5
Solving Polynomial
Equations
Chapter Overview
A goal throughout all of algebra has been to
solve equations.
• In this chapter, we will develop methods to
solve polynomial equations of any degree.
5.1
The Remainder and
Factor Theorems;
Synthetic Division
Objectives
1. Understand the Definition of a Zero of a
Polynomial
2. Use the Remainder Theorem
3. Use the Factor Theorem
4. Use Synthetic Division to Divide Polynomials
5. Use Synthetic Division to Evaluate Polynomials
6. Use Synthetic Division to Solve Polynomial
Equations
1. Understand the Definition of
a Zero of a Polynomial
Polynomial Equations
A polynomial equation is an equation that
can be written in the form P(x) = 0, where
P(x) = anxn + an–1xn–1 + an–2xn–2 + … + a1x + a0
where n is a natural number and the
polynomial is of degree n.
Zero of a Polynomial Function
A zero of the polynomial P(x) is any
number r for which P(r) = 0.
Roots and Zeros
In general, the roots of the polynomial
equation P(x) = 0 are the zeros of the
polynomial P(x) .
2. Use the Remainder Theorem
Example 1
Let P(x) = 3x3 – 5x2 + 3x – 10.
a. Find P(1).
b. Divide P(x) by x – 1.
Example 1(a) – Solution
To find P(1), we will substitute 1 for x in the
polynomial.
P 1  3 1  5 1  3 1  10
3
2
 3  5  3  10
 9
Example 1(b) – Solution
To divide P(x) by x – 1, we proceed as
follows:
2
3 x  2x  1
x  1 3 x 3  5 x 2  3 x  10
3x3  3x 2
 2x 2  3 x
 2x 2  2x
 x  10
x 1
 9
Example 1 – Solution
Note that the remainder is equal to P(1).
The Remainder Theorem
If P(x) is a polynomial, r is any number, and
P(x) is divided by x – r, the remainder is P(r).
3. Use the Factor Theorem
The Factor Theorem
If P(x) is a polynomial, r is any number, then
• If P(r) = 0, then x – r is a factor of P(x).
• If x – r is a factor of P(x), then P(r) = 0.
Alternate Form of The Factor Theorem
If r is a zero of the polynomial P(x) = 0,
then x – r is a factor of P(x).
If x – r is a factor of P(x),
then r is a zero of the polynomial.
Example 4
Determine whether x + 2 is a factor of
P(x) = x4 – 7x2 – 6x
Example 4 – Solution
By the factor theorem, x + 2, or x – (–2),
is a factor of P(x) if P(–2) = 0.
•
So we find the value of P(–2).
P  x   x 4  7x 2  6x
P  2    2   7  2   6  2 
4
2
 16  28  12
0
•
Since P(–2) = 0, we know that x – (–2), or
x + 2, is a factor of P(x).
4. Use Synthetic Division to
Divide Polynomials
Synthetic Division
Synthetic division is an easy way to divide
higher-degree polynomials by binomials of
the form x – r.
• And, it is much faster than long division.
Example 5
Use synthetic division to divide
10x + 3x4 – 8x3 + 3
by x – 2.
Example 5 – Solution
We first write the terms of P(x) in descending
powers of x:
3x4 – 8x3 + 10x + 3
• We then write the coefficients of the dividend, with its
terms in descending powers of x and writing 0 for the
coefficient of the missing x2 term, and the 2 from the
divisor in the following form:
2 3 8 0 10 3
Example 5 – Solution
Then we follow these steps:
Example 5 – Solution
Example 5 – Solution
Thus,
10 x  3 x 4  8 x 3  3
7
3
2
 3 x  2x  4 x  2 
x 2
x 2
5. Use Synthetic Division to
Evaluate Polynomials
Example 6
Use synthetic division to find P(–2) when
P(x) = 5x3 + 3x2 – 21x – 1
Example 6 – Solution
Because of the remainder theorem,
P(–2) is the remainder when P(x) is
divided by x – (–2).
• Earlier in the section, we would have found
the remainder by using long division.
• We now can simplify the work and find the
remainder by using synthetic division.
Example 6 – Solution
2 5
5
2 5
5
3 21 1
10
14
7
7
3 21 1
10
14 14
7
7 13
• Because the remainder is 13, P(–2) = 13.
6. Use Synthetic Division to Solve
Polynomial Equations
Example 8
Let P(x) = 3x3 – 5x2 + 3x – 10.
• Completely solve the polynomial equation
P(x) = 0 given that 2 is one solution.
Example 8 – Solution
Since 2 is a solution of the equation
P(x) = 0, we know that 2 is a zero of P(x).
• We will use synthetic division and divide P(x)
by x – 2, obtaining a remainder of 0.
• Then we will use the result of the synthetic
division to help factor the polynomial and
solve the equation.
Example 8 – Solution
We use synthetic division to divide P(x)
by x – 2.
2 3 5 3 10
3
6 2
10
1 5
0
• We then write the quotient and factor it.
3 x 3  5 x 2  3 x  10   x  2   3 x 2  x  5 
Example 8 – Solution
Finally, we solve the polynomial equation
P(x) = 0.
3 x 3  5 x 2  3 x  10  0
 x  2 3x
2
 x  5  0
x  2  0 or 3 x  x  5  0
2
1  1  4(3)(5)
x
2(3)
2
x2
1  i 59
x
6
Example 8 – Solution
The solution set is

1
59
1
59 
i,  
i
2,  
6
6
6
6 

Multiplicity of Zeros
If r is a zero of P(x) that occurs n times,
we say that r is a zero of multiplicity n.
• For example,
P  x   x 3  x 2  21x  45
  x  3  x  3  x  5 
has zeros 3, 3, and –5.
• Because 3 occurs twice as a zero, we say
that 3 is a zero of multiplicity 2.