Math 121, Practice for Chapter 3
Hints and Answers
1. (a)
3x2
2x2 + 6x − 5
6x4
−(6x4
− 5x
+ 8x3
+ 18x3
− 10x3
−(−10x3
−1
− 47x2 + 19x
+5
2
− 15x )
− 37x2 + 19x
− 30x2 + 25x)
− 2x2
− 6x + 5
2
−(− 2x
− 6x + 5)
0
Thus the answer is 3x2 − 5x − 1; notice there is no remainder.
(b) No; the degree of the numerator is 2 larger than the degree in the denominator.
(c)
x3 + 2x2
x
x4
−(x4
−2
+ 0x3
+ 2x3 )
− 2x3
−(−2x3
Thus the answer is x − 2 +
− 6x2
+ 0x − 3
− 6x2
− 4x2 )
− 2x2
- 3
−2x2 − 3
x3 + 2x2
or x − 2 −
2x2 + 3
x3 + 2x2
(d) Yes, the slant asymptote is y = x − 2.
2. The synthetic division with c = −3 is as follows
−3 1
9
0
−3 −18
1
6 −18
−5
10
54 −147
49 −137
Therefore, the answer is:
x3 + 6x2 − 18x + 49 +
3. (a) Use synthetic division with c = 6 as follows
6
8 −8 −31
4
36 264 1536 9030
6 44 256 1505 9034
6
The remainder is 9034, and so P (6) = 9034.
1
−137
.
x+3
(b) The remainder is Q(−1) = 3(−1)105 +(−1)58 −4(−1)27 +2(−1)2 −7 = −3+1+4+2−7 = −7.
(c) Yes: 5(−1)101 + 6(−1)44 − 7(−1)2 + 2(−1) + 8 = −5 + 6 − 7 − 2 + 8 = 0. According to the
factor theorem, (x + 1) is a factor of 5x101 + 6x44 − 7x2 + 2x + 8.
4. (a) The polynomial P has even degree with a negative leading coefficient, therefore its
graph is down to the far left and down to the far right.
(b) The polynomial Q has odd degree with positive leading coefficient. Therefore, its graph is
down to the far left and up to the far right.
(c) The polynomial R has odd degree with negative leading coefficient. Therefore, the graph
is up to the far left and down to the far right.
(d)The polynomial S has even degree with positive leading coefficient, therefore its graph is
up to the far left and up to the far right.
5. (a) The x-intercepts are −3, −1, 2, 3. The graph crosses x-axis at (−1, 0) and (−3, 0) because
the corresponding zeros have odd multiplicities, while the graph intersects but does not cross
the x-axis at (3, 0) and (2, 0) because the corresponding zeros have even multiplicities.
(b) The degree of P (x) is 10, and its leading coefficient is positive. Therefore, the graph goes
up to the far left, and to the far right.
(c) For Q, the x-intercepts are (3000, 0), (2000, 0), (−1000, 0) and (−3000, 0). The graph
touches but does not cross at zeros of even multiplicity, and crosses at zeros of odd multiplicity.
Thus the graph crosses the x-axis at x = −1000 and x = 3000, whereas it merely intersects
but does not cross the x-axis at x = −3000 and x = 2000.
6. (a) P (1) = −68 and P (7) = 1132. Therefore, the appropriate response is (i).
(b) Because P (0) = −58 and P (2) = −48, this means P (0) and P (2) have the same sign, so
we do not know if P has at least one zero between 0 and 2.
7. (a) The possible rational zeros are plus or minus all factors of 6 divided by all factors of 2:
3
1
±1, ±2, ±3, ±6, ± , ±
2
2
notice that ± 22 and ± 26 were not listed above because they are redundant.
(b) P (0) = −6 and P (1) = 18 have opposite signs, therefore, the Zero Location Theorem states
that there is a zero between 0 and 1.
(c) Observe that P (x) has 3 sign variations, so there are 3 or 1 positive real zeros. Now
P (−x) = −2x5 − 3x4 + 12x2 − 13x − 6 has 2 sign variations so there are 2 or 0 negative real
zeros.
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8. (a) By Descarte’s rule of signs, we know P has no positive real zeros, so we divide out the
zeros −3 and − 12 using synthetic division.
Starting with c = − 21 , the first synthetic division is as follows:
− 12
2
7
5
7
−1 −3 −1
2
6
2
6
3
-3
0
Thus, P (x) = (x + 21 )(2x3 + 6x2 + 2x + 6). Now using synthetic division with c = −3 on the
factor 2x3 + 6x2 + 2x + 6 we obtain
−3 2
6 2
6
The zeros of P (x) are thus −3, − 12 , i, −i (since 2x2 + 2 = 0 implies
−6 0 −6
2
0 2
0
x = ±i. Therefore, we write P as a product of linear factors as follows
P (x) = 2(x + 3)(x + 1/2)(x + i)(x − i).
(b) Because zeros of polynomials with real coefficients come in conjugate pairs, we know that
both 2+3i and 2−3i are zeros of Q(x). We now use synthetic division to find Q(x)÷(x−(2+3i)).
2 + 3i
−4
14
−4
13
2 + 3i −13
2 + 3i −13
1 −2 + 3i
1 −2 + 3i
0
1
Now use synthetic division to find (x3 + (−2 + 3i)x2 + x − 2 + 3i) ÷ (x − (2 − 3i)).
2 − 3i
1 −2 + 3i 1 −2 + 3i
2 − 3i 0
2 − 3i
1
0 1
0
Now x2 + 1 has zeros ±i. Thus the zeros of Q(x) are 2 + 3i, 2 − 3i, i, −i. Therefore,
Q(x) = (x − (2 + 3i))(x − (2 − 3i))(x − i)(x + i)
9. (a) The zeros of the polynomial of smallest degree would be 2 + 3i, 2 − 3i, −2 and 3 (we
know 2 − 3i is also a zero because complex zeros come in conjugate pairs). Therefore, we let
P (x) = (x − (2 + 3i))(x − (2 − 3i))(x + 2)(x − 3)
= (x2 − 4x + 13)(x2 − x − 6)
= x4 − 5x3 + 11x2 + 11x − 78.
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(b) The polynomial Q(x) will have the form
P (x) = k(x − (2 + 3i))(x − (2 − 3i))(x + 2)(x − 3)
where we will find k so that Q(1) = 120. Therefore, Q(x) = k(x4 − 5x3 + 11x2 + 11x − 78),
and so Q(1) = −60k. Now −60k = 120 and so k = −2. Consequently, Q(x) = −2x4 + 10x3 −
22x2 − 22x + 156.
10. (a) For f , the x-intercepts are x = ±1, the y-intercept is y = 1/8, there is an horizontal
asymptote y = 2, and there are vertical asymptotes x = 4 and x = −4. Also, f (x) → +∞ as
x → −4− and f (x) → −∞ as x → −4+ ; f (x) → −∞ as x → 4− and f (x) → +∞ as x → 4+ .
There is no slant asymptote.
(b) For g, the x-intercepts are x = ±1, the y-intercept is y = −2/3, there is no horizontal
asymptote, there is a vertical asymptote x = −3, and f (x) → −∞ as x → −3− and f (x) →
+∞ as x → −3+ . There is a slant asymptote since the degree in the numerator is one larger
than the denominator. To find the asymptote, we do the division and find
g(x) = 2x − 6 +
16
x+3
Therefore, the slant asymptote is y = 2x − 6.
(c) For h, the x-intercepts are x = ±2, the y-intercept is y = 16/9, there is no horizontal
asymptote, the vertical asymptotes are x = 3 and x = −3, and f (x) → +∞ as x → −3− ,
f (x) → −∞ as x → −3+ , f (x) → −∞ as x → 3− and f (x) → +∞ as x → 3+ . There is no
slant asymptote.
11. (a) The graph below is f graphed along with its horizontal asymptote y = 2 and vertical
asymptotes x = −4 and x = 4.
(b) The graph below is g graphed along with its slant asymptote y = 2x − 6 and vertical
asymptote x = −3.
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12. (a) h = x, w = 2x + 2, l = 3w − 1 = 3(2x + 2) − 1 = 6x + 5. Therefore, V =
x(2x + 2)(6x + 5) = 12x3 + 22x2 + 10x.
(b) Using synthetic division:
4 12
22 10
0
48 280 1160
12 70 290 1160
The remainder is 1160, and so the volume is 1160 units3 when the height is 4 units. You can
check this by plugging the number directly into V : V (4) = 4(2(4) + 2)(6(4) + 5) = 4(10)(29) =
1160.
(c) Using synthetic division:
8
1
1 10
-8
8 72 656
1 9 82 648
The remainder is 648, and thus the volume of the solid is 648 cm3 . You can check by plugging
this directly into the formula: V (8) = 83 + 82 + 10(8) − 8 which is more difficult to compute
by hand.
13. (a) Yes: let P (x) = xn − 1, then P (1) = 1n − 1 = 0. Therefore the factor theorem implies
x − 1 is a factor of P (x).
(b) Let P (x) = xn + 1, when n is odd, (−1)n = −1, thus P (−1) = (−1)n + 1 = −1 + 1 = 0.
According to the factor theorem x + 1 is a factor of xn + 1 when n is odd.
(c) No: when n is even (−1)n = 1, thus (−1)n + 1 = 2. According to the factor theorem, x + 1
is not a factor of xn + 1 when n is even.
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14. There is a horizontal asymptote y = bamn if n = m, and y = 0 is a horizontal asymptote
if m > n (no slant asymptotes in these cases). If n = m + 1, then there is a slant asymptote
(no horizontal asymptote). In all other cases n > m + 1, there are neither horizontal nor slant
asymptotes.
15. (a) We know that x − 2 and x + 5 are factors of P (x) (by using the Factor Theorem).
(b) According to the remainder theorem, the remainder of P (x) ÷ (x + 7) = P (−7) = −14,
and the remainder of P (x) ÷ (x − 7) = P (7) = 3.
16. (a) Q and G are of even degree.
(b) F and P are of odd degree.
(c) Only Q has leading coefficient an > 0.
(d) The polynomials F , G and P each have negative leading coefficient.
(e) The zeros of F are −1, 0, 1, 2; The zeros of G are −1, 0, 2; the zeros of P are −1, 2; and the
zeros of Q are −2, 0, 1.
(f) The x-intercepts of F are (−1, 0), (0, 0), (1, 0), and (2, 0).
The x-intercepts of G are (−1, 0), (0, 0) and (2, 0).
The x-intercepts of P are (−1, 0) and (2, 0).
The x-intercepts of Q are (−2, 0), (0, 0) and (1, 0).
(g) The y-intercept of F is (0, 0); the y-intercept of G is (0, 4); the y-intercept of P is (0, 2)
and the y-intercept of Q is (0, 0).
(h) The zero(s) of even multiplicity: of F is 1; of G is q of P is −1 and of Q is 1. (These are
the zeros where the graphs touch but do not cross the x-axis.)
(i) The zero(s) of odd multiplicity: of F are −1, 0, 2; of G are −1, 2; of P is 2; of Q are −2, 0.
(These are the zeros where the graphs cross the x-axis.)
17. The degree of F is 1 + 1 + 2 + 1 is 5; the degree of G is 1 + 2 + 1 is 4; the degree of P is
2 + 1 is 3; and the degree of Q is 1 + 1 + 2 is 4.
18. First, G(x) = a4 (x+1)(x−1)2 (x−2) and so G(0) = a4 (1)(−1)2 (−2) thus a4 (1)(−1)2 (−2) =
4 since the y-intercept is 4. Thus −2a4 = 4 and so a4 = −2. Therefore,
G(x) = −2(x + 1)(x − 1)2 (x − 2).
Similarly, P (x) = a3 (x + 1)2 (x − 2), and so P (0) = a3 (1)2 (−2) = 2. Thus a3 = −1 and
P (x) = −(x + 1)2 (x − 2).
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19. (a) The degree of R is 5.
(b) The leading coefficient of R is −2.
(c) The graph of R goes up to the far left and down to the far right.
(d) The zero −1 has multiplicity 3 so the graph of R crosses the x-axis at the point (−1, 0).
The zero 2 has multiplicity 2 so the graph of R touches but does not cross the x-axis at (2, 0).
(e) The x-intercepts are (−1, 0) and (2, 0); the y-intercept is (0, −8) since R(0) = −8.
(f) A computer generated graph of R is as follows. Notice it should have the following features:
The graph is up to the far left, and heads down and crosses the x-axis at −1 and then the
y-axis at −8 and bends up to touch the x-axis at 2 and then head down to the far right.
20. (a) The degree of S is 6.
(b) The leading coefficient is 2.
(c) The graph goes up to the far right and up to the far left.
(d) The zeros −2 has multiplicity 1 so the graph crosses the x-axis at (−2, 0). The zero −1
has multiplicity 3 so the graph crosses the x-axis at (−1, 0). The zero 1 has multiplicity 2 so
the graph touches but does not cross the x-axis at (1, 0).
(e) The x-intercepts are: (−2, 0), (−1, 0) and (1, 0). The y-intercept is (0, 4) since S(0) = 4.
(f) Note the graph of S is up to the far left, so the graph heads down and crosses the x-axis
at (−2, 0) then it will bend back up to cross the x-axis at (−1, 0) and continues up through
the y-intercept (0, 4) and bend back down to touch (but not cross) the x-axis at (1, 0) and
head up to the far right. Plotting more points will better reveal the shape as in the computer
generated graph given below.
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21. This is similar to question 8, but the hint is graphical. That is, from the graph we see
that −2 and 1/2 are zeros of P . We first dvide out c = 1/2:
1
2
2
7
1
2 8
14 11 −10
4 9
10
18 20
0
Now use synthetic division to find (2x3 + 8x2 + 18x + 20) ÷ (x + 2).
−2 2
8 18
20
−4 −8 −20
2
4 10
0
Thus P (x) = (x − 1/2)(x + 2)(2x2 + 4x + 10) = 2(x − 1/2)(x
+ 2)(x2 + 2x + 5). Using the
√
−2 ± −16
quadratic formula, the zeros of x2 + 2x + 5 are x =
= −1 ± 2i. Thus the zeros of
2
P are −2, 1/2, −1 − 2i and −1 + 2i, and
P (x) = 2(x + 2)(x − 1/2)(x + 1 + 2i)(x + 1 − 2i).
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22. Consider the function f (x) =
(a) The domain of f is:
2x + 2
.
x−1
(−∞, 1) ∪ (1, ∞)
(b) Coordinates of x-intercept(s) of f :
(−1, 0)
(c) Coordinates of y-intercept(s) of f :
(0, −2)
(d) Equations of the horizontal asymptote(s) of f :
(e) Equations of the veritcal asymptote(s) of f :
(f) Equations of the slant asymptote(s) of f :
y=2
x=1
There are no slant asymptotes.
(g) The graphs (ii) and (ivi) are not possibilities because their y-intercepts are not (0, −2),
out of (i) and (iii), notice that (iii) is the correct graph because it has x-intercept (1, 0), and
horizontal asymptote y = 2 and vertical asymptote x = 1 which we have included in the graph
below.
(i)
(ii)
(iii)
(iv)
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23. First write f (x) = −
(3x + 1)(x − 2)
3x + 1
(3x + 1)(x − 2)
=
=
, x 6= 2.
2
−(x − x − 2)
((x − 2)(x + 1)
x+1
all real numbers with x 6= −1, x 6= 2 i.e., (−∞, −1) ∪ (−1, 2) ∪ (2, ∞)
(b) Coordinates of x-intercept(s) of f :
− 13 , 0
(a) The domain of f is:
(c) Coordinates of y-intercept(s) of f :
(0, 1)
(d) Equations of the horizontal asymptote(s) of f :
(e) Equations of the veritcal asymptote(s) of f :
(f) Equations of slant asymptote(s) of f :
y=3
x = −1
there are no slant asymptotes
(g) Note that f only has one vertical asymptote, so neither (ii) nor (iii) which have two
vertical asymptotes are correct. Then the correct answer is (i) which has y-intercept (1, 0) and
a horizontal asymptote y = 3. The correct graph with the asymptotes is displayed below:
24. Use division to write f (x) =
(a) The domain of f is:
2
2x2 − 7x + 4
= 2x − 3 −
.
x−2
x−2
{x | x 6= 2} = (−∞, 2) ∪ (2, ∞)
(b) Coordinates of y-intercept(s) of f :
(c) Equations of slant asymptote(s) of f :
(0, −2)
y = 2x − 3
(d) There is no horizontal asymptote, the vertical asymptote is x = 2
(e) The graphs (ii) and (iii) are not correct, because they have x = −2 as the vertical asymptote.
The intercept of (iv) is not correct. Therefore, (i) is the correct graph.
25. The answer cannot be (a) or (d) because their vertical asymptote is x = 2. It cannot be
(e) because that function has an x-intercept at (−1, 0), and it cannot be (b) which is negative
when x < −2. The only possible answer is (c).
26. The answer cannot be (a) or (b) which have vertical asymptote x = −2. It cannot be (d)
which is negative for all x 6= 2. It cannot be (e) which has an intercept at (−4, 0). The only
possible answer is (c).
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27. The only possibilities are (a) and (d) because they have vertical asymptote x = −2; it
cannot be (a) because that function has a horizontal, not a slant asymptote. Thus, the only
possible answer is (d).
28. The only possibilities are (a) and (b) because they have vertical asymptote x = −3. The
horizontal asymptote for (a) is y = 1, while the horizontal asymptote for (b) is y = −2/3, thus
the only possible answer is (b).
For further practice: see Test 3 from Autumn 2004 and later quarters which are available
at
http://faculty.lasierra.edu/∼jvanderw/classes/testbanks/m121tb.htm
A good source of practice is also the Chapter Review Questions and Test in the text.
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