Name                           

Math 152B Midterm I

Please do all of the following problems.  Credit earned will be based on the steps that you show that lead to the final solution.  Good Luck!

Problem 1:  Factor the expression completely:

A)   x⁴ - 9x²

Solution

First find the GCF, which is x².  Factoring this out gives

        x² (x² - 9)

The second factor is a difference of squares with a = x and b = 3.

        x²(x - 3)(x + 3)

B)    x² + 3x - 54

Solution

There are 3 terms here, so we use reverse FOIL.  The question is, "what two numbers multiply to -54 and add to 3.  Notice that

        (9)(-6) = 54 and 9 + (-6)  =  3

We can write the problem as

         =  x² - 6x + 9x - 54

Now factor by grouping

        =  (x² - 6x) + (9x - 54)

Factor out the GCFs

        =  x(x - 6) + 9(x - 6)

and factor out the GCF again

        =  (x - 6)(x + 9)

C)    6x² + 11x - 10

Solution

We use the AC method. 

        AC = (6)(-10)  =  -60

Now find two numbers that multiply to -60 and add to 11.  Notice that (15)(-4)  =  -60 and 15 + (-4)  =  11.  We write

        6x² - 4x + 15x - 10

        =  2x(3x - 2) + 5(3x - 2)

        =  (3x - 2)(2x + 5)

D)    128xyz³ + 54x⁴y

Solution

Pull out the GCF of 2xy:

        =  2xy(64z³ + 27x³)

The last factor is a sum of cubes with a = 4z and b = 3x.  We get

        =  2xy(4z + 3x)(16z² - 12xz + 9x³)

E)     3st + 18t - 6u - su

Solution

Factor by grouping:

        =  (3st + 18t) + (-6u - su)

        =  3t(s + 6) - u(6 + s)

        =  (s + 6)(3t - u)

Problem 2:  Simplify the following rational expression          

                 x³ - 8    
A.                                           
                x² - 4   

Solution

        First factor the numerator and denominator.  The numerator is a difference of cubes and the denominator is a difference of squares

                  (x - 2)(x² + 2x + 4)    
         =                                                         
                      (x - 2)(x + 2)   

Now cancel the (x - 2) factors

                  x² + 2x + 4   
         =                                             
                      x + 2

                 -15r⁸s⁵    
B.                                           
                 21rs⁸     

We work on the numbers first:

             -15           -5
                      =                              
              21              7

Now work on the "r" factors

              r⁸         
                      =  r⁷                   
                 r      

Put these together to get

                      5 r⁷
           =   -           
                      s³

Problem 3:  Solve the following equations

A.   x³ - 2x² + x = 0

Solution

We factor the left hand side.  First pull out the GCF:  x

        x(x² - 2x + 1)  =  0

Now notice that the second factor is a perfect square trinomial (square of a difference) with a = x and b = 1.  We get

        x(x - 1)²  =  0

Now use the zero factor property:

        x  =  0      or      x - 1  =  0

so

        x  =  0     or     x  =  1

B.    4x² = 15 - 4x

Solution

First set the equation equal to 0 by adding 4x - 15 to both sides:

        4x² + 4x - 15  =  0

Now use the AC method

        AC  =  -60

Notice that

        (10)(-6)  =  60     and     10 + (-6)  =  60

so we write

        4x² - 6x + 10x - 15  =  0

Now factor by grouping

        2x(2x - 3) + 5(2x - 3)  =  0

        (2x + 5)(2x - 3)  =  0

Now use the zero factor property

        2x + 5  =  0    or    2x - 3  =  0

subtract 5 from both sides on the first and add 3 to both sides on the second

        2x  =  -5    or    2x  =  3

Now divide both sides by 2 for both equations

        x  =  -5/2    or    x  =  3/2

Problem 4:  The pressure p in pounds per square foot of a wind is directly proportional to the square of the velocity v of he wind.  If a 10-mi/hr wind produces a pressure of 0.3 lb/ft², what pressure will a 100-mi/hr wind produce?  

Solution

Since this is a "directly proportional" problem, we have

        p  =  kv²

where k is some constant.

The first phrase of the second sentence gives us

        v  =  10    p  =  0.3

Plugging these in gives

        0.3  =  (k)(10)²  =  100k

Divide both sides by 100 to get

        k  =  0.3/100  =  0.003

Putting this back into the equation gives

        p  =  0.003v

Now the question is to find p when w  =  100.  We plug in to get

        p  =  (0.003)(100)²  =  30

So the pressure will be 30 lbs/ft² when the wind is 100 miles per hour.

Problem 5:  The temperature F in degrees Fahrenheit is related to the temperature C in degrees Celsius by the equation

    F = 9/5 C + 32

Solve this equation for C.

  Solution

First subtract 32 from both sides to get

        F - 32  =  9/5 C

Now to get C by itself multiply by 5/9.

        (5/9)(F - 32)  =  (5/9)(9/5)C

Distribute the 5/9

        5/9 F - 5/9 (32)  =  C

or

        5/9 F - 160/9  =  C

or

        C  =  5/9 F - 160/9

Extra Credit:  Write down one thing that your instructor can do to make the class better.

(Any constructive remark will be worth full credit)