2 – 25x = –50
x3 – 2x2 – 25x + 50 = 0
Set the equation equal to 0.
(x + 5)(x – 2)(x – 5) = 0
Factor.
x + 5 = 0, x – 2 = 0, or x – 5 = 0
x = –5, x = 2, or x = 5
Solve for x.
The roots are –5, 2, and 5.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Sometimes a polynomial equation has a factor
that appears more than once. This creates a
multiple root. In 3x5 + 18x4 + 27x3 = 0 has two
multiple roots, 0 and –3. For example, the root 0
is a factor three times because 3x3 = 0.
The multiplicity of root r is the number of times
that x – r is a factor of P(x). When a real root has
even multiplicity, the graph of y = P(x) touches the
x-axis but does not cross it. When a real root has
odd multiplicity greater than 1, the graph “bends”
as it crosses the x-axis.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
You cannot always determine the multiplicity of a
root from a graph. It is easiest to determine
multiplicity when the polynomial is in factored
form.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Example 2A: Identifying Multiplicity
Identify the roots of each equation. State the
multiplicity of each root.
x3 + 6x2 + 12x + 8 = 0
x3 + 6x2 + 12x + 8 = (x + 2)(x + 2)(x + 2)
x + 2 is a factor three
times. The root –2 has
a multiplicity of 3.
Check Use a graph. A
calculator graph shows
a bend near (–2, 0). 
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Example 2B: Identifying Multiplicity
Identify the roots of each equation. State the
multiplicity of each root.
x4 + 8x3 + 18x2 – 27 = 0
x4 + 8x3 + 18x2 – 27 = (x – 1)(x + 3)(x + 3)(x + 3)
x – 1 is a factor once, and x
+ 3 is a factor three times.
The root 1 has a multiplicity
of 1. The root –3 has a
multiplicity of 3.
Check Use a graph. A
calculator graph shows
a bend near (–3, 0) and
crosses at (1, 0). 
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Check It Out! Example 2a
Identify the roots of each equation. State the
multiplicity of each root.
x4 – 8x3 + 24x2 – 32x + 16 = 0
x4 – 8x3 + 24x2 – 32x + 16 = (x – 2)(x – 2)(x – 2)(x – 2)
x – 2 is a factor four
times. The root 2 has
a multiplicity of 4.
Check Use a graph. A
calculator graph shows
a bend near (2, 0). 
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Check It Out! Example 2b
Identify the roots of each equation. State the
multiplicity of each root.
2x6 – 22x5 + 48x4 + 72x3 = 0
2x6 – 22x5 + 48x4 + 72x3 = 2x3(x + 1)(x – 6)(x – 6)
x is a factor three times, x + 1 is a factor once,
and x – 6 is a factor two times.
The root 0 has a multiplicity of 3. The root –1
has a multiplicity of 1. The root 6 has a
multiplicity of 2.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Example 3: Marketing Application
The design of a box specifies that its length is 4
inches greater than its width. The height is 1
inch less than the width. The volume of the box
is 12 cubic inches. What is the width of the box?
Step 1 Write an equation to model the volume of
the box.
Let x represent the width in inches. Then the length
is x + 4, and the height is x – 1.
x(x + 4)(x – 1) = 12 V = lwh.
x3 + 3x2 – 4x = 12 Multiply the left side.
x3 + 3x2 – 4x – 12 = 0
Holt McDougal Algebra 2
Set the equation equal to 0.
6-5
Finding Real Roots of
Polynomial Equations
Example 3 Continued
Factor the polynomial. The synthetic substitution
of 2 results in a remainder of 0, so 2 is a root
and the polynomial in factored form is (x – 2)(x2
+ 5x + 6).
(x – 2)(x2 + 5x + 6) = 0
Set the equation equal to 0.
(x – 2)(x + 2)(x + 3) = 0
Factor x2 + 5x + 6.
x = 2, x = –2, or x = –3
Set each factor equal to 0,
and solve.
The width must be positive, so the width
should be 2 inches.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Polynomial equations may also have irrational roots.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
The Irrational Root Theorem say that irrational
roots come in conjugate pairs. For example, if you
know that 1 +
is a root of x3 – x2 – 3x – 1 = 0,
then you know that 1 –
is also a root.
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Example 4: Identifying All of the Real Roots of a
Polynomial Equation
Identify all the real roots of 2x3 – 9x2 + 2 = 0.
Graph y = 2x3 – 9x2 + 2 to find the x-intercepts.
Holt McDougal Algebra 2
Finding Real Roots of
Polynomial Equations
6-5
Example 4 Continued
Test the possible rational root
1
2
2 –9
.
1
2
Test 1 . The remainder is
2 1
0, so (x – ) is a factor.
2
0 2
1 –4 –2
2 –8 –4 0
1
The polynomial factors into (x –
)(2x2 – 8x – 4).
2
Solve 2x2 – 8x – 4 = 0 to find the remaining roots.
2(x2 – 4x – 2) = 0
x=
4±
16+8
=2 
2
Holt McDougal Algebra 2
6
Factor out the GCF, 2
Use the quadratic formula to
identify the irrational roots.
6-5
Finding Real Roots of
Polynomial Equations
Example 4 Continued
1
The roots are
, 2 + 6 , and 2 - 6 .
2
Holt McDougal Algebra 2
6-5
Finding Real Roots of
Polynomial Equations
Lesson Quiz
Solve by factoring.
1. x3 + 9 = x2 + 9x
–3, 3, 1
Identify the roots of each equation. State the
multiplicity of each root.
0 and 2 each with
2. 5x4 – 20x3 + 20x2 = 0
multiplicity 2
3. x3 – 12x2 + 48x – 64 = 0
4 with multiplicity 3
4. A box is 2 inches longer than its height. The width
is 2 inches less than the height. The volume of the
box is 15 cubic inches. How tall is the box? 3 in.
5. Identify all the real roots of x3 + 5x2 – 3x – 3 = 0.
1, -3 + 6, -3 - 6
Holt McDougal Algebra 2