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7.4 and 7.5
Solving and Zeros of Polynomials
Day 1
As with some quadratic equations, factoring a
polynomial equation is one way to find its real
roots.
Recall the Zero Product Property. 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.
Fundamental Theorem of Algebra
• Every polynomial function with degree n
greater than or equal to 1 has exactly n
complex zeros, including multiplicities
The polynomial 3x5 + 18x4 + 27x3 = 0 has two
multiple roots, 0 and –3. 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.
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.
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.
Example 1A Continued
Check Use a graph. The
roots appear to be
located at x = 0, x = –3,
and x = 2. 
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.
Solve for x.
Example 2A: Identifying Multiplicity
Identify the roots of each equation. State the
multiplicity of each root.
x3 + 6x2 + 12x + 8 = 0
Since this is difficult to
factor, use a graph. A
calculator graph shows
a bend near (–2, 0). 
The root –2 has a multiplicity of 3. Therefore
x + 2 is a factor three times.
Check to see if that is true!
(x + 2)(x + 2)(x + 2)= x3 + 6x2 + 12x + 8

Example 2B: Identifying Multiplicity
Identify the roots of each equation. State the
multiplicity of each root.
x4 + 8x3 + 18x2 – 27 = 0
A calculator graph shows a
bend near (–3, 0) and
crosses at (1, 0).
The root 1 has a multiplicity of 1. The root –3 has
a multiplicity of 3. Therefore (x – 1) is a factor
once, and (x + 3) is a factor three times.
Check:
(x – 1)(x + 3)(x + 3)(x + 3)=x4 + 8x3 + 18x2 – 27

Location Principle
If P is a polynomial function and P(x1) and P(x2) have
opposite signs, then there is a real number r between
x1 and x2 that is a zero of P, that is, P(r) = 0
P( x)  x  3x  5x  13x  6
4
3
2
Lesson Quiz
Identify the roots of each equation. State the
multiplicity of each root.
0 and 2 each with
1. 5x4 – 20x3 + 20x2 = 0
multiplicity 2
2. x3 – 12x2 + 48x – 64 = 0
3. x3 + 9 = x2 + 9x
4 with multiplicity 3
–3, 3, 1
Homework
• 7.4 ws