Math 209
150 points
Work the following problems. Show all your own work for maximum credit. An incorrect answer,
with no associated calculations has no basis for partial credit. Each problem is worth 10 points.
Post your completed work as an attachment in your individual forum.
1. Perform the indicated operations:
 (- 8x2 + 5x - 12) + (- 3x2 - 9x + 18)
Combining like terms we get
-8x2 -3x2 + 5x - 9x -12 + 18
-11x2-4x+6
 (a+b)(a2 +ab + b2)
Using FOIL we get
a3+a2b + ab2+ a2b + ab2 + b3
a3+2a2b+2ab2 + b3

-3a3b7
9a2b10
using rules of exponent am/an = am-n we get
-a3-2/3b10-7
-a/3b3
 (3x + 5)(2x – 6)
Using FOIL we get
6x2-18x + 10x -30
6x2-8x -30
 (- 5w3r2) * (2w4 r8)
Using the law of exponents am *an = am+n we get
-10w7r10
2. The newest jet fighter, the F-22 Raptor, cruises at 1.5 times the speed of sound. The speed of
sound is 1.24x103 km/hr. How many hours does it take for the F-22 to fly the 5.8x103 km from
New York to Paris? Demonstrate the use of scientific method in your calculations.
Distance = speed x time
Time = distance/speed
Time = 5.8 x 103/1.24 x 103
Time = 4.677 x 103-3 => 4.68 hrs
3. Jason and Mary have a new baby. They want to invest $20,000 at the baby’s birth in order to have
$100,000 for the baby’s college education. What interest rate do they need to earn in order to
achieve this in 18 years?
Use the following formula to determine the average interest rate:
r = (S/P)1/n – 1
where S = the final amount
P = the initial investment
n = number of years
Using S = 100,000, P = 20,000 and n = 18 we get
r = (100000/20000)1/18-1
r = (5)1/18-1
r =1.0935-1
r = 0.0935
So the rate of interest will be 9.35% or approximately 9.4%
4. A retailer of DVD players knows that at a price of q dollars per player, he can sell (900 – 3q)
players per month. Write a polynomial, in function notation, that represents the monthly income
from the sale of DVD players.
Monthly income = (900-3q)*q
=> monthly income = 900q – 3q2
5. A red ball and a green ball are simultaneously tossed into the air. The red ball is given an initial
velocity of 96 feet per second, and its height, t seconds after it is tossed is( -16t2 + 96t )feet. The
green ball is given an initial velocity of 80 feet per second, and its height, t seconds after it is
tossed is (-16t2 + 80t ) feet.
5.1. Write the algebraic expression, in simplest form, that represents the difference in heights
between the two balls at any time, t.
Difference in heights = -16t2+96t – (-16t2+80t)
 difference = -16t2+96t +16t2-80t
 difference = 16t
5.2. How much higher is the red ball after 2 seconds?
Put t = 2
Difference = 16(2) = 32 ft
6. What are the six steps in the strategy for completely factoring polynomials?
Factor out the GCF.
Note the number of terms (this determines the factoring method.)
If there are:
4 terms: Factor by grouping
3 terms: Factor by trial and error or ac method
2 terms: Must be a difference of squares, difference of cubes, or sum of cubes (know
formulas).
Factor completely (factor each factor again if possible.)
Check: Make sure that no factor contains a GCF; multiply the factors to be sure the product
matches the original one.
7. Factor the following expressions, completely:
 9x+12y
Here the GCF is 3, so factoring out GCF we get
3(3x + 4)
 15x3y2 – 20xy2 - 20x2y2
GCF is 5xy
5xy2(3x2 -4-4x)
5xy2(3x2-4x-4)
5xy2(3x2-6x + 2x -4)
5xy2(3x(x-2) + 2(x-2))
5xy2(3x+2)(x-2)
 y3 – 27
(y)3 –(3)3
Using the rule a3-b3 = (a-b)(a2+ab+b2)
(y-3)(y2+3y+9)
 3x2 +x -10
3x2+6x-5x-10
3x(x+2)-5(x+2)
(3x-5)(x+2)
 X16 -1
(x4)2-(1)2
(x4-1)(x4+1)
(x2-1)(x2+1)(x4+1)
(x-1)(x+1)(x2+1)(x4+1)
8. The volume of a cubic container is x3 cubic inches. The height of the container was decreased and
the length was increased so that the volume is now modeled by the expression
x3 + 4x2 -21x
By how many feet were the height and length changed?
We need to factor the expression
X(x2+4x-21)
X(x2+7x-3x-21)
X(x(x+7)-3(x+7))
X(x+7)(x-3)
So the length was increased by 7 inches and the height was decreased by 3 inches
9. A car is traveling on a road that is perpendicular to a railroad track. When the car is 60 yards from
the crossing, the car’s new collision alert system warns the driver that a train is, diagonally, 100
yards from him and traveling toward the same crossing. How far is the train from the crossing?
60 yds
car
?
100
yds yds
Train
Let the distance between the crossing and train is x
Using pythagorus rule we get
X2 = 1002 – 602
X2 = 10000-3600
X2 = 6400
X = 800
So the distance is 800 yards
10. Reduce the rational expression to simplest form.
h 3- t3
h-t
Factoring the numerator we get
(h  t )( h 2  ht  t 2 )
ht
2
h  ht  t 2
11. Perform the indicated operation:
x3 – 1 ÷ 9x2 + 9x + 9
x2 + 1
x2 - x
x3  1 9x 2  9x  9

x2 1
x2  x
factoring and taking reciprocal
( x  1)( x 2  x  1)
x( x  1)

2
x 1
9( x 2  x  1)
x( x  1) 2
9( x 2  1)
.
12. Name 4 methods for solving quadratic equations. Give an example of each.
1. FACTORING
Set the equation equal to zero. If the quadratic side is factorable, factor, then set each factor
equal to zero.
Example: x2 = −5x − 6
Move all terms to one side x2 + 5x + 6 = 0
Factor (x + 3)(x + 2) = 0
Set each factor to zero and solve x + 3 = 0 x + 2 = 0
x = −3 x = −2
2. PRINCIPLE OF SQUARE ROOTS
If the quadratic equation involves a SQUARE and a CONSTANT (no first degree term),
position the square on one side and the constant on the other side. Then take the square root
of both sides. (Remember, you cannot take the square root of a negative number, so if this
process leads to taking the square root of a negative number, there are no real solutions.)
Example 1: x2 −16 = 0
Move the constant to the right side x2 = 16
Take the square root of both sides x2 = ± 16
x = ±4, which means x = 4 and x = −4
Example 2: 2(x + 3)^2 −14 = 0
Move the constant to the other side 2(x + 3)^2 = 14
Isolate the square (x + 3)^2 = 7 (divide both sides by 2)
Take the square root of both sides (x + 3)= ±
x+3=±
x = -3±
7
7
7
This represents the exact answer.
Decimal approximations can be found using a calculator.
3. COMPLETING THE SQUARE
If the quadratic equation is of the form ax^2 + bx + c = 0, where a ≠ 0 and the quadratic
expression is not factorable, try completing the square.
Example: x^2 + 6x −11 = 0
**Important: If a ≠ 1, divide all terms by “a” before proceeding to the next steps.
Move the constant to the right side x^2+6x = 11
Find half of b, which means : 6/2 = 3
Find (b/2)^2 = (3)^2 = 9
Add (b/2)^2 to both sides : x^2 +6x + 9 = 11+9
Factor the quadratic side: (x+3)^2 = 20
Take square root on both sides
x+3=± 2 5
x = -3± 2 5
This way we can use the completing square method
4. QUADRATIC FORMULA
Any quadratic equation of the form ax^2 + bx + c = 0, where a ≠ 0 can be solved for both real
and imaginary solutions using the quadratic formula:
Example: x^2 + 6x −11 = 0 (a = 1, b = 6, c = −11)
The quadratic formula we get
 b  b 2  4ac
x
2a
x
 6  (6) 2  4(1)( 11)
2(1)
 6  80
2
64 5
x
2
x  3  2 5
x
13. Solve for x in the following equations:
 x2 –x -12 = 0
Factoring we get
(x-4)(x+3) = 0
=> x = 4 , x = -3
X = 4, -3
 x2 + 8x -4 = 0
x^2 + 2(4x) = 4
Add 16 to both sides
x^2 + 2(4x) + 16 = 20
(x+4)^2 = 20
Taking square root on both sides we get
x +4 =  2 5
x = -4  2 5
So the solution is
x = -4 + 2 5 , -4 - 2 5
 m2 -6m -7 = 0
Factoring we get
m^2 -7m + m -7 = 0
(m-7)(m+1) = 0
m = 7 , -1
14. The diagonal of a square is 3 meters longer than a side. Find the length of a side.
Diagonal of a square, d = s 2 where s is the side
Now given that
d=s+3
So s 2 =s + 3
s 2 - s=3
s( 2 -1) = 3
s = 3/( 2 -1)
Rationalizing the denominator we get
s = 3( 2 +1)/(2-1)
s = 3( 2 +1) meters
Approximately
s = 3(1.414+1) = 3(2.414) = 7.243m
15. Thomas is going to make an open-top box by cutting equal squares from the four corners on an
11 inch by 14 inch sheet of cardboard and folding up the sides. If the area of the base is to be 80
square inches, then what size square should be cut from each corner (What is the value of x)?
Length of the base (after folding the sides) = 14-2x
Width of the base (after folding the sides) = 11-2x
Area of the base = (14-2x)(11-2x) = 80
 using FOIL we get
 154 -50x + 4x2 = 80
 4x2 -50x + 154-80 =0
 4x2-50x+ 74 = 0
 Divide both sides by 2 we get
 2x2-25x + 37 =0
 Using quadratic formula we get
x
25  (625)  4(2)(37)
2 ( 2)
25  (329)
4
25  18.14
x
4
x  10.78,1.72
 x
Now the corner can’t be 10.78 inches because that’s too large
So the corner will be 1.72 inch long