Download (1) Possible Rational Roots of a Polynomial P(x) = a nxn + ...a 1x + a

F ORMULA S HEET
(1) Possible Rational Roots of a Polynomial P (x) = an xn + ...a1 x + a0 :
factors of a0
±
factors of an
(2) Quadratic Equation:
√
−b ± b2 − 4ac
2a
(3) Imaginary Numbers:
i=
√
−1, i2 = −1
(4) Logarithmic and Exponential Relationship:
loga (b) = c ⇐⇒ ac = b
1
NAME: SOLUTIONS
COLLEGE ALGEBRA -ANALYTIC GEOMETRY MIDTERM 2
INSTRUCTOR: HAROLD SULTAN
1. I NSTRUCTIONS
(1) Timing: You have exactly 1 hour 55 minutes for this exam.
(2) There are 6 questions in total, with the following scoring breakdown:
Question Total Points Your points
1
10
2
10
3
10
4
10
5
10
6
10
Total
60
(3) Please show your work and JUSTIFY all answers unless otherwise specified. Partial credit will be awarded
(4) Many of questions have multiple parts. Each question is on its own page. Please
feel free to work on the back of the page or scrap paper if necessary.
(5) Challenging Questions are marked (**)
(6) Good Luck!
Date: March 29, 2011.
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COLLEGE ALGEBRA -ANALYTIC GEOMETRY MIDTERM 2
3
2. Q UESTIONS
Question 2.1.
Consider the polynomial P (x) = 4x5 − 24x4 + 47x3 − 30x2
(a) (2 points) Explicitly LIST all the different possible rational roots (zeros) of P (x)
without repetition of roots. How many possible rational roots of P (x) are there in
total?
ANSWER
1
3
5
3
5
15
15
1
± , ± , ± , ±1, ± , ± , ±2, ± , ±3, ± , ±5, ±6, ± , ±10, ±15, ±30
4
2
4
4
2
2
4
2
In total there are 32 possible rational roots
(b) (2 points) In light of Descartes Rules what can you say about the possibilities for
the number of positive and negative real roots.
ANSWER
P(x) has three sign changes so there are either 1 or 3 real positive roots. On the
other hand, the polynomial P (−x) has no sign changes so there are no negative
real roots.
(c) (2 points) Is 32 a root of P (x)?
ANSWER
By synthetic devision, 23 is a root of P (x). In fact,
P (x)
= 4x4 − 18x3 + 20x2
(x − 23 )
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INSTRUCTOR: HAROLD SULTAN
(d) (2 points) Factor the polynomial P (x) completely?
ANSWER
P (x)
3
(x − )(4x4 − 18x3 + 20x2 )
2
3
= (x − )2x2 (2x2 − 9x + 10)
2
3
= 2(x − )x2 (2x2 − 4x − 5x + 10)
2
3
= 2(x − )x2 (2x(x − 2) − 5(x − 2))
2
3
= 2(x − )x2 (2x − 5)(x − 2)
2
=
(e) (2 points) Sketch a graph the polynomial P (x). Explicitly mark the x-intercepts
and y-intercepts.
2
-0.5
0.5
-2
-4
-6
-8
1.0
1.5
2.0
2.5
COLLEGE ALGEBRA -ANALYTIC GEOMETRY MIDTERM 2
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Question 2.2.
Consider the rational function
4x3 − 4x
2x2 + 3x + 1
(a) (2 points) Factor the numerator and denominator of R(x)
ANSWER:
4x3 − 4x
4x(x − 1)(x + 1)
4x(x − 1)
R(x) = 2
=
=
2x + 3x + 1
(x + 1)(2x + 1)
2x + 1
R(x) =
(b) (2 points) Find the x and y-intercepts of R(x)
ANSWER:
The x-intercepts are the roots of the numerator which after factoring and simplifying as in the previous part are 0 and 1.
To find the y-intercept we plug in x = 0 to R(x) and solve, specifically,
R(0) =
4(03 ) − 4(0)
0
= =0
2(02 ) + 3(0) + 1
1
so the y-intercept is 0.
(c) (2 points) Find the vertical asymptotes of R(x), and determine whether R(x) →
+∞ or R(x) → −∞ as x approaches each asymptote from the left and right.
ANSWER:
The vertical asymptotes are the roots of the denominator. After factoring and simplifying as in part (a) we see that there is a single vertical asymptote of −1
2 .
−1 −
Moreover, as x → ( 2 ) , plugging in we see that R(x) → −∞, while as
+
x → ( −1
2 ) , plugging in we see that R(x) → +∞
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INSTRUCTOR: HAROLD SULTAN
(d) (2 points) Find the slant horizontal asymptote of R(x).
ANSWER:
Since the degree of the numerator (3) is larger than the degree of the denominator
(2) in R(x) strictly speaking there is no horizontal asymptote. However, there is a
slant horizontal asymptote. Specifically, after long division on R(x) we have:
4x3 − 4x
3x + 3
= (2x − 3) + 2
2x2 + 3x + 1
2x + 3x + 1
It follows that the slant horizontal asymptote of R(x) is the line y = 2x − 3.
(e) (2 points) Graph R(x).
R(x) =
10
-4
-3
-2
-1
1
-10
-20
2
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COLLEGE ALGEBRA -ANALYTIC GEOMETRY MIDTERM 2
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Question 2.3.
(a) (10 points) Find a real, degree four polynomial with imaginary roots −3i and 1+i,
whose graph has y-intercept 12
Answer:
Since we are looking for a real polynomial with imaginary zeros of −3i and 1+i, we know
that the polynomial must have imaginary zeros of 3i and 1 − i as well because they are
conjugates of imaginary zeros. Since our desired polynomial is degree four and we know
four roots, it follows that we know all the roots by the fundamental theorem of algebra.
Hence our polynomial must be of the form:
P (x)
=
a(x − 3i)(x + 3i)(x − (1 + i))(x − (1 − i))
=
a(x2 + 9)(x2 − 2x + 2)
where a above is any real constant. Since we are given that the y-intercept is 12 we know
that P (0) = 12. Plugging we have
a((0)2 + 9)((0)2 − 2(0) + 2) = 18a
2
=⇒ 12 = 18a =⇒ a =
3
Putting things together our final answer is
2
P (x) = (x2 + 9)(x2 − 2x + 2)
3
12 = P (0)
=
8
INSTRUCTOR: HAROLD SULTAN
Question 2.4.
(a) (5 points) Simplify the following complex expression:
1 + 2i 1
+
2−i
i
[Your final answer should be a complex number in the form a + bi]
ANSWER: First consider,
1 + 2i
(1 + 2i)(2 + i)
2 + 5i − 2
5i
=
=
=
=i
2−i
(2 − i)(2 + i)
5
5
Next consider,
1
1 × (−i)
−i
=
=
= −i
i
i × (−i)
1
Putting things together we have
1 + 2i 1
+ = i − i = 0 = 0 + 0i
2−i
i
(b) (5 points) Evaluate the following expression:
log27 (18) − log27 (6) + log27 (3)
[Your final answer should be a fraction]
Answer:
log27 (18) − log27 (6) + log27 (3)
18
) + log27 (3)
6
18 × 3
= log27 (
)
6
= log27 (9)
log3 (9)
=
log3 (27)
2
=
3
=
log27 (
COLLEGE ALGEBRA -ANALYTIC GEOMETRY MIDTERM 2
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Question 2.5.
(a) (5 points) Solve for x in the following equation:
ln(x + 4) = ln(x) + ln(4)
Answer:
Using a law of logs to combine the right hand side of the equation, we can rephrase
the question as:
ln(x + 4) = ln(4x)
However, it is now clear that the above expression is true only if x + 4 = 4x
and x + 4, 4x > 0 (this last condition is necessary as the function ln can only be
applied to positive numbers). Solving the linear equation x + 4 = 4x we have
4
x=
3
This is our only answer, and moreover it is not a pseudo answer as 4x and x + 4
are greater than zero.
(b) (5 points) (*) Solve for x in the following equation:
2
−2
log5 (x)
=
log10 (10)
16
Answer: Begin by noting that log10 (10) = 1. Hence, we have to solve
1
2(−2/ log5 (x)) =
16
Taking log base two of both sides, we have
1
log2 2(−2/ log5 (x)) = log2 ( )
16
−2
= −4
=⇒
log5 (x)
−2
=⇒
= log5 (x)
−4
1
=⇒
= log5 (x)
2
√
=⇒ x = 51/2 = 5
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INSTRUCTOR: HAROLD SULTAN
Question 2.6.
(a) (4 points) Graph of the function f (x) = − log2 (x + 1)
Answer:
the graph f (x) is obtained by beginning with the usual graph of log2 (x), shifting
horizontally to the left by 1, and then reflecting about the x-axis. The resulting
graph is below. Two features to note are that the graph contains the point (0, 0)
and that the domain is [−1, ∞).
3
2
1
-2
-1
1
2
3
4
5
-1
-2
(b) (4 points) Graph of the function g(x) = log 12 (x + 1) + 1
Answer:
the graph g(x) is obtained by beginning with the usual graph of log1/2 (x), shifting
horizontally to the left by 1, and then shifting vertically up by 1. The resulting
graph is below. Two features to note are that the graph contains the point (0, 1)
and that the domain is [−1, ∞).
4
3
2
1
-2
-1
1
2
3
4
5
-1
(c) (**)
(2 points) Graph
of the function h(x) = f (x) − g(x) = − log2 (x + 1) −
log 21 (x + 1) + 1 Justify your answer
Answer:
COLLEGE ALGEBRA -ANALYTIC GEOMETRY MIDTERM 2
h(x)
=
11
− log2 (x + 1) − log 12 (x + 1) + 1 = − log2 (x + 1) − log 21 (x + 1) − 1
log2 (x + 1)
log2 (x + 1)
− 1 = − log2 (x + 1) −
−1
log2 (1/2)
−1
= − log2 (x + 1) + log2 (x + 1) − 1 = −1
= − log2 (x + 1) −
so the graph h(x) is the horizontal line y = −1 drawn below
-3
-2
-1
1
-0.5
-1.0
-1.5
-2.0
2
3