Download Math 110 Placement Exam Solutions FRACTION ARITHMETIC 1

Math 110 Placement Exam Solutions
FRACTION ARITHMETIC
1 1
1. Find the sum. +
2 3
1
3
2
1
+ = + =
Solution Convert to a common denominator. Then add numerators.
2
3
6
6
3+2
5
=
6
6
Google Help: addition of fractions
2. Find the difference.
1 2
−
2 3
3 4
1 2
Solution Convert to a common denominator. Then subtract numerators. − = − =
2 3
6 6
3−4
−1
1
=
=−
6
6
6
Google Help: subtraction of fractions
1
2
3. Find the difference. 3 − 1
2
3
Solution Convert the mixed numbers to fractions. Convert to a common denominator. Then
subtract numerators. Convert to a mixed number.
2
7 5
21 10
21 − 10
11
5
1
−
=
=
=1
3 −1 = − =
2
3
2 3
6
6
6
6
6
Google Help: subtraction of mixed numbers
2
1
4. Find the product. 3 × 1
2
3
Solution Convert the mixed numbers to fractions. Compute the product by multiplying
numerators and multiplying denominators. Convert to a mixed number.
1
2
7 5
7·5
35
5
3 ×1 = × =
=
=5
2
3
2 3
2·3
6
6
Google Help: product of fractions and product of mixed numbers
1
1
5. Find the quotient. 6 ÷ 1
2
4
Solution Convert the mixed numbers to fractions. Compute the quotient by multiplying the
first number by the reciprocal of the second. Convert to a mixed number.
1
1
13 5
13 4
52
26
1
6 ÷1 =
÷ =
× =
=
=5
2
4
2
4
2
5
10
5
5
Google Help: division of fractions and division of mixed numbers
WORKING WITH POLYNOMIAL EXPRESSIONS
6. Simplify the expression 2(9 − x) − (2x + 1).
Solution Use the distributive property to write 2(9 − x) as 18 − 2x. Then
2(9 − x) − (2x + 1) = 18 − 2x − 2x − 1.
Collect like terms and simplify to get the final answer.
18 − 2x − 2x − 1 = (18 − 1) + (−2x − 2x) = 17 − 4x
Google Help: distributive property and addition of polynomials
7. Simplify (4x3 − 2x2 + 1) − (2x3 + x − 3).
Solution Use the distributive property to remove the parentheses, collect like terms, and
simplify.
(4x3 − 2x2 + 1) − (2x3 + x − 3) = 4x3 − 2x2 + 1 − 2x3 − x + 3 =
(4x3 − 2x3 ) − 2x2 − x + (1 + 3) = 2x3 − 2x2 − x + 4
Google Help: subtraction of polynomials
8. Find the product (3x + 5)(4x − 7).
Solution Take each term inside the first set of parentheses times each term inside the second
set of parentheses.
(3x + 5)(4x − 7) = 3x · 4x + 3x · (−7) + 5 · 4x + 5 · (−7)
Multiply out each of the terms, collect like terms, and simplify.
3x · 4x + 3x · (−7) + 5 · 4x + 5 · (−7) = 12x2 − 21x + 20x − 35 = 12x2 − x − 35
Google Help: multiplication of polynomials
9. Find the product 3y 3 (4y + 3)(y − 2).
Solution First multiply the last two terms, then multiply by the first term.
3y 3 (4y + 3)(y − 2) = 3y 3 (4y 2 − 5y − 6) = 12y 5 − 15y 4 − 18y 3
Google Help: multiplication of polynomials
10. Find the product (x + y)(2x + y)2 .
Solution (x + y)(2x + y)2 = (x + y)(4x2 + 4xy + y 2 ) = x · 4x2 + x · 4xy + x · y 2 + y · 4x2 + y ·
4xy + y · y 2 = 4x3 + 4x2 y + xy 2 + 4x2 y + 4xy 2 + y 3 = 4x3 + (4x2 y + 4x2 y) + (xy 2 + 4xy 2 ) + y 3 =
4x3 + 8x2 y + 5xy 2 + y 3
Google Help: multiplication of polynomials
11. Divide 5s3 + 5s2 + 90 by s + 3.
Solution Remember the missing term 0s in the dividend. The term 5s3 divided by s yields
5s2 . This is the first term in the quotient. The process is similar to grade school long division.
The details are given below.
5s2 − 10s + 30
s + 3)5s3 + 5s2 + 0s + 90
5s3 + 15s2
−10s2 + 0s
−10s2 − 30s
30s + 90
30s + 90
0
Google Help: polynomial long division
FACTORING POLYNOMIAL EXPRESSIONS
12. Factor the polynomial x2 − 5x + 6.
Solution Since the last term is positive, look at the factor pairs for 6 and find a factor pair
where the sum of the factors is 5 (found in the middle term). 6 = 1 · 6 and 6 = 2 · 3. Since 2
and 3 sum to 5 the solution is (x − 2)(x − 3). This can be checked by multiplication.
Google Help: factoring trinomials
13. Factor the polynomial x2 + x − 12.
Solution Since the last term is negative, look at the factor pairs for 12 and find a factor pair
where the difference of the factors is 1 (found in the middle term). 12 = 1 · 12 = 2 · 6 = 3 · 4.
Since the difference of 3 and 4 is 1, the solution is (x − 3)(x + 4). This can be checked by
multiplication.
Google Help: factoring trinomials
14. Factor the polynomial 6p2 − 5p − 6.
Solution Since the coefficient of the first term is 6, this problem is more difficult than the
previous two. Although there is a more systematic way to factor this polynomial, it is simple
enough to notice that the factors of 6 are 1 and 6 or 2 and 3. Check all the possibilities such
as (p + 6)(6p + 1). Such products have the correct first and last term. By checking all such
possibilities, the correct answer (3p + 2)(2p − 3) is found.
Google Help: factoring trinomials
15. Find the number that should be added to x2 − 8x to make the expression a perfect square.
!
"
! "2
! "2
b
b 2
b
2
2
because x +
= x +bx+
Solution In general when considering x +bx, add
2
2
2
! "2 ! "2
b
−8
In our case b = −8 so
=
= (−4)2 = 16. Check that x2 − 8x + 16 = (x − 4)2
2
2
Google Help: completing the square
POLYNOMIAL EQUATIONS INCLUDING QUADRATIC EQUATIONS
16. Solve for x by factoring if x2 − 5x + 6 = 0. Hint: See Problem 12.
Solution By problem 12, x2 − 5x + 6 = (x − 2)(x − 3) = 0. If the product of two numbers is
zero, then one of the numbers must be zero. So x − 2 = 0 or x − 3 = 0. So x = 2 or x = 3.
Google Help: Solve quadratic equations by factoring
17. Solve for x by factoring if x2 + x − 12 = 0. Hint: See Problem 13.
Solution By problem 13, x2 + x − 12 = (x − 3)(x + 4) = 0. So x − 3 = 0 or x + 4 = 0. So
x = 3 or x = −4.
Google Help: Solve quadratic equations by factoring
18. Solve for p by factoring if 6p3 − 5p2 − 6p = 0. Hint: See Problem 14.
Solution First factor out a p to get 6p3 − 5p2 − 6p = p(6p2 − 5p − 6) = 0. Using problem 14,
factor further to get p(6p2 − 5p − 6) = p(3p + 2)(2p − 3) = 0. If the product of three numbers
is zero, then one of the numbers must be zero. So p = 0 or 3p + 2 = 0 or 2p − 3 = 0. So p = 0
or p = −2/3 or p = 3/2.
Google Help: Solve equations by factoring
19. Solve the equation x2 − 8x + 9 = 0 by completing the square. Hint: See Problem 15.
Solution If x2 − 8x + 9 = 0, then x2 − 8x = −9. Complete the square by adding 16 to both
sides to get√x2 − 8x + 16 = 16 − 9 or (x − 4)√2 = 7. Taking the square root of both sides yields
x − 4 = ± 7. Solving for x gives x = 4 ± 7.
Google Help: completing the square
20. Use the quadratic formula to solve for x in the equation x2 + 4x + 2 = 0.
Solution If a, b, and c are constants with a $= 0, then the solutions to ax2 + bx + c = 0 are
given by
√
−b ± b2 − 4ac
x=
.
2a
√
√
√
−4 ± 2 2
−4 ± 8
2
=
= −2 ± 2.
Applying this to x + 4x + 2 = 0, we get x =
2
2
LINEAR EQUATIONS
21. Find the slope of the line in the x-y plane that goes through the points (1, 2) and (4, −2).
Solution Given any two distinct points on a line in the x-y plane, the slope of the line is
the change in y divided by the change in x. If the coordinates of the two points are given by
y2 − y1
(x1 , y1 ) and (x2 , y2 ), then the slope m is given by the formula m =
. For this problem
x2 − x1
4
−2 − 2
=− .
(x1 , y1 ) = (1, 2) and (x2 , y2 ) = (4, −2). The slope is given by m =
4−1
3
Google Help: Find the slope of a line
22. Find the equation of the line in the x-y plane that goes through the point (1, 2) and has slope
−2. Solve for y.
Solution If a line in the x-y plane has slope m, and goes through the point (x1 , y1 ), then
the equation of the line can be found by taking an arbitrary point on the line (x, y), different
from (x1 , y1 ), and computing the slope using (x1 , y1 ) and (x, y). The equation is given by
x − x1
= m. Multiplying both sides by x − x1 gives the so called point-slope form for the
y − y1
equation of a line y − y1 = m(x − x1 ). In this problem we get y − 2 = −2(x − 1). Solving for
y gives y = −2x + 4.
Google Help: point-slope form
23. Find the equation of the line in the x-y plane that goes through the points (1, 2) and (3, 3).
Write the solution in the form Ax + By + C = 0.
3−2
1
Solution The slope of the line is
= . Using the slope and the point (1, 2), we get the
3−1
2
1
equation y − 2 = (x − 1). Multiplying both sides by 2 gives 2y − 4 = x − 1. Simplifying
2
gives x − 2y + 3 = 0.
Google Help: find equation of line from two points
24. Find the slope of the line in the x-y plane whose equation is given by 2x + 3y + 4 = 0.
Solution Solve for y. y = − 32 x − 34 . The slope is the coefficient of x which is − 23 .
Google Help: slope-intercept form
25. Find the equation of the line in the x-y plane that goes through the point (−1, 1) and is
parallel to the line whose equation is given by 2x + 3y + 4 = 0.
Solution From problem 19 the slope of the line is − 23 . The parallel line has the same slope
and passes through (−1, 1). Using the point-slope form of the line gives y − 1 = − 32 (x + 1).
This equation simplifies to 2x + 3y − 1 = 0.
26. Select the equation whose graph is the given line.
Solution If the slope of a line is m and
its y-intercept is b, then the
equation of the line is y = mx + b.
3
The slope is and the y-intercept
2
is 1. So the equation of the line is
3
y = x = 1.
2
Google Help: slope-intercept form
−3
y
4
3
2
1
−2
−1
−1
−2
−3
1
2
3
4
x
EXPONENTS, RADICALS, RATIONAL EXPRESSIONS, AND COMPLEX NUMBERS
27. Simplify
9x10 y 4
. Use only positive exponents.
−3x5 y 6
Solution
28. Simplify
9x10 y 4
(9/3)x10−5
3x5
=
−
=
−
Google Help: laws of exponents
−3x5 y 6
y 6−4
y2
x + 3/4
.
1/2 − 1/7
Solution Multiply numerator and denominator of
and 7. The least such common multiple is 28.
x + 3/4
. by a common multiple of 4, 2,
1/2 − 1/7
28x + 21
28x + 21
28 x + 3/4
·
=
=
28 1/2 − 1/7
14 − 4
10
Google Help: complex fractions
√
3
29. Simplify 56
√
# $1
1
3
Solution 56 = 56 3 = 5(6· 3 ) = 52 = 25
Google Help: radicals and exponents .
30. Simplify (2bc2 d3 )2 (2b3 c5 d)3 .
Solution (2bc2 d3 )2 (2b3 c5 d)3 = 22 b2 c4 d6 23 b9 c15 d3 = 25 b11 c19 d9 = 32b11 c19 d9
Google Help: laws of exponents
√
√
31. Simplify the expression 5 81 − 5 25.
Solution 5 · 9 − 5 · 5 = 45 − 25 = 20
Google Help: radicals and exponents
32. Find the product of the complex numbers 2 + i and 3 − i. Recall that i2 = −1.
Solution (2 + i)(3 − i) = 6 − 2i + 3i − i2 = 6 + i + 1 = 7 + i
Google Help: complex numbers
FUNCTIONS AND GRAPHS
33. A function is given by g(x) = −2x2 + 3x − 5. Find g(−1).
Solution If g(x) = −2x2 + 3x − 5, then g(−1) = −2(−1)2 + 3(−1) − 5 = −2 − 3 − 5 = −10
Google Help: evaluate a function
34. A function is given by g(x) = x2 + x + 1. Find g(x + 1).
Solution If g(x) = x2 + x + 1, then g(x + 1) = (x + 1)2 + (x + 1) + 1 = x2 + 2x + 1 + x + 1 + 1 =
x2 + 3x + 3
Google Help: evaluate a function
35. The point in the x-y plane is given. Find the coordinates of the point.
Solution Through the given point draw lines through the point parallel to the x-axis and
the y-axis. The x coordinate of the point is −3 where one of the lines crosses the x-axis, and
the y coordinate of the point is 2 where the other line crosses the y-axis. So the coordinates
are given by the ordered pair (−3, 2).
Google Help: plotting points
y
4
3
2
1
−5
−4
−3
−2
1
−1
−1
2
3
4
3
4
x
36. The graph of a function f is given. Use the graph of the function to find f (4).
Solution Find the point on the curve with
x-coordinate 4. The
y-coordinate of this point is
f (4). So f (4) = 2
y
4
3
Google Help: graph of a function
2
1
−5
−4
−3
−2
−1
−1
1
2
x