3-5 Roots and Zeros
Solve each equation. State the number and type of roots.
2. x 2 – 4x + 40 = 0
1. 5x + 12 = 0
SOLUTION:
SOLUTION:
The polynomial has degree 1, so there is one root in the set of complex
numbers. The equation has one real root,
.
The polynomial has degree 2, so there are two roots in the set of
complex numbers. The equation has two imaginary roots, 2 + 6i and 2
– 6i.
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3-5 Roots and Zeros
3. x 5 + 4x 3 = 0
4. x 4 – 625 = 0
SOLUTION:
SOLUTION:
The polynomial has degree 5, so there are five roots in the set of
complex numbers. Because x 3 is a factor, x = 0 is a root with
multiplicity 3. The equation has one real repeated root, 0, and two
imaginary roots, 2i and −2i.
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The polynomial has degree 4, so there are four roots in the set of
complex numbers. The equation has two real roots, −5 and 5, and two
imaginary roots, 5i and −5i.
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3-5 Roots and Zeros
5. 4x 2 – 4x – 1 = 0
6. x 5 – 81x = 0
SOLUTION:
SOLUTION:
The polynomial has degree 2, so there are two roots in the set of
complex numbers. The equation has two real roots,
.
and
The polynomial has degree 5, so there are five roots in the set of
complex numbers. The equation has three real roots, 0, −3, and 3, and
two imaginary roots, −3i and 3i.
7. 2x 2 + x – 6 = 0
SOLUTION:
The polynomial has degree 2, so there are two roots in the set of
complex numbers. The equation has two real roots, −2 and
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.
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3-5 Roots and Zeros
9. x 3 + 1 = 0
8. 4x 2 + 1 = 0
SOLUTION:
SOLUTION:
The polynomial has degree 2, so there are two roots in the set of
complex numbers. The equation has two imaginary roots,
and
The polynomial has degree 3, so there are three roots in the set of
complex numbers. The equation has one real root, −1, and two
imaginary roots,
and
.
.
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3-5 Roots and Zeros
10. 2x 2 – 5x + 14 = 0
11. −3x 2 – 5x + 8 = 0
SOLUTION:
SOLUTION:
The polynomial has degree 2, so there are two roots in the set of
The polynomial has degree 2, so there are two roots in the set of
complex numbers. The equation has two real roots,
and 1.
complex numbers. The equation has two imaginary roots,
and
.
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3-5 Roots and Zeros
12. 8x 3 – 27 = 0
13. 16x 4 – 625 = 0
SOLUTION:
SOLUTION:
The polynomial has degree 3, so there are three roots in the set of
complex numbers. The equation has one real root,
imaginary roots,
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and
.
, and two
The polynomial has degree 4, so there are four roots in the set of
complex numbers. The equation has two real roots,
two imaginary roots,
and
and
, and
.
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3-5 Roots and Zeros
14. x 3 – 6x 2 + 7x = 0
15. x 5 – 8x 3 + 16x = 0
SOLUTION:
SOLUTION:
The polynomial has degree 3, so there are three roots in the set of
complex numbers. The equation has three real roots, 0,
,
.
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The polynomial has degree 5, so there are five roots in the set of
complex numbers. Because (x 2 – 4)2 is a factor, x = ±2 are both roots
with multiplicity of 2. The equation has five real roots, −2, −2, 0, 2, and
2.
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3-5 Roots and Zeros
16. x 5 + 2x 3 + x = 0
SOLUTION:
Count the number of changes in sign for the coefficients of g(x).
There are 2 sign changes, so there are 2 or 0 positive real zeros.
Find the possible number of negative real zeros.
The polynomial has degree 5, so there are five roots in the set of
complex numbers. Because (x 2 + 1)2 is a factor, x = ±i are both roots
with multiplicity of 2. The equation has one real root, 0, and four
imaginary roots, −i, −i, i, and i.
State the possible number of positive real zeros, negative real
zeros, and imaginary zeros of each function.
17. g(x) = 3x 3 – 4x 2 – 17x + 6
SOLUTION:
Because g(x) has degree 3, it has three zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Find the possible number of positive real zeros.
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Count the number of changes in sign for the coefficients of g(−x).
There is 1 sign change, so there is 1 negative real zero.
Find the number of imaginary zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
2
1
0
2+1+0=3
0
1
2
0+1+2=3
18. h(x) = 4x 3 – 12x 2 – x + 3
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3-5 Roots and Zeros
SOLUTION:
Because h(x) has degree 3, it has three zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Find the possible number of positive real zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
2
1
0
2+1+0=3
0
1
2
0+1+2=3
19. f(x) = x 3 – 8x 2 + 2x – 4
SOLUTION:
Because f(x) has degree 3, it has three zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Count the number of changes in sign for the coefficients of h(x).
Find the possible number of positive real zeros.
There are 2 sign changes, so there are 2 or 0 positive real zeros.
Find the possible number of negative real zeros.
Count the number of changes in sign for the coefficients of f(x).
There are 3 sign changes, so there are 3 or 1 positive real zeros.
Count the number of changes in sign for the coefficients of h(−x).
Find the possible number of negative real zeros.
There is 1 sign change, so there is 1 negative real zero.
Find the number of imaginary zeros.
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3-5 Roots and Zeros
Count the number of changes in sign for the coefficients of f(−x).
There are 0 sign changes, so there are 0 negative real zero.
Find the number of imaginary zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
3
0
0
3+0+0=3
1
0
2
1+0+2=3
20. p(x) = x 3 – x 2 + 4x – 6
SOLUTION:
Because p(x) has degree 3, it has three zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Find the possible number of positive real zeros.
Count the number of changes in sign for the coefficients of p(−x).
There are 0 sign changes, so there are 0 negative real zero.
Find the number of imaginary zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
3
0
0
3+0+0=3
1
0
2
1+0+2=3
21. q(x) = x 4 + 7x 2 + 3x – 9
SOLUTION:
Because q(x) has degree 4, it has four zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Find the possible number of positive real zeros.
Count the number of changes in sign for the coefficients of p(x).
There are 3 sign changes, so there are 3 or 1 positive real zeros.
Find the possible number of negative real zeros.
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3-5 Roots and Zeros
Count the number of changes in sign for the coefficients of q(x).
There is 1 sign change, so there is 1 positive real zero.
Find the possible number of negative real zeros.
Count the number of changes in sign for the coefficients of f(x).
There are 2 sign changes, so there are 2 or 0 positive real zeros.
Find the possible number of negative real zeros.
Count the number of changes in sign for the coefficients of q(−x).
There is 1 sign change, so there is 1 negative real zero.
Find the number of imaginary zeros.
Count the number of changes in sign for the coefficients of f(−x).
Positive Real
Zeros
1
Negative Real
Imaginary Zeros Total Zeros
Zeros
1
2
1+1+2=4
22. f(x) = x 4 – x 3 – 5x 2 + 6x + 1
SOLUTION:
Because f(x) has degree 4, it has four zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Find the possible number of positive real zeros.
There are 2 sign changes, so there are 2 or 0 negative real zeros.
Find the number of imaginary zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
2
2
0
2+2+0=4
2
0
2
2+0+2=4
0
2
2
0+2+2=4
0
0
4
0+0+4=4
23. f(x) = x 4 – 5x 3 + 2x 2 + 5x + 7
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3-5 Roots and Zeros
SOLUTION:
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
Because f(x) has degree 4, it has four zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
2
2
0
2+2+0=4
2
0
2
2+0+2=4
0
2
2
0+2+2=4
Find the possible number of positive real zeros.
0
0
4
0+0+4=4
24. f(x) = 2x 3 – 7x 2 – 2x + 12
SOLUTION:
Count the number of changes in sign for the coefficients of f(x).
Because f(x) has degree 3, it has three zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
There are 2 sign changes, so there are 2 or 0 positive real zeros.
Find the possible number of positive real zeros.
Find the possible number of negative real zeros.
Count the number of changes in sign for the coefficients of f(x).
Count the number of changes in sign for the coefficients of f(−x).
There are 2 sign changes, so there are 2 or 0 positive real zeros.
There are 2 sign changes, so there are 2 or 0 negative real zeros.
Find the possible number of negative real zeros.
Find the number of imaginary zeros.
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3-5 Roots and Zeros
Count the number of changes in sign for the coefficients of f(x).
There are 2 sign changes, so there are 2 or 0 positive real zeros.
Find the possible number of negative real zeros.
Count the number of changes in sign for the coefficients of f(−x).
There is 1 sign change, so there is 1 negative real zero.
Find the number of imaginary zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
2
1
0
2+1+0=3
0
1
2
0+1+2=3
25. f(x) = –3x 5 + 5x 4 + 4x 2 – 8
SOLUTION:
Because f(x) has degree 5, it has five zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Find the possible number of positive real zeros.
Count the number of changes in sign for the coefficients of f(−x).
There is 1 sign change, so there is 1 negative real zero.
Find the number of imaginary zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
2
1
2
2+1+2=5
0
1
4
0+1+4=5
26. f(x) = x 4 – 2x 2 – 5x + 19
SOLUTION:
Because f(x) has degree 4, it has four zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Find the possible number of positive real zeros.
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3-5 Roots and Zeros
27. f(x) = 4x 6 – 5x 4 – x 2 + 24
SOLUTION:
Count the number of changes in sign for the coefficients of f(x).
Because f(x) has degree 6, it has six zeros, either real or imaginary. Use
Descartes' Rule of Signs to determine the possible number and types of
real zeros.
There are 2 sign changes, so there are 2 or 0 positive real zeros.
Find the possible number of positive real zeros.
Find the possible number of negative real zeros.
Count the number of changes in sign for the coefficients of f(x).
Count the number of changes in sign for the coefficients of f(−x).
There are 2 sign changes, so there are 2 or 0 positive real zeros.
There are 2 sign changes, so there are 2 or 0 negative real zeros.
Find the possible number of negative real zeros.
Find the number of imaginary zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
2
2
0
2+2+0=4
2
0
2
2+0+2=4
0
2
2
0+2+2=4
0
0
4
0+0+4=4
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Count the number of changes in sign for the coefficients of f(−x).
There are 2 sign changes, so there are 2 or 0 negative real zeros.
Find the number of imaginary zeros.
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3-5 Roots and Zeros
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
2
2
2
2+2+2=6
2
0
4
2+0+4=6
0
2
4
0+2+4=6
0
0
6
0+0+6=6
28. f(x) = −x 5 + 14x 3 + 18x – 36
Count the number of changes in sign for the coefficients of f(−x).
There is 1 sign changes, so there is 1 negative real zero.
Find the number of imaginary zeros.
SOLUTION:
Because f(x) has degree 5, it has five zeros, either real or imaginary.
Use Descartes' Rule of Signs to determine the possible number and
types of real zeros.
Positive Real
Zeros
Negative Real
Imaginary Zeros Total Zeros
Zeros
2
1
2
2+1+2=5
0
1
4
0+1+4=5
Find the possible number of positive real zeros.
Find all of the zeros of each function and use them to sketch a
rough graph.
29. h(x) = x 3 – 5x 2 + 5x + 3
SOLUTION:
Since h(x) has degree 3, the function has 3 zeros.
Count the number of changes in sign for the coefficients of f(x).
Examine the number of sign changes for h(x) and h(−x).
There are 2 sign changes, so there are 2 or 0 positive real zeros.
Find the possible number of negative real zeros.
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3-5 Roots and Zeros
Use this information and points with coordinates found in the table
above to sketch the graph.
Because there are 2 sign changes for the coefficients of h(x), the
function has 2 or 0 positive real zeros. Because there is 1 sign change
for the coefficients of h(−x), h(x) has 1 negative real zero. Thus, h(x)
has 3 real zeros, or 1 real zero and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate h(x) for real values of x.
x
1
2
3
1
1
1
1
−5
−4
−3
−2
5
1
−1
−1
3
4
1
0
30. g(x) = x 3 – 6x 2 + 13x – 10
SOLUTION:
3 is a zero of the function and the depressed polynomial is x 2 – 2x – 1.
Since it is quadratic, use the Quadratic Formula. The zeros of h(x)
= x 2 – 2x – 1 are
and
.
Since g(x) has degree 3, the function has 3 zeros.
Examine the number of sign changes for g(x) and g(−x).
The function has zeros at 1,
, and
.
The function has three real zeros at x = 3,
, so the function goes through (3, 0),
, and
, and
.
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, h(x) → −∞ and as x → ∞, h(x) → ∞.
Because there are 3 sign changes for the coefficients of g(x), the
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3-5 Roots and Zeros
function has 3 or 1 positive real zeros. Because there are 0 sign change
for the coefficients of g(−x), g(x) has 0 negative real zeros. Thus, g(x)
has 3 real zeros, or 1 real zero and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate g(x) for real values of x.
x
0
1
2
1
1
1
1
−6
−6
−5
−4
13
13
8
5
−10
−10
−2
0
2 is a zero of the function and the depressed polynomial is x 2 – 4x + 5.
Since it is quadratic, use the Quadratic Formula. The zeros of g(x)
= x 2 – 4x + 5 are 2 + i and 2 – i.
31. h(x) = x 3 + 4x 2 + x – 6
SOLUTION:
Since h(x) has degree 3, the function has 3 zeros.
The function has zeros at 2, 2 + i, and 2 – i.
Examine the number of sign changes for h(x) and h(−x).
The function has one real zero at x = 2, so the function goes through (2,
0) and does not cross the x-axis at any other place.
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, g(x) → −∞ and as x → ∞, g(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
Because there is 1 sign change for the coefficients of h(x), the function
has 1 positive real zero. Because there are 2 sign changes for the
coefficients of h(−x), h(x) has 2 or 0 negative real zeros. Thus, h(x) has
3 real zeros, or 1 real zero and 2 imaginary zeros.
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3-5 Roots and Zeros
List some possible values, and then use synthetic substitution to
evaluate h(x) for real values of x.
x
−3
−2
−1
0
1
1
1
1
1
1
1
4
1
2
3
4
5
1
−2
−3
−2
1
6
−6
0
0
−4
−6
0
The function has zeros at 1, −2, and −3.
32. q(x) = x 3 + 3x 2 – 6x – 8
The function has three real zeros at x = 1, x = −2, and x = −3, so the
function goes through (1, 0), (−2, 0), and (−3, 0).
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, h(x) → −∞ and as x → ∞, h(x) → ∞.
SOLUTION:
Since q(x) has degree 3, the function has 3 zeros.
Examine the number of sign changes for q(x) and q(−x).
Use this information and points with coordinates found in the table
above to sketch the graph.
Because there is 1 sign change for the coefficients of q(x), the function
has 1 positive real zero. Because there are 2 sign changes for the
coefficients of q(−x), q(x) has 2 or 0 negative real zeros. Thus, q(x) has
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3-5 Roots and Zeros
3 real zeros, or 1 real zero and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate q(x) for real values of x.
x
−4
−3
−2
−1
0
1
2
1
1
1
1
1
1
1
1
3
−1
0
1
2
3
4
5
−6
−2
−6
−8
−8
−6
−2
4
−8
0
10
8
0
−8
−10
0
33. g(x) = x 4 – 3x 3 – 5x 2 + 3x + 4
SOLUTION:
The function has zeros at 2, −1, and −4.
Since g(x) has degree 4, the function has 4 zeros.
The function has three real zeros at x = 2, x = −1, and x = −4, so the
function goes through (2, 0), (−1, 0), and (−4, 0).
Examine the number of sign changes for g(x) and g(−x).
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, q(x) → −∞ and as x → ∞, q(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
Because there are 2 sign changes for the coefficients of g(x), the
function has 2 or 0 positive real zeros. Because there are 2 sign changes
for the coefficients of g(−x), g(x) has 2 or 0 negative real zeros.
Thus, g(x) has 4 real zeros, 2 real zeros and 2 imaginary zeros, or 4
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3-5 Roots and Zeros
imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate g(x) for real values of x.
x
−1
0
1
2
3
4
1
1
1
1
1
1
1
−3
−4
−3
−2
−1
0
1
−5
−1
−5
−7
−7
−5
−1
3
4
3
−4
−11
−12
−1
Use this information and points with coordinates found in the table
above to sketch the graph.
4
0
4
0
−18
−32
0
−1, 1, and 4 are zeros of the function. Since these are 3 real roots, there
must be one more real zero at −1, 1, 4. After factoring out x + 1, the
depressed polynomial is x 3 – 4x 2 – x + 4. Determine if −1 is another
zero for this depressed polynomial.
34. f(x) = x 4 – 21x 2 + 80
SOLUTION:
x
−1
1
1
−4
−5
−1
4
4
0
−1 is a zero, so it is a double root of the function.
Since f(x) has degree 4, the function has 4 zeros.
Examine the number of sign changes for f(x) and f(−x).
The function has zeros at −1, −1, 1, and 4.
The function has four real zeros at x = −1 (multiplicity 2), x = 1, and
x = 4, so the function touches (−1, 0) and goes through (1, 0), and (4,
0) and does not cross the x-axis at any other place.
Because the degree is even and the leading coefficient is positive, the
end behavior is that as x → −∞, g(x) → ∞ and as x → ∞, g(x) → ∞.
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3-5 Roots and Zeros
Because there are 2 sign changes for the coefficients of f(x), the function
has 2 or 0 positive real zeros. Because there are 2 sign changes for the
coefficients of f(−x), f(x) has 2 or 0 negative real zeros. Thus, f(x) has
4 real zeros, or 2 real zeros and 2 imaginary zeros, or 4 imaginary
zeros.
5 are
and
.
The function has zeros at −4, 4,
, and
.
The function has four real zeros at x = −4, x = 4,
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
, so the function goes through (−4, 0), (4, 0),
, and
, and
.
x
−4
−3
−2
−1
0
1
2
3
4
1
1
1
1
1
1
1
1
1
1
0
−4
−3
−2
−1
0
1
2
3
4
−21
−5
−12
−17
−20
−21
−20
−17
−12
−5
0
20
36
34
0
0
−20
−34
−36
−20
80
0
−28
12
8
80
60
12
−28
0
Because the degree is even and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → ∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
−4 and 4 are zeros of the function. Since these are the only real roots,
there must be two imaginary roots. After factoring out x – 4, the
depressed polynomial is x 3 – 4x 2 – 5x + 20. Use synthetic substitution
and the other real zero, 4.
x
4
1
1
−4
0
−5
−5
20
0
The depressed polynomial is x 2 – 5. Since it is quadratic and does not
have an x-term, use the Square Root Property. The zeros of f(x) = x 2 –
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35. f(x) = x 3 + 7x 2 + 4x – 12
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3-5 Roots and Zeros
SOLUTION:
The function has zeros at −6, −2, and 1.
Since f(x) has degree 3, the function has 3 zeros.
Examine the number of sign changes for f(x) and f(−x).
The function has three real zeros at x = −6, x = −2, and x = 1, so the
function goes through (−6, 0), (−2, 0), and (1, 0).
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → −∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
Because there is 1 sign change for the coefficients of f(x), the function
has 1 positive real zero. Because there are 2 sign changes for the
coefficients of f(−x), f(x) has 2 or 0 negative real zeros. Thus, f(x) has
3 real zeros, or 1 real zero and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
x
−6
−5
−4
−3
−2
−1
0
1
1
1
1
1
1
1
1
1
1
7
1
2
3
4
5
6
7
8
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4
−2
−6
−8
−8
−6
−2
4
12
−12
0
18
38
24
0
−10
−12
0
36. f(x) = x 3 + x 2 – 17x + 15
SOLUTION:
Since f(x) has degree 3, the function has 3 zeros.
Examine the number of sign changes for f(x) and f(−x).
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3-5 Roots and Zeros
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → −∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
Because there are 2 sign changes for the coefficients of f(x), the function
has 2 or 0 positive real zeros. Because there is 1 sign change for the
coefficients of f(−x), f(x) has 1 negative real zero. Thus, f(x) has 3 real
zeros, or 1 real zero and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
x
−5
−4
−3
−2
−1
0
1
2
3
1
1
1
1
1
1
1
1
1
1
1
−4
−3
−2
−1
0
1
2
3
4
−17
3
−5
−11
−14
−17
−17
−15
−11
−5
15
0
35
−18
43
32
15
0
−7
0
37. f(x) = x 4 – 3x 3 – 3x 2 – 75x – 700
SOLUTION:
Since f(x) has degree 4, the function has 4 zeros.
Examine the number of sign changes for f(x) and f(−x).
The function has zeros at −5, 1, and 3.
The function has three real zeros at x = −5, x = 1, and x = 3, so the
function goes through (−5, 0), (1, 0), and (3, 0).
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3-5 Roots and Zeros
x
7
Because there is 1 sign change for the coefficients of f(x), the function
has 1 positive real zero. Because there are 3 sign changes for the
coefficients of f(−x), f(x) has 3 or 1 negative real zeros. Thus, f(x) has
4 real zeros, or 2 real zeros and 2 imaginary zeros.
1
1
−7
0
25
25
−175
0
The depressed polynomial is x 2 + 25. Since it is quadratic and does not
have an x-term, use the Square Root Property. The zeros of f(x) = x 2 +
25 are −5i and 5i.
The function has zeros at −4, 7, −5i, and 5i.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
x
−4
−3
−2
−1
0
1
2
3
4
5
6
7
1
1
1
1
1
1
1
1
1
1
1
1
1
−3
−7
−6
−5
−4
−3
−2
−1
0
1
2
3
4
−3
25
15
7
1
−3
−5
−5
−3
1
7
−21
25
−75 −700
−175
0
−150 −250
−89 −522
−76 −700
−75 −700
−80 −780
−85 −870
−84 −952
−71 −984
−40 −900
−201 −1906
100
0
The function has two real zeros at x = −4 and x = 7, so the function
goes through (−4, 0) and (7, 0).
Because the degree is even and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → ∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
−4 and 7 are zeros of the function. Since these are the only real roots,
there must be two imaginary roots. After factoring out x – 4, the
depressed polynomial is x 3 – 7x 2 + 25x – 175. Use synthetic
substitution and the other real zero, 7.
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3-5 Roots and Zeros
coefficients of f(−x), f(x) has 4, 2, or 0 negative real zeros. Thus, f(x)
has 4 real zeros, or 2 real zeros and 2 imaginary zeros, or 4 imaginary
zeros.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
x
−5
−4
−3
38. f(x) = x 4 + 6x 3 + 37x 2 + 384x + 576
SOLUTION:
Since f(x) has degree 4, the function has 4 zeros.
1
1
1
1
6
1
2
3
73
68
65
64
384
44
124
192
576
356
80
0
−3 is a zero of the function. Since there must be 4, 2, or 0 negative real
roots, there might be another real zero at −3. After factoring out x + 3,
the depressed polynomial is x 3 + 3x 2 + 64x + 192. Use synthetic
substitution and the real zero, −3.
x
−3
1
1
3
0
64
64
192
0
Examine the number of sign changes for f(x) and f(−x).
−3 is a zero, so it is a double root of the function.
The depressed polynomial is x 2 + 64. Since it is quadratic and does not
have an x-term, use the Square Root Property. The zeros of f(x) = x 2 +
64 are −8i and 8i.
The function has zeros at −3, −3, −8i, and 8i.
Because there are 0 sign changes for the coefficients of f(x), the function
has 0 positive real zeros. Because there are 4 sign changes for the
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The function has two real zeros at x = −3 (multiplicity 2), so the function
touches (−3, 0).
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3-5 Roots and Zeros
Because the degree is even and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → ∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
Because there are 4 sign changes for the coefficients of f(x), the function
has 4, 2, or 0 positive real zeros. Because there are 0 sign changes for
the coefficients of f(−x), f(x) has 0 negative real zeros. Thus, f(x) has 4
real zeros, or 2 real zeros and 2 imaginary zeros, or 4 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
x
2
3
4
39. f(x) = x 4 – 8x 3 + 20x 2 – 32x + 64
1
1
1
1
−8
−6
−5
−4
20
8
5
4
−32
−16
−17
−16
64
32
13
0
4 is a zero of the function. Since there must be 4, 2, or 0 positive real
roots, there might be another real zero at 4. After factoring out x – 4,
the depressed polynomial is x 3 – 4x 2 + 4x – 16. Use synthetic
substitution and the real zero, 4.
SOLUTION:
Since f(x) has degree 4, the function has 4 zeros.
Examine the number of sign changes for f(x) and f(−x).
x
4
1
1
−4
0
4
4
−16
0
4 is a zero, so it is a double root of the function.
The depressed polynomial is x 2 + 4. Since it is quadratic and does not
have an x-term, use the Square Root Property. The zeros of f(x) = x 2 +
4 are −2i and 2i.
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3-5 Roots and Zeros
The function has zeros at 4, 4, −2i, and 2i.
The function has two real zeros at x = 4 (multiplicity 2), so the function
touches (4, 0).
Because the degree is even and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → ∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
Because there is 1 sign change for the coefficients of f(x), the function
has 1 positive real zero. Because there is 1 sign change for the
coefficients of f(−x), f(x) has 1 negative real zero. Notice, the function
does not have a constant term, so one zero is 0. Thus, f(x) has 3 real
zeros and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
40. f(x) = x 5 – 8x 3 – 9x
SOLUTION:
Since f(x) has degree 5, the function has 5 zeros.
x
−3
−2
−1
0
1
2
3
1
1
1
1
1
1
1
1
0
−3
−2
−1
0
1
2
3
−8
1
−4
−7
−8
−7
−4
1
0
−3
8
7
0
−7
−8
3
−9
0
−25
−16
−9
−16
−25
0
0
0
50
16
0
16
−50
0
−3, 0, and 3 are zeros of the function. After factoring out x + 3, the
depressed polynomial is x 4 – 3x 3 + x 2 – 3x. Use synthetic substitution
and the real zero, 0.
Examine the number of sign changes for f(x) and f(−x).
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3-5 Roots and Zeros
x
0
1
1
−3
−3
1
1
−3
−3
After factoring out x, the depressed polynomial is x 3 – 3x 2 + x – 3. Use
synthetic substitution and the real zero, 3.
x
3
1
1
−3
0
1
1
−3
0
The depressed polynomial is x 2 + 1. Since it is quadratic and does not
have an x-term, use the Square Root Property. The zeros of f(x) = x 2 +
1 are −i and i.
The function has zeros at −3, 0, 3, −i, and i.
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → −∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
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3-5 Roots and Zeros
Write a polynomial that could be represented by each graph.
42. ​
41. ​
SOLUTION:
SOLUTION:
The graph crosses the x-axis 2 times, so the function is at least of
degree 2. It crosses the x-axis at x = −3 and x = 2, so its factors are x
+ 3 and x – 2.
The graph crosses the x-axis 3 times, so the function is at least of
degree 3. It crosses the x-axis at x = −2, x = 1, and x = 3, so its
factors are x + 2, x – 1, and x – 3.
To determine a polynomial, find the product of the factors.
To determine a polynomial, find the product of the factors.
A polynomial that could be represented by the graph is y = x 2 + x – 6.
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A polynomial that could be represented by the graph is y = x 3 – 2x 2 –
5x + 6.
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3-5 Roots and Zeros
43. ​
a fish is at sea level at −3, −2, −1, 1, 2, and 3 seconds from noon.
Graph a polynomial function that could represent the location of the fish
compared to sea level y, in centimeters, x seconds from noon.
SOLUTION:
−3, −2, −1, 1, 2, and 3 are zeros of the function. Use the zeros to write
factors x + 3, x + 2, x + 1, x – 1, x – 2, and x – 3. Use the factors to
write an equation of the function.
SOLUTION:
The graph crosses the x-axis 4 times, so the function is at least of
degree 4. It crosses the x-axis at x = −1, x = 1, x = 2, and x = 4, so its
factors are x + 1, x – 1, x – 2, and x – 4.
The function that represents the given information is y = x 6 – 14x 4 +
49x 2 – 36.
Graph the function.
To determine a polynomial, find the product of the factors.
A polynomial that could be represented by the graph is y = x 4 – 6x 3 +
7x 2 + 6x – 8.
44. FISH Some fish jump out of the water. When a fish is out of the water,
its location is above sea level. When a fish dives back into the water, its
location is below sea level. A biologist can use polynomial functions to
model the location of fish compared to sea level. A biologist noticed that
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The graph passes through the zeros and shows reasonable locations of
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3-5 Roots and Zeros
the fish above and below sea level.
45. BUSINESS After introducing a new product, a company’s profit is
modeled by a polynomial function. In 2012 and 2017, the company’s
profit on the product was $0. Graph a polynomial function that could
represent the amount of profit p(x), in thousands of dollars, x years
since 2010.
SOLUTION:
In 2012 and 2017, the company’s profit on the product was $0.
Because x is the years since 2010, the zeros 2012 – 2010, or 2, and
2017 – 2010, or 7. Use the zeros to write factors x – 2 and x – 7. Use
the factors to write an equation of the function.
The graph passes through the zeros, but does not show reasonable
profits for years since 2010.
Suppose the company multiplies this function by 31.25.
The function that represents the given information is y = 31.25x 2 –
281.25x + 437.5.
Multiply the function by –1. The new function that represents the given
information is y = –31.25x 2 + 281.25x – 437.5.
Graph the function.
Graph the function.
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3-5 Roots and Zeros
Write a polynomial function of least degree with integral
coefficients that has the given zeros.
46. 5, −2, −1
SOLUTION:
The graph crosses the x-axis 3 times, so the function is at least of
degree 3. It crosses the x-axis at x = 5, x = −2, and x = −1, so its
factors are x – 5, x + 2, and x + 1.
To determine a polynomial, find the product of the factors.
The graph passes through the zeros and shows that the profit is negative
before 2012 and after 2017, which show reasonable profits for years
since 2010. This makes sense in the context of the situation.
A polynomial that could be represented by the graph is y = x 3 – 2x 2 –
13x – 10.
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3-5 Roots and Zeros
47. –4, –3, 5
48. –1, –1, 2i
SOLUTION:
SOLUTION:
The graph crosses the x-axis 3 times, so the function is at least of
degree 3. It crosses the x-axis at x = –4, x = –3, and x = 5, so its
factors are x + 4, x + 3, and x – 5.
–1, –1, and 2i are zeros. Because 2i is a zero, –2i is also a zero, so the
function is at least of degree 4. It crosses the x-axis at x = –1, x = –1,
x = 2i, and x = –2i, so its factors are x + 1, x + 1, x – 2i, and x + 2i.
To determine a polynomial, find the product of the factors.
To determine a polynomial, find the product of the factors.
A polynomial that could be represented by the graph is y = x 3 + 2x 2 –
23x – 60.
A polynomial that could be represented by the graph is y = x 4 + 2x 3 +
5x 2 + 8x + 4.
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3-5 Roots and Zeros
49. –3, 1, –3i
50. 0, –5, 3 + i
SOLUTION:
SOLUTION:
–3, 1, and –3i are zeros. Because –3i is a zero, 3i is also a zero, so the
function is at least of degree 4. It crosses the x-axis at x = –3, x = 1,
x = 3i, and x = –3i, so its factors are x + 3, x – 1, x – 3i, and x + 3i.
0, –5, and 3 + i are zeros. Because 3 + i is a zero, 3 – i is also a zero,
so the function is at least of degree 4. It crosses the x-axis at x = 0, x =
–5, x = 3 + i, and x = 3 – i, so its factors are x, x + 5, [x – (3 + i)],
and [x – (3 – i)].
To determine a polynomial, find the product of the factors.
To determine a polynomial, find the product of the factors.
A polynomial that could be represented by the graph is y = x 4 + 2x 3 +
6x 2 + 18x – 27.
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A polynomial that could be represented by the graph is y = x 4 – x 3 –
20x 2 + 50x.
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3-5 Roots and Zeros
51. –2, –3, 4 – 3i
SOLUTION:
–2, –3, and 4 – 3i are zeros. Because 4 – 3i is a zero, 4 + 3i is also a
zero, so the function is at least of degree 4. It crosses the x-axis at x =
–2, x = –3, x = 4 – 3i, and x = 4 + 3i, so its factors are x + 2, x + 3,
[x – (4 – 3i)], and [x – (4 + 3i)].
To determine a polynomial, find the product of the factors.
SOLUTION:
Since f(x) has degree 3, the function has 3 zeros.
Examine the number of sign changes for f(x) and f(−x).
f(x) = x 3 – 5x 2 – 2x + 24
f(−x) = −x 3 – 5x 2 + 2x + 24
Because there are 2 sign changes for the coefficients of f(x), the function
has 2 or 0 positive real zeros. Because there is 1 sign change for the
coefficients of f(−x), f(x) has 1 negative real zero. Thus, f(x) has 3 real
zeros, or 1 real zero and 2 imaginary zeros.
A polynomial that could be represented by the graph is y = x 4 – 3x 3 –
9x 2 + 77x + 150.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
Sketch the graph of each function using its zeros.
52. f(x) = x 3 – 5x 2 – 2x + 24
x
−2
−1
0
1
2
3
4
1
1
1
1
1
1
1
1
−5
−7
−6
−5
−4
−3
−2
−1
−2
12
4
−2
−6
−8
−8
−6
24
0
20
24
18
8
0
0
The function has zeros at −2, 3, and 4.
The function has three real zeros at x = −2, x = 3, and x = 4, so the
function goes through (−2, 0), (3, 0), and (4, 0).
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3-5 Roots and Zeros
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → −∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
SOLUTION:
Since f(x) has degree 3, the function has 3 zeros.
Examine the number of sign changes for f(x) and f(−x).
f(x) = 4x 3 + 2x 2 – 4x – 2
53. f(x) = 4x 3 + 2x 2 – 4x – 2
f(−x) = –4x 3 + 2x 2 + 4x – 2
Because there is 1 sign change for the coefficients of f(x), the function
has 1 positive real zero. Because there are 2 sign changes for the
coefficients of f(−x), f(x) has 2 or 0 negative real zeros. Thus, f(x) has
3 real zeros, or 1 real zero and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
x
−1
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4
4
2
−2
−4
−2
−2
0
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3-5 Roots and Zeros
−0.5
0
0.5
1
4
4
4
4
0
2
4
6
−4
−4
−2
2
0
−2
−3
0
The function has zeros at −1, −0.5, and 1.
The function has three real zeros at x = −1, x = −0.5, and x = 1, so the
function goes through (−1, 0), (−0.5, 0), and (1, 0).
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → −∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
SOLUTION:
Since f(x) has degree 4, the function has 4 zeros.
Examine the number of sign changes for f(x) and f(−x).
f(x) = x 4 – 6x 3 + 7x 2 + 6x – 8
f(−x) = x 4 + 6x 3 + 7x 2 – 6x – 8
Because there are 3 sign changes for the coefficients of f(x), the function
has 3 or 1 positive real zeros. Because there is 1 sign change for the
coefficients of f(−x), f(x) has 1 negative real zero. Thus, f(x) has 4 real
zeros, or 2 real zeros and 2 imaginary zeros, or 4 imaginary zeros.
54. f(x) = x 4 – 6x 3 + 7x 2 + 6x – 8
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
x
−1
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1
1
−6
−7
7
14
6
−8
−8
0
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3-5 Roots and Zeros
0
1
2
3
4
1
1
1
1
1
−6
−5
−4
−3
−2
7
2
−1
−2
−2
6
8
4
0
2
−8
0
0
−8
0
The function has zeros at −1, 1, 2, and 4.
The function has four real zeros at x = −1, x = 1, x = 2, and x = 4, so
the function goes through (−1, 0), (1, 0), (2, 0), and (4, 0).
Because the degree is even and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → ∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
SOLUTION:
Since f(x) has degree 4, the function has 4 zeros.
Examine the number of sign changes for f(x) and f(−x).
f(x) = x 4 – 6x 3 + 9x 2 + 4x – 12
f(−x) = x 4 + 6x 3 + 9x 2 – 4x – 12
Because there are 3 sign changes for the coefficients of f(x), the function
has 3 or 1 positive real zeros. Because there is 1 sign change for the
coefficients of f(−x), f(x) has 1 negative real zero. Thus, f(x) has 4 real
zeros, or 2 real zeros and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate f(x) for real values of x.
55. f(x) = x 4 – 6x 3 + 9x 2 + 4x – 12
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x
−1
1
1
−6
−7
9
16
4
−12
−12
0
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3-5 Roots and Zeros
0
1
2
3
1
1
1
1
−6
−6
−4
−3
9
3
1
0
4
7
6
4
−12
−5
0
0
−1, 2, and 3 are zeros of the function. Since there must be 3 or 1
positive real roots and 2 positive real roots have been found there might
be another real zero at 2 or 3. After factoring out x + 1, the depressed
polynomial is x 3 – 7x 2 + 16x – 12. Use synthetic substitution and the
real zero, 2.
x
2
1
1
−7
−5
16
6
−12
0
2 is a zero, so it is a double root of the function.
The function has zeros at −1, 2, 2, and 3.
56. USE A SOURCE Linear algebra is the study of linear equations. In
linear algebra, the coefficients of linear equations are often organized
into rectangular arrays called matrices. Research the eigenvalues of a
matrix and how they relate to the roots of a polynomial function. What
fields use linear algebra, matrices, and eigenvalues?
SOLUTION:
The function has four real zeros at x = −1, x = 2 (multiplicity
2), and x = 3, so the function goes through (−1, 0) and (3, 0), and
touches (2, 0).
The characteristic polynomial of a square matrix is a polynomial which is
invariant under matrix similarity and has the eigenvalues as roots.
Because the degree is even and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → ∞ and as x → ∞, f(x) → ∞.
Linear algebra matrices, and eigenvalues are used in quantum
mechanics, computer engineering, geology, and other sciences.
Use this information and points with coordinates found in the table
above to sketch the graph.
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57. SPACE The technology for a rocket that will safely return to Earth for
refueling and reuse is currently being developed. The three sections of
the booster that will power the flight of the payload are cylindrical with a
total volume of about 234π cubic meters. If the second stage section of
the booster is x meters tall, then the interstage section x + 3 meters tall,
and the first stage section is 5x + 6.5 meters tall. The radius of the
booster is x – 5 meters.
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3-5 Roots and Zeros
V(−x) = –7x 3 – 60.5x 2 – 80x + 3.5
a. Write and solve an equation to represent the total volume of the
booster.
b. What are the dimensions of the first stage section of the booster?
Explain your reasoning.
SOLUTION:
a. The volume of a cylinder can be found using the formula V = πr2h,
where r is the radius length and h is the height. The radius of the booster
is x – 5 meters. The height of the booster is x + (x + 3) + (5x + 6.5), or
7x + 9.5 meters. So, the equation (x – 5)2(7x + 9.5)π = 234π
represents the total volume of the booster. Simplify the equation.
Because there are 2 sign changes for the coefficients of V(x), the
function has 2 or 0 positive real zeros. Because there is 1 sign change
for the coefficients of V(−x), V(x) has 1 negative real zero. Thus, V(x)
has 3 real zeros, or 1 real zero and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
evaluate V(x) for real values of x.
x
6
7
7
7
7
−60.5
−18.5
−11.5
Examine the number of sign changes for V(x) and V(−x).
V(x) = 7x 3 – 60.5x 2 + 80x + 3.5
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3.5
−182.5
0
7 is a zero of the function and the depressed polynomial is 7x 2 – 11.5x
– 0.5. Since it is quadratic, use the Quadratic Formula. The zeros
of V(x) = 7x 2 – 11.5x – 0.5 are
The function has zeros at 7,
Since the function has degree 3, the function has 3 zeros.
80
−31
−0.5
and
, and
.
.
b. The height of the first stage of the booster is 5x + 6.5 = 5(7) + 6.5 =
41.5 m. The radius x – 5 = 7 – 5 = 2 m. x = 7 is the only reasonable
solution in the context of the situation. The other possible values
of x result in negative measures or very small measures.
58. CREATE Consider two polynomial functions, f(x) and g(x).
a. Write a polynomial function f(x) of least degree with integral
Page 40
3-5 Roots and Zeros
coefficients and zeros that include −1 – 4i and
. Explain how
you found the function.
b. Write another polynomial function g(x) with integral coefficients that
has the same degree and zeros. How did you find this function?
c. Are you able to sketch the graphs of f(x) and g(x) based on the
zeros? Explain your reasoning. Then sketch the graphs of f(x) and g(x).
SOLUTION:
a. Because −1 – 4i is a zero, −1 + 4i is also a zero. Because
is a zero, then
is also a zero. The factors are [x – (–1 – 4i)],
[x – (–1 + 4i)],
, and
.
When x = 0, f(0) = (02 + 2(0) + 17)(9(0)2 – 12(0) + 5) = 85.
When x = 1, f(1) = (12 + 2(1) + 17)(9(1)2 – 12(1) + 5) = 40.
When x = 2, f(2) = (22 + 2(2) + 17)(9(2)2 – 12(2) + 5) = 425.
For f(x), graph the points (0, 85), (1, 40), and (2, 425). Connect the
points with a smooth curve.
Because the degree of f(x) is even and the leading coefficient is positive,
the end behavior is that as x → −∞, f(x) → ∞ and as x → ∞, f(x) →
∞.
When x = 0, g(0) = 2(02 + 2(0) + 17)(9(0)2 – 12(0) + 5) = 170.
When x = 1, g(1) = 2(12 + 2(1) + 17)(9(1)2 – 12(1) + 5) = 80.
When x = 2, g(2) = 2(22 + 2(2) + 17)(9(2)2 – 12(2) + 5) = 850.
For g(x), graph the points (0, 170), (1, 80), and (2, 850). Connect the
points with a smooth curve.
Because the degree of g(x) is even and the leading coefficient is
positive, the end behavior is that as x → −∞, g(x) → ∞ and as x →
∞, g(x) → ∞.
To determine a polynomial, find the product of the factors.
b. Another polynomial function with the same zeros and degree can be
obtained by multiplying f(x) by any nonzero whole number. For
instance, g(x) = 2(x 2 + 2x + 17)(9x 2 – 12x + 5).
c. No, the graphs of f(x) and g(x) cannot be sketched from only
imaginary zeros because they do not show on a graph.
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3-5 Roots and Zeros
evaluate P(x) for real values of x.
x
–1
0
1
2
3
4
5
59. ANALYZE Use the zeros to draw the graph of P(x) = x 3 – 7x 2 +
7x + 15 by hand. Discuss the accuracy of your graph, and what could
be done to improve the accuracy.
SOLUTION:
Since P(x) has degree 3, the function has 3 zeros.
Examine the number of sign changes for P(x) and P(–x).
1
1
1
1
1
1
1
1
–7
–8
–7
–6
–5
–4
–3
–2
7
15
7
1
3
–5
–5
–3
15
0
15
6
9
0
–5
0
The function has zeros at –1, 3, and 5.
The function has three real zeros at x = –1, x = 3, and x = 5, so the
function goes through (–1, 0), (3, 0), and (5, 0).
Because the degree is odd and the leading coefficient is positive, the
end behavior is that as x → –∞, f(x) → –∞ and as x → ∞, f(x) → ∞.
Use this information and points with coordinates found in the table
above to sketch the graph.
P(x) = x 3 – 7x 2 + 7x + 15
P(–x) = –x 3 – 7x 2 – 7x + 15
Because there are 2 sign changes for the coefficients of P(x), the
function has 2 or 0 positive real zeros. Because there is 1 sign change
for the coefficients of P(–x), P(x) has 1 negative real zero. Thus, P(x)
has 3 real zeros, or 1 real zero and 2 imaginary zeros.
List some possible values, and then use synthetic substitution to
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60. PERSEVERE Let the polynomial function f(x) have real coefficients,
be of degree 5, and have zeros 4 + 3i, 2 − 7i, and 6 + bi, where b is a
real number.
a. What can be determined about b? Explain your reasoning.
b. Write a possible equation for f(x).
SOLUTION:
The graph is accurate as far as the location of the zeros, but does not
consider any vertical dilation. It could be improved by finding more
points between the roots.
a. By the Complex Conjugates Theorem, because 4 + 3i is a zero, 4 –
3i is also a zero, because 2 – 7i is a zero, 2 + 7i is also a zero, and
because 6 + bi is a zero, 6 – bi is also a zero of f(x). Because f(x) has
degree 5, it can have at most 5 zeros. This means that b must be zero,
so that the zero 6 + bi = 6 + 0i or 6 is a real zero.
b. The factors are [x – (4 – 3i)], [x – (4 + 3i)], [x – (2 – 7i)], [x – (2 +
7i)], and (x – 6).
To determine a polynomial, find the product of the factors.
61. CREATE Sketch the graph of a polynomial function with:
a. 3 real, 2 imaginary zeros
b. 4 real zeros
c. 2 imaginary zeros
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3-5 Roots and Zeros
SOLUTION:
a. Because there are 3 real zeros and 2 imaginary zeros, the function is
of degree 5. Because there are 3 real zeros, the function will cross the
x-axis at 3 points. Assume the leading coefficient is positive. Because
the degree is odd and the leading coefficient is positive, the
end behavior is that as x → −∞, f(x) → −∞ and as x → ∞, f(x) → ∞.
c. Because there are 2 imaginary zeros, the function is of degree 2.
Because there are 2 imaginary zeros, the function will not cross the xaxis. Assume the leading coefficient is positive. Because the degree is
even and the leading coefficient is positive, the end behavior is that
as x → −∞, f(x) → ∞ and as x → ∞, f(x) → ∞.
b. Because there are 4 real zeros, the function is of degree 4. Because
there are 4 real zeros, the function will cross the x-axis at 4 points.
Assume the leading coefficient is negative. Because the degree is even
and the leading coefficient is negative, the end behavior is that as x →
−∞, f(x) → −∞ and as x → ∞, f(x) → –∞.
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3-5 Roots and Zeros
62. PERSEVERE Write an equation in factored form of a polynomial
function of degree 5 with 2 imaginary zeros, 1 real nonintegral zero, and
2 irrational zeros. Explain.
SOLUTION:
63. WHICH ONE DOESN’T BELONG Determine which equation is
not like the others. Justify your conclusion.
r4 + 1 = 0
Because the function is of degree 5, there will be 5 factors. Let one
imaginary zero be 2i, then the conjugate of 2i, or –2i, is also a zero. Let
the one real nonintegral zero be
. Let the one irrational zero be
, then the conjugate of
, is also a zero.
, or
The factors are x + 2i, x – 2i, (3x + 5),
, and
r3 + 1 = 0
r2 – 1 = 0
r3 – 8 = 0
SOLUTION:
.
A possible function of degree 5 with 2 imaginary zeros, 1 real
nonintegral zero, and 2 irrational zeros is
.
r4 + 1 = 0 does not belong because for the other three. The equation
has imaginary solutions and all of the others have real solutions.
64. ANALYZE Provide a counterexample for each statement.
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3-5 Roots and Zeros
a. All polynomial functions of degree greater than 2 have at least 1
negative real root.
b. All polynomial functions of degree greater than 2 have at least 1
positive real root.
SOLUTION:
have at least 1 positive real root is f(x) = x 3 + 6x 2 + 9x.
65. WRITE Explain to a friend how you would use Descartes’ Rule
of Signs to determine the number of possible positive real roots and
the number of possible negative real roots of the polynomial
function f(x) = x 4 – 2x 3 + 6x 2 + 5x – 12.
SOLUTION:
a. A polynomial function of degree 4 has a degree greater than 2. A
polynomial of degree 4 could have 4 imaginary zeros, such as
,
,
, and
, which does not include 1 negative real zero.
To determine a polynomial, find the product of the factors.
To determine the number of positive real roots, determine how many
times the signs change in the polynomial, moving left to right.
In the function f(x) = x 4 – 2x 3 + 6x 2 + 5x – 12, there are 3 changes in
sign. Therefore, there may be 3 or 1 positive real roots.
To determine the number of negative real roots, first evaluate the
polynomial for −x: f(−x) = x 4 + 2x 3 + 6x 2 – 5x – 12.
So, one polynomial functions of degree greater than 2 that does not
have at least 1 negative real root is f(x) = x 4 + 4x 2 + 4.
All of the terms with an odd-degree variable would change signs. Then
count the number of sign changes moving left to right. There is 1 sign
change. Therefore there may be 1 negative real root.
b. A polynomial function of degree 3 has a degree greater than 2. A
polynomial of degree 3 could have 3 real zeros, such as 0, –3, and –3,
which does not include 1 positive real root.
To determine a polynomial, find the product of the factors.
So, one polynomial functions of degree greater than 2 that does not
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3-5 Roots and Zeros
66. FIND THE ERROR The graph shows a polynomial function.
Brianne says the function is a 4th degree polynomial. Amrita says the
function is a 2nd degree polynomial. Is either of them correct? Explain
your reasoning.
SOLUTION:
The graph crosses the x-axis twice, so there are two real zeros, one
positive and one negative. This means Amrita could be correct. There
could be two other zeros of the function that are imaginary. This means
Brianne could also be correct.
So, both Brianne and Amrita could be correct.
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