5-3 Solving Polynomial Equations
Quick Review
One way to solve a polynomial equation is by factoring. First write the equation in the form
P
(
x
)
=
0
,
p open x close equals 0 comma where
P
(
x
)
p open x close is the polynomial. Then factor the polynomial. Last, use the Zero-Product Property to find the solutions, or roots. The solutions may be real or imaginary. Real solutions and approximations of irrational solutions can also be found by using a graphing calculator.
Example
Solve
x
3
+
4
x
2
=
12
x
x cubed , plus 4 , x squared , equals 12 x by factoring.
x
3
+
4
x
2
−
12
x
=
0
Set equal to
0.
x
(
x
−
2
)
(
x
+
6
)
=
0
Factor the left side
.
x
=
0
,
x
−
2
=
0
,
x
+
6
=
0
Zero Product Property
.
x
=
0
,
x
=
2
,
x
=
−
6
Solve each equation
.
table with 4 rows and 2 columns , row1 column 1 , x cubed , plus 4 , x squared , minus 12 x equals 0 , column 2 cap setequalto . 0. , row2 column 1 , x . open , x minus 2 , close . open , x plus 6 , close . equals 0 , column 2 cap factortheleftside . . , row3 column 1 , x equals 0 comma x minus 2 equals 0 comma x plus 6 equals 0 , column 2 cap zerocap productcap property . . , row4 column 1 , x equals 0 comma x equals 2 comma x equals negative 6 , column 2 cap solveeachequation . . , end table
The solutions are 0, 2, and
−
6
.
negative 6 .
Exercises
Find the real or imaginary solutions of each equation by factoring.
-
x
2
−
11
x
=
−
24
x squared , minus 11 x equals negative 24
-
4
x
2
=
−
4
x
−
1
4 x squared , equals negative 4 x minus 1
-
3
x
3
+
3
x
2
=
27
x
3 x cubed , plus 3 , x squared , equals 27 x
-
2
x
2
+
3
=
4
x
2 x squared , plus 3 equals 4 x
Find the real roots of each equation by graphing.
-
x
4
+
3
x
2
−
2
x
+
5
=
0
x to the fourth , plus 3 , x squared , minus 2 x plus 5 equals 0
-
x
2
+
3
=
x
3
−
5
x squared , plus 3 equals , x cubed , minus 5
- The height and width of a rectangular prism are each 2 inches shorter than the length of the prism. The volume of the prism is 40 cubic inches. Approximate the dimensions of the prism to the nearest hundredth.
5-4 Dividing Polynomials
Quick Review
You can divide a polynomial by one of its factors to find another factor. When you divide by a linear factor, you can simplify this division by writing only the coefficients of each term. This is called synthetic division. The Remainder Theorem says that
P
(
a
)
p open eh close is the remainder when you divide
P
(
x
)
p open x close by
x
−
a
.
x minus eh .
Example
Let
P
(
x
)
=
3
x
2
−
13
x
+
15
.
p open x close equals , 3 x squared , minus 13 x plus 15 . What is
P
(
3
)
?
p open 3 close question mark
According to the Remainder Theorem,
P
(
3
)
p open 3 close is the remainder when you divide
P
(
x
)
p open x close by
x
−
3
.
x minus 3 .
3
3
−
13
15
Put the opposite of the constant in
9
−
12
the divisor at the top left
.
3
−
4
3
¯
table with 3 rows and 2 columns , row1 column 1 , 3 3 minus 13 15 , column 2 cap puttheoppositeoftheconstantin , row2 column 1 , 9 minus 12 , column 2 thedivisoratthetopleft . . , row3 column 1 , 3 minus 4 3 bar , column 2 , end table
The quotient is
3
x
−
4
3 x minus 4 with remainder 3, so
P
(
3
)
=
3
.
p open 3 close equals 3 .
Exercises
Divide using long division. Check your answers.
-
(
x
3
+
7
x
2
+
15
x
+
9
)
÷
(
x
+
1
)
open , x cubed , plus , 7 x squared , plus 15 x plus 9 close divides open x plus 1 close
-
(
2
x
3
−
7
x
2
−
7
x
+
13
)
÷
(
x
−
4
)
open 2 , x cubed , minus , 7 x squared , minus 7 x plus 13 close divides open x minus 4 close
Determine whether each binomial is a factor of
x
3
+
x
2
−
10
x
+
8
.
x cubed , plus , x squared , minus 10 x plus 8 .
-
x
−
2
x minus , 2
-
x
−
4
x minus , 4
Divide using synthetic division.
-
(
x
3
+
5
x
2
−
x
−
5
)
÷
(
x
+
5
)
open , x cubed , plus , 5 x squared , minus x minus 5 close divides open x plus 5 close
-
(
2
x
3
+
14
x
2
−
58
x
)
÷
(
x
+
10
)
open 2 , x cubed , plus , 14 x squared , minus 58 x close divides open x plus 10 close
-
(
5
x
3
+
8
x
2
−
60
)
÷
(
x
−
2
)
open 5 , x cubed , plus , 8 x squared , minus 60 close divides open x minus 2 close
Use the Remainder Theorem to determine the value of
P
(
a
)
.
p open eh close .
-
P
(
x
)
=
2
x
3
+
5
x
2
+
7
x
−
4
,
a
=
−
2
p open x close equals , 2 x cubed , plus , 5 x squared , plus 7 x minus 4 comma eh equals negative 2
-
P
(
x
)
=
x
3
−
4
x
2
+
2
x
+
3
,
a
=
1
p open x close equals , x cubed , minus , 4 x squared , plus 2 x plus 3 comma eh equals 1
