Download Math 1320, Section 10 Exam I, Due February 1st 100 Points Please

Exam I, Due February 1st
Math 1320, Section 10
100 Points
Please answer each question. To receive full credit you must show all work and give
answers in simplest form. Graphing calculators are not allowed.
Name and Date:
1. (2 points) Evaluate the expression | − 2| − | − 5|.
Solution: | − 2| − | − 5| = 2 − 5 = −3
2. (3 points) Find the distance between a = 5/4 and b = 1/7.
1 5 4
35 4 − 35 31 31
= −
=
Solution: d(a, b) = |b − a| = − = − = 7 4
28 28
28 28 28
3. Evaluate the expressions for the given value of x.
(a) (2 points) x2 − 3x + 5, x = −2
Solution: (−2)2 − 3(−2) + 5 = 4 + 6 + 5 = 15
(b) (2 points) −90x2 , x = − 13
2
1
1
90
Solution: −90 −
= −90
= − = −10
3
9
9
4. Simplify each radical expression.
p
(a) (3 points) 72x2 y −4
s
p
Solution: 72x2 y −4 =
√
√
√
√
√
72x2
72x2
36 · 2 · x2
6|x| 2
6|x| 2
6|x| 2
= p
=
= p
=
=
y4
|y| · |y|
|y|2
y2
y4
y2 · y2
√
√
(b) (3 points) 5 200 − 3 18
√
√
√
√
√
√
√
√
√
Solution: 5 200−3 18 = 5 100 · 2−3 9 · 2 = 5·10 2−3·3 2 = 50 2−9 2 = 41 2
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5. (3 points) For the following expression, (a) write the polynomial in standard form, (b)
identify the degree and leading coefficient of the polynomial, and (c) state whether the
polynomial is a monomial, a binomial, or a trinomial.
−x2 + 6x7 + 1
Solution:
(a) A polynomial is in standard form when its exponents are written in descending
order. The standard form of the given polynomial is
6x7 − x2 + 1
(b) The degree is 7 and the leading coefficient is 6.
(c) There are three terms in our polynomial (6x7 , −x2 , and 1) so it is a trinomial.
6. (4 points) Find the sum of the polynomials.
(3x4 − 6x2 + 11x + 18) + (x3 − 4x2 − 7x + 3)
Solution: Since we do not have to distribute anything inside the parenthesis we can
drop them to get
(3x4 − 6x2 + 11x + 18) + (x3 − 4x2 − 7x + 3) = 3x4 − 6x2 + 11x + 18 + x3 − 4x2 − 7x + 3
We then move the like terms near each other and simplify.
3x4 + x3 − 6x2 − 4x2 + 11x − 7x + 18 + 3 = 3x4 + x3 − 10x2 + 4x + 21
7. (4 points) Find the product (3x − 1)(2x + 5). (Hint: Remember that you can FOIL the
product of two binomials)
Solution: Start by using the FOIL method.
I
L
F
O
z }| { z }| { z }| { z }| {
(3x − 1)(2x + 5) = 3x · 2x + 3x · 5 + (−1) · 2x + (−1) · 5
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Now simplify and combine the like terms
I
L
F
O
z }| { z }| { z }| { z }| {
3x · 2x + 3x · 5 + (−1) · 2x + (−1) · 5 = 6x2 + 15x − 2x − 5 = 6x2 + 13x − 5
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8. A take-out fast-food company is constructing an open box by cutting squares from the
corners of a piece of cardboard that is 18 centimeters by 24 centimeters. The edge of
each cut-out square is x centimeters.
(a) (3 points) Draw a picture to model the situation like the one on page 34 of the
textbook. (Note: You need to use 24 centimeters as your width)
(b) (3 points) Find the volume of the box in terms of x.
Solution: V = (24 − 2x)(18 − 2x)x
(c) (3 points) Find the volume of the box when x = 4 cm.
Solution: Plugging in x = 4 into the answer from part (b) we get
V = (24 − 2(4))(18 − 2(4))(4) = (24 − 8)(18 − 8)(4) = (16)(10)(4) = 640 cm3
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9. Factor the expressions.
(a) (4 points) x3 − 9x
Solution: Start by factoring out an x to get
x3 − 9x = x(x2 − 9)
Now x2 − 9 is a difference of squares so it factors as
x2 − 9 = (x + 3)(x − 3)
Using this we have
x3 − 9x = x(x2 − 9) = x(x + 3)(x − 3)
(b) (4 points) 2x2 − 5x − 12
Solution: You could notice that 2(−4) + 3 = −5 to get the factorization
2x2 − 5x − 12 = (2x + 3)(x − 4)
If you use the process outlined in class you would do the following:
i. Multiply 2(−12) = −24.
ii. Find all possible (integer) factors of −24; they are: (−1 and 24), (1 and
−24), (−2 and 12), (2 and −12), (−3 and 8), (3 and −8), (−4 and 6), and
(4 and −6).
iii. Pick the pair that sum to −5. In this case we must choose the pair (3 and
−8).
iv. Use this pair to write −5x as −5x = 3x − 8x. We now have
2x2 − 5x − 12 = 2x2 + 3x − 8x − 12
Grouping the first two terms together and the second pair of terms together
we get
(2x2 + 3x) + (−8x − 12)
Remember: always put + between the parenthesis and keep the negative
sign attached to the coefficient inside the parenthesis. As you can see above,
the term inside the parenthesis is −8x NOT 8x. In other words, the following
is WRONG:
(2x2 + 3x) − (8x − 12)
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v. Factoring an x from the first set of parenthesis and −4 from the second set
we get
(2x2 + 3x) + (−8x − 12) = x(2x + 3) − 4(2x + 3)
Factoring the common (2x + 3) term we get
2x2 − 5x − 12 = (2x + 3)(x − 4)
(c) (4 points) x3 + 64
Solution: This is a sum of two cubes. Notice that 64 = 43 . Using the formula
on page 38 we have
x3 + 64 = x3 + 43 = (x + 4)(x2 − 4x + 16)
(d) (4 points) x3 − x2 + 2x − 2 (Hint: For expressions of this kind it is best to factor
by grouping)
Solution:
x3 − x2 + 2x − 2 = (x3 − x2 ) + (2x − 2) = x2 (x − 1) + 2(x − 1) = (x − 1)(x2 + 2)
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10. (3 points) Write an expression in factored form for the area of the shaded portion of the
figure.
Solution: This was done in class.
area of shaded region = area of square − area of circle = 4r2 − πr2 = (4 − π)r2
11. Find the domain of each expression.
(a) (4 points) 2x2 − 7x − 4
Solution: The domain of any polynomial is all real numbers, i.e. R.
(b) (4 points)
x2 − 4x + 1
x2 − 6x + 9
Solution: Since the numerator is a polynomial we do not need to worry about
its domain, because it is R. However, we cannot divide by zero, so we must
factor the denominator to determine the values for x which could make it zero;
it factors as
x2 − 6x + 9 = (x − 3)(x − 3)
We do not want our denominator zero, i.e. (x − 3)(x − 3) 6= 0. This tells us that
x 6= 3. Hence, the domain for the expression is all real numbers, except 3.
(c) (4 points) √
1
x−4
√
Solution: First consider the domain of x − 4, don’t worry that it’s part of
the denominator right now. Remember that whatever is under the root must
be greater than or equal to zero. This means
x−4≥0
Adding 4 to both sides
x ≥ 4. Now look
√ we get√
√ at what happens when x = 4.
When x = 4 we get 4 − 4 = 0 = 0. Since x − 4 is in the denominator, we
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cannot allow it to take on the value of 0. Hence, we must exclude x = 4 from
our domain. This means that the domain for the expression
√
1
x−4
is all real numbers strictly greater than 4, i.e. x > 4.
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12. Write the rational expression in simplest form
4y − 16y 2
(a) (3 points)
7 − 28y
Solution:
(b) (3 points)
4y − 16y 2
4y(1 − 4y)
4y
=
=
7 − 28y
7(1 − 4y)
7
x2 + 8x − 20
x2 + 11x + 10
Solution:
x2 + 8x − 20
(x + 10)(x − 2)
x−2
=
=
2
x + 11x + 10
(x + 10)(x + 1)
x+1
13. (5 points) Perform the multiplication and simplify
t2 + t − 6
t2 − 1
·
t2 − 4t + 4
t−1
Solution:
t2 − 1
t2 + t − 6
(t + 1)(t − 1) (t + 3)(t − 2)
(t + 1)(t + 3)
·
=
·
=
2
t − 4t + 4
t−1
(t − 2)(t − 2)
t−1
t−2
14. (5 points) Perform the division and simplify
x2
x
÷ 2
x−1 x −1
Solution: First flip the second fraction, then proceed as you would for a multiplication problem
x
x2
x
x2 − 1
x
(x + 1)(x − 1)
x+1
=
÷ 2
·
=
·
=
x−1 x −1
x−1
x2
x−1
x2
x
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15. (5 points) Perform the addition.
4
x
+
2x + 1 x + 2
Solution: The least common denominator is (2x + 1)(x + 2). Multiplying the first
fraction (on top and bottom) by (x+2), and the second fraction (on top and bottom)
by (2x + 1) we get
4
x
4(x + 2)
x(2x + 1)
+
=
+
2x + 1 x + 2
(2x + 1)(x + 2) (2x + 1)(x + 2)
Multiplying everything out we get
4(x + 2)
x(2x + 1)
4x + 8 + 2x2 + x
2x2 + 5x + 8
+
=
=
(2x + 1)(x + 2) (2x + 1)(x + 2)
(2x + 1)(x + 2)
(2x + 1)(x + 2)
16. (5 points) Factor the expression by removing the common factor with the smaller exponent.
x2 (x2 + 4)−5 − (x2 + 4)−4
Solution:
x2 (x2 + 4)−5 − (x2 + 4)−4 = (x2 + 4)−5 (x2 − (x2 + 4)) =
−4
x2 − x2 − 4
= 2
2
5
(x + 4)
(x + 4)5
17. (8 points) Find (a) the distance and (b) the midpoint of the line segment joining the
following pair of points
(3, 7) & (−1, 6)
Solution:
p
√
√
√
(3 − (−1))2 + (7 − 6)2 = 42 + 12 = 16 + 1 = 17
7+6
13
(b) Midpoint = 3+(−1)
,
=
1,
2
2
2
(a) Distance =
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Extra Credit
1. From what language does the word ”algebra” come?
2. Who is known as the father of algebra?
3. Explain why we can’t ”divide by zero.”
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