Section 6.2
Zeros of Polynomials
577
6.2 Exercises
In Exercises 1-6, use direct substitution to show that the given value is a
zero of the given polynomial.
16.
p(x) = 3x3 + x2 − 12x − 4
17.
p(x) = 2x3 + 5x2 − 2x − 5
1.
p(x) = x3 − 3x2 − 13x + 15, x = −3
18.
p(x) = 2x3 − 5x2 − 18x + 45
2.
p(x) = x3 − 2x2 − 13x − 10, x = −2
19.
p(x) = x4 + 4x3 − 9x2 − 36x
3.
p(x) = x4 − x3 − 12x2 , x = 4
20.
p(x) = −x4 + 4x3 + x2 − 4x
4.
p(x) = x4 − 2x3 − 3x2 , x = −1
21.
p(x) = −2x4 − 10x3 + 8x2 + 40x
5.
p(x) = x4 + x2 − 20, x = −2
22.
p(x) = 3x4 + 6x3 − 75x2 − 150x
23.
p(x) = 2x3 − 7x2 − 15x
24.
p(x) = 2x3 − x2 − 10x
25.
p(x) = −6x3 + 4x2 + 16x
26.
p(x) = 9x3 + 3x2 − 30x
27.
p(x) = −2x7 − 10x6 + 8x5 + 40x4
28.
p(x) = 6x5 − 21x4 − 45x3
6. p(x) = x4 + x3 − 19x2 + 11x + 30,
x = −1
In Exercises 7-28, identify all of the
zeros of the given polynomial without
the aid of a calculator. Use an algebraic technique and show all work (factor when necessary) needed to obtain the
zeros.
1
7.
p(x) = (x − 2)(x + 4)(x − 5)
8.
p(x) = (x − 1)(x − 3)(x + 8)
9.
p(x) = −2(x − 3)(x + 4)(x − 2)
10.
p(x) = −3(x + 1)(x − 1)(x − 8)
11.
p(x) = x(x − 3)(2x + 1)
12.
p(x) = −3x(x + 5)(3x − 2)
13.
p(x) = −2(x + 3)(3x − 5)(2x + 1)
14.
p(x) = 3(x − 2)(2x + 5)(3x − 4)
15.
p(x) = 3x3 + 5x2 − 12x − 20
Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/
Version: Fall 2007
578
Chapter 6
Polynomial Functions
In Exercises 29-34, the graph of a polynomial is given. Perform each of the following tasks.
32.
y
10
i. Copy the image onto your homework
paper. Label and scale your axes,
then label each x-intercept with its
coordinates.
ii. Identify the zeros of the polynomial.
x
10
29.
y
10
33.
y
10
x
10
x
10
30.
y
10
34.
y
10
x
10
x
10
31.
y
10
x
10
Version: Fall 2007
Section 6.2
Zeros of Polynomials
579
For each of the polynomials in Exercises 3546, perform each of the following tasks.
i. Factor the polynomial to obtain the
zeros. Show your work.
ii. Set up a coordinate system on graph
paper. Label and scale the horizontal
axis. Use the zeros and end-behavior
to help sketch the graph of the polynomial without the use of a calculator.
iii. Verify your result with a graphing calculator.
35.
p(x) = 5x3 + x2 − 45x − 9
36.
p(x) = 4x3 + 3x2 − 64x − 48
37.
p(x) = 4x3 − 12x2 − 9x + 27
38.
p(x) = x3 + x2 − 16x − 16
39.
p(x) = x4 + 2x3 − 25x2 − 50x
40.
p(x) = −x4 − 5x3 + 4x2 + 20x
41.
p(x) = −3x4 − 9x3 + 3x2 + 9x
42.
p(x) = 4x4 − 29x2 + 25
43.
p(x) = −x3 − x2 + 20x
44.
p(x) = 2x3 − 7x2 − 30x
45.
p(x) = 2x3 + 3x2 − 35x
46.
p(x) = −2x3 − 11x2 + 21x
Version: Fall 2007
Chapter 6
Polynomial Functions
6.2 Solutions
1.
p(−3) = (−3)3 − 3(−3)2 − 13(−3) + 15 = −27 − 27 + 39 + 15 = 0
3.
p(4) = 44 − 43 − 12(4)2 = 256 − 64 − 192 = 0
5.
p(−2) = (−2)4 + (−2)2 − 20 = 16 + 4 − 20 = 0
7.
Set p(x) = 0 in p(x) = (x − 2)(x + 4)(x − 5),
0 = (x − 2)(x + 4)(x − 5),
then use the zero product property to write
x−2=0
or
x+4=0
x − 5 = 0.
or
Solving, the zeros are x = 2, −4, and 5.
9.
Set p(x) = 0 in p(x) = −2(x − 3)(x + 4)(x − 2),
0 = −2(x − 3)(x + 4)(x − 2),
then use the zero product property to write
x−3=0
or
x+4=0
x − 2 = 0.
or
Solving, the zeros are x = 3, −4, and 2.
11.
Set p(x) = 0 in p(x) = x(x − 3)(2x + 1),
0 = x(x − 3)(2x + 1),
then use the zero product property to write
x=0
or
x−3=0
or
2x + 1 = 0.
Solving, the zeros are x = 0, 3, and −1/2.
13.
Set p(x) = 0 in p(x) = −2(x + 3)(3x − 5)(2x + 1),
0 = −2(x + 3)(3x − 5)(2x + 1),
then use the zero product property to write
x+3=0
or
3x − 5 = 0
Solving, the zeros are x = −3, 5/3, and −1/2.
Version: Fall 2007
or
2x + 1 = 0.
Section 6.2
Zeros of Polynomials
15. Factor p(x) = 3x3 + 5x2 − 12x − 20 by grouping, then finish the factoring using
the difference of squares pattern.
p(x) = 3x3 + 5x2 − 12x − 20
p(x) = x2 (3x + 5) − 4(3x + 5)
p(x) = (x2 − 4)(3x + 5)
p(x) = (x + 2)(x − 2)(3x + 5)
To find the zeros, set p(x) = 0,
0 = (x + 2)(x − 2)(3x + 5),
then use the zero product property to write
x+2=0
or
x−2=0
or
3x + 5 = 0.
Solving, the zeros are x = −2, 2, or −5/3.
17. Factor p(x) = 2x3 + 5x2 − 2x − 5 by grouping, then finish the factoring using the
difference of squares pattern.
p(x) = 2x3 + 5x2 − 2x − 5
p(x) = x2 (2x + 5) − 1(2x + 5)
p(x) = (x2 − 1)(2x + 5)
p(x) = (x + 1)(x − 1)(2x + 5)
To find the zeros, set p(x) = 0,
0 = (x + 1)(x − 1)(2x + 5),
then use the zero product property to write
x+1=0
or
x−1=0
or
2x + 5 = 0.
Solving, the zeros are x = −1, 1, or −5/2.
19. Start with p(x) = x4 + 4x3 − 9x2 − 36x. Factor out the gcf (x in this case), then
factor by grouping. In the last step, use the difference of squares pattern to complete
the factorization.
p(x) = x[x3 + 4x2 − 9x − 36]
p(x) = x[x2 (x + 4) − 9(x + 4)]
p(x) = x(x2 − 9)(x + 4)
p(x) = x(x + 3)(x − 3)(x + 4)
Set
0 = x(x + 3)(x − 3)(x + 4)
Version: Fall 2007
Chapter 6
Polynomial Functions
and use the zero product property to write
x=0
or
x+3=0
or
x−3=0
or
x + 4 = 0.
Solving, the zeros are x = 0, −3, 3, and −4.
21. Start with p(x) = −2x4 − 10x3 + 8x2 + 40x. Factor out the gcf (−2x in this
case), then factor by grouping. In the last step, use the difference of squares pattern
to complete the factorization.
p(x) = −2x[x3 + 5x2 − 4x − 20]
p(x) = −2x[x2 (x + 5) − 4(x + 5)]
p(x) = −2x(x2 − 4)(x + 5)
p(x) = −2x(x + 2)(x − 2)(x + 5)
Set
0 = −2x(x + 2)(x − 2)(x + 5)
and use the zero product property to write
−2x = 0
or
x+2=0
or
x−2=0
or
x + 5 = 0.
Solving, the zeros are x = 0, −2, 2, and −5.
23. Start with p(x) = 2x3 − 7x2 − 15x. Factor out the gcf (x in this case), then use
the ac-method to complete the factorization.
p(x) = x[2x2 − 7x − 15]
p(x) = x[2x2 − 10x + 3x − 15]
p(x) = x[2x(x − 5) + 3(x − 5)]
p(x) = x(2x + 3)(x − 5)
Set
0 = x(2x + 3)(x − 5)
and use the zero product property to write
x=0
or
2x + 3 = 0
or
x − 5 = 0.
Solving, the zeros are x = 0, −3/2, and 5.
25. Start with p(x) = −6x3 + 4x2 + 16x. Factor out the gcf (−2x in this case), then
use the ac-method to complete the factorization.
p(x) = −2x[3x2 − 2x − 8]
p(x) = −2x[3x2 − 6x + 4x − 8]
p(x) = −2x[3x(x − 2) + 4(x − 2)]
p(x) = −2x(3x + 4)(x − 2)
Version: Fall 2007
Section 6.2
Zeros of Polynomials
Set
0 = −2x(3x + 4)(x − 2)
and use the zero product property to write
−2x = 0
or
3x + 4 = 0
or
x − 2 = 0.
Solving, the zeros are x = 0, −4/3, and 2.
27. Start with p(x) = −2x7 − 10x6 + 8x5 + 40x4 . Factor out the gcf (−2x4 in this
case), then use grouping and difference of squares to complete the factorization.
p(x) = −2x4 [x3 + 5x2 − 4x − 20]
p(x) = −2x4 [x2 (x + 5) − 4(x + 5)]
p(x) = −2x4 (x2 − 4)(x + 5)
p(x) = −2x4 (x + 2)(x − 2)(x + 5)
Set
0 = −2x4 (x + 2)(x − 2)(x + 5)
and use the zero product property to write
−2x4 = 0
or
x+2=0
or
x−2=0
or
x + 5 = 0.
Solving, the zeros are x = 0, −2, 2, and −5.
29.
The graph of the polynomial
y
10
x
10
intercepts the x-axis at (−4, 0), (1, 0), and (2, 0). Hence, the zeros of the polynomial
are −4, 1, and 2.
Version: Fall 2007
Chapter 6
31.
Polynomial Functions
The graph of the polynomial
y
10
x
10
intercepts the x-axis at (−4, 0), (0, 0), and (5, 0). Hence, the zeros of the polynomial
are −4, 0, and 5.
33.
The graph of the polynomial
y
10
x
10
intercepts the x-axis at (−3, 0), (0, 0), (2, 0), and (6, 0). Hence, the zeros of the polynomial are −3, 0, 2, and 6.
35. Factor p(x) = 5x3 + x2 − 45x − 9 by grouping, then complete the factorization
with the difference of squares pattern.
p(x) = x2 (5x + 1) − 9(5x + 1)
p(x) = (x2 − 9)(5x + 1)
p(x) = (x + 3)(x − 3)(5x + 1)
Using the zero product property, the zeros are −3, 3, and −1/5. Hence, the graph
of the polynomial must intercept the x-axis at (−3, 0), (3, 0), and (−1/5, 0). Further,
the leading term of the polynomial is 5x3 , so the polynomial must have the same endbehavior as y = 5x3 , namely, it must rise from negative infinity, wiggle through its
x-intercepts, then rise to positive infinity. The sketch with the appropriate zeros and
end behavior follows.
Version: Fall 2007
Section 6.2
Zeros of Polynomials
y
(−3,0)
(−1/5,0)
(3,0)
x
Checking on the calculator.
37. Factor p(x) = 4x3 − 12x2 − 9x + 27 by grouping, then complete the factorization
with the difference of squares pattern.
p(x) = 4x2 (x − 3) − 9(x − 3)
p(x) = (4x2 − 9)(x − 3)
p(x) = (2x + 3)(2x − 3)(x − 3)
Using the zero product property, the zeros are −3/2, 3/2, and 3. Hence, the graph
of the polynomial must intercept the x-axis at (−3/2, 0), (3/2, 0), and (3, 0). Further,
the leading term of the polynomial is 4x3 , so the polynomial must have the same endbehavior as y = 4x3 , namely, it must rise from negative infinity, wiggle through its
x-intercepts, then rise to positive infinity. The sketch with the appropriate zeros and
end behavior follows.
y
(−3/2,0)
(3/2,0)
(3,0)
x
Version: Fall 2007
Chapter 6
Polynomial Functions
Checking on the calculator.
39. Start with p(x) = x4 + 2x3 − 25x2 − 50x, then factor out the gcf (x in this case).
Then, factor by grouping and complete the factorization with the difference of squares
pattern.
p(x) = x[x3 + 2x2 − 25x − 50]
p(x) = x[x2 (x + 2) − 25(x + 2)]
p(x) = x(x2 − 25)(x + 2)
p(x) = x(x + 5)(x − 5)(x + 2)
Using the zero product property, the zeros are 0, −5, 5, and −2. Hence, the graph of
the polynomial must intercept the x-axis at (0, 0), (−5, 0), (5, 0), and (−2, 0). Further,
the leading term of the polynomial is x4 , so the polynomial must have the same endbehavior as y = x4 , namely, it must fall from positive infinity, wiggle through its
x-intercepts, then rise to positive infinity. The sketch with the appropriate zeros and
end behavior follows.
y
(−2,0)
(−5,0)
Checking on the calculator.
Version: Fall 2007
(0,0)
(5,0)
x
Section 6.2
Zeros of Polynomials
41. Start with p(x) = −3x4 − 9x3 + 3x2 + 9x, then factor out the gcf (−3x in this
case). Then, factor by grouping and complete the factorization with the difference of
squares pattern.
p(x) = −3x[x3 + 3x2 − x − 3]
p(x) = −3x[x2 (x + 3) − 1(x + 3)]
p(x) = −3x(x2 − 1)(x + 3)
p(x) = −3x(x + 1)(x − 1)(x + 3)
Using the zero product property, the zeros are 0, −1, 1, and −3. Hence, the graph of
the polynomial must intercept the x-axis at (0, 0), (−1, 0), (1, 0), and (−3, 0). Further,
the leading term of the polynomial is −3x4 , so the polynomial must have the same
end-behavior as y = −3x4 , namely, it must rise from negative infinity, wiggle through
its x-intercepts, then fall back to negative infinity. The sketch with the appropriate
zeros and end behavior follows.
y
(−1,0)
(−3,0)
(0,0)
(1,0)
x
Checking on the calculator.
43. Start with p(x) = −x3 − x2 + 20x, then factor out the gcf (−x in this case). Then,
complete the factorization with the ac-method.
p(x) = −x[x2 + x − 20]
p(x) = −x(x + 5)(x − 4)
Using the zero product property, the zeros are 0, −5, and 4. Hence, the graph of the
polynomial must intercept the x-axis at (0, 0), (−5, 0), and (4, 0). Further, the leading
term of the polynomial is −x3 , so the polynomial must have the same end-behavior as
y = −x3 , namely, it must fall from positive infinity, wiggle through its x-intercepts,
then fall to negative infinity. The sketch with the appropriate zeros and end behavior
follows.
Version: Fall 2007
Chapter 6
Polynomial Functions
y
(0,0)
(4,0)
(−5,0)
x
Checking on the calculator.
45. Start with p(x) = 2x3 + 3x2 − 35x, then factor out the gcf (x in this case). Then,
complete the factorization with the ac-method.
p(x) = x[2x2 + 3x − 35]
p(x) = x[2x2 − 7x + 10x − 35]
p(x) = x[x(2x − 7) + 5(2x − 7)]
p(x) = x(x + 5)(2x − 7)
Using the zero product property, the zeros are 0, −5, and 7/2. Hence, the graph of
the polynomial must intercept the x-axis at (0, 0), (−5, 0), and (7/2, 0). Further, the
leading term of the polynomial is 2x3 , so the polynomial must have the same endbehavior as y = 2x3 , namely, it must rise from negative infinity, wiggle through its
x-intercepts, then rise to positive infinity. The sketch with the appropriate zeros and
end behavior follows.
y
(0,0)
(−5,0)
Version: Fall 2007
(7/2,0)
x
Section 6.2
Zeros of Polynomials
Checking on the calculator.
Version: Fall 2007