Name
. Math 152B Final Please do all
of the following problems. All work
and all answers must be done on your own paper.
Credit earned will be based on the steps that you show that lead to the
final solution. Good Luck! Problem
1: Factor the following A)
x5y2 - x3y4
x5y2 - x3y4 = x3y2(x2 - y2) Pulling out the GCD = x3y2(x - y)(x + y) Difference of Squares
B)
6x2 + 5x - 6
AC = 6(-6) = -36 We use the AC method (-4,9) Finding two numbers that add to -36 and multiply to 5 6x2 + 5x - 6 = 6x2 - 4x + 9x - 6 Using 5x = -4x + 9x = 2x(3x - 2) + 3(3x - 2) Factoring by grouping = (3x - 2)(2x + 3) Pulling out the GCD of 3x - 2
C)
x3
x3
= (x + 1)(x2 - 4) Pulling out the GCD = (x + 1)(x - 2)(x + 2) Difference of Squares
Problem
2: Perform the indicated operations and simplify A)
x3 + 3x2 + 2x
x(x2 + 3x + 2)
x(x + 2)(x +
1)
x (x + 2)
x(x + 2)(x +
1)
1
x + 1 1
x + 1
10(h - 5)
5(h - 10)
10(h - 5) - 5(h - 10)
10h - 50 - 5h + 50
5h
Problem
3: Solve A)
x + 2 = x2 Squaring both sides x2 - x - 2 = 0 Moving all terms to the right and then switching left and right. (x - 2)(x + 1) = 0 Factoring x - 2 = 0 or x + 1 = 0 The zero product property x = 2 or x = -1 Solving Notice that x = 2 works in the original equation, but x = -1 does not work, since a root can never produce a negative number. Hence the solution is x = 2
B)
x3 + 2x2 + x = 0
x(x2 + 2x + 1) = 0 Factor out the GCF x(x + 1)2 = 0 Square of a sum x = 0 or (x + 1)2 = 0 Zero Product Property x = 0 or x = -1 Solving
Problem
4:
A motorboat travels 10 mi/hr in still water.
If the boat takes the same amount of time to travel 2 miles upstream as
it does to travel 3 miles downstream, find the speed of the current. Let x = the speed of the current We set up a distance rate time table, noting that the rate upstream is the rate in still water minus the speed of the current and the rate downstream is the rate in still water minus plus the speed of the current.
Since d = rt t = d/r Using the fact that the times are equal gives
2
3 2(10 + x) = 3(10 - x) 20 + 2x = 30 - 3x Multiplying out 5x = 10 Adding 3x and subtracting 10 from both sides x = 2 Dividing both sides by 5 The speed of the current is 2 miles per hour.
Problem
5: A) Find the point of intersection of the two lines
4x - 3y = 15
and 5x - 6y = 21 We use the method of elimination:
8x - 6y = 30 Multiplying
the first equation by 2 x = 3 Dividing both sides by 3 4(3) - 3y = 15 Substituting x = 3 into the first equation 12 - 3y = 15 Multiplying -3y = 3 Subtracting 12 from both sides ` y = -1 Dividing by -3 Hence the solution is the point (3,-1)
Problem
6:
Simplify. Write your answer
without negative exponents.
First use the distributive rule, multiplying the 3/2 through each exponent
x -1 y -2 = x -1 - 3 y -2 - 6 Using the quotient rule for exponents = x -4 y -8 Subtracting
1
Problem
7: Simplify A)
= 2
Problem
8: Solve. Any
complex solutions should be written in the form a + bi. A)
x2 - 3x
-15 = 0
The equation does not factor, so we use the quadratic formula:
a = 1 b = -3 c
= -15
B)
2x2 + 4 = x
2x2 - x + 4 = 0 Setting everything equal to zero. The equation does not factor, so we use the quadratic formula: a = 2 b = -1 c = 4
C)
3x2
- x - 1 = 0
The equation does not factor, so we use the quadratic formula: a = 3 b = -1 c = -1
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Problem 9 Complete the square
Problem 10 Solve the absolute value inequality. 3 - | 2x - 5| < 2 SolutionFirst we solve the equality
3 - |2x
- 5| = 2
-|2x - 5| =
-1 Subtract 3
from both sides
|2x - 5| =
1 Multiply
both sides by -1
2x - 5 = 1
or 2x
- 5 = -1
Absolute value equations split up
as an "OR" statement
2x = 6
or 2x
= 4
Add 5 to both sides of the equations
x = 3
or x
= 2
Divide both sides of the equations by 2
Now we find test points to the left of 2, between 2 and 3, and to the right of 3. We have 3 - |2(0) - 5| = 3 - |-5| = 3 - 5 = -2 < 2 Thus the numbers to the left of 2 are included.
3 - |2(2.5) - 5| = 3 - |5 - 5| = 3 - 0 = 3 > 2 Thus the numbers between 2 and 3 are not included. 3 - |2(4) - 5| = 3 - |8 - 5| = 3 - 3 = 0 < 2 Thus the numbers to the right of 3 are included. The final solution is
Problem 11 Answer the following true or false. Please explain your reasoning.
True, Since the inequality is a "<", the solution is between the endpoints.
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