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MATH 1010-4: INTERMEDIATE ALGEBRA
Third EXAM
Tuesday, November 11th, 2008
- (Polynomials) Consider the following questions for the given polyno-
mial: −x + 3x^2 − 3 + x^3
(a) What is the Standard Form of this polynomial? (5 pts)
(b) Find out the degree, leading coefficient, and constant term of this
polynomial. (5 pts)
(c) Completely factor this polynomial. (Using “common factor”,
“grouping”, and “special form: a^2 − b^2 = (a + b)(a − b).” Note that you should have three different factors in your answer.) (10 pts)
Answer:
(a) Start from the highest degree term: x^3 + 3x^2 − x − 3.
(b) Degree = 3; Leading coefficient = 1; Constant term = -3.
(c)
x^3 + 3x^2 − x − 3 = (x^3 + 3x^2 ) + (−x − 3)
= x
2 (x + 3) + (−1)(x + 3)
= (x + 3)(x^2 − 1)
= (x + 3)(x + 1)(x − 1).
- (Word Problem I) An object is thrown upward from a height of 32
feet with an initial velocity of 16 feet per second. The height h (in
feet) of the object after t seconds is modeled by the equation
h(t) = − 16 t^2 + 16t + 32.
How long will it take for the object to reach the ground? (10 pts)
Answer: The object reaches the ground exactly when the height
“h(t)=0.” So we solve the edquation:
h(t) = − 16 t^2 + 16t + 32 = 0
⇒ − 16(t
2 − t − 2) = 0
⇒ − 16(t − 2)(t + 1) = 0
⇒ Either t − 2 = 0 or t + 1 = 0.
⇒ Either t = 2 or t = − 1.
⇒ So t = 2 as negative answer doesn
′ t f it physics.
It will take 2 seconds for the obeject to reach the ground.
So
2 x+3 +^
3 x− 2 2 x− 1 −^
1 2 x+
5(x+1) (x−2)(x+3) 3(x+1) (x−1)(2x+1)
5(x + 1)
(x − 2)(x + 3)
×
(x − 1)(2x + 1)
3(x + 1)
(cancel out (x + 1))
5(x − 1)(2x + 1)
3(x − 2)(x + 3)
(b)
1 st term :
6 t √ 2 t
6 t)(
2 t)
(
2 t)(
2 t)
12 t
t
2 t
3 t
t
2 t
(cancel out 2 t)
t
= 5
3 t.
2 nd term :
108 t
4
36 ∗ 3 t
4
3 t
4
3 t
2
So
6 t √ 2 t
108 t
4
3 t +
3 t
2
3 t
3 t
3 t.
Compare to a + b
3 t, we have a=0 and b = 132.
- (Word Problem II) Ray and his grader are going to grade your 3rd
exam. From the former experience, we know that it takes triple time
for his grader to complete the grading by himself as Ray does. If they finish all the work together in 3 hours, then how many hours it would
take Ray himself to grade all the exams? (10 pts) (Hint: Ray’s Rate + Grader’s Rate = Rate for Cooperation )
Answer:
Rate =
T otal work
T ime Spent
Total work = 1
Time for Ray to finish grading by himself = x (hours) Time for grader to finish grading by himself = 3x (hours)
Time for cooperation to finishing grading = 3 (hours) Ray’s Rate = (^1) x
Grader’s Rate = (^31) x Cooperation Rate = (^13)
By hint, we get the equation
x
3 x
⇒ (3x)(
x
3 x
)(3x)
⇒ 3 + 1 = x
⇒ 4 = x
So it takes Ray 4 hours to finish grading by himself.
So
Compare to a + b
6 , we have a=1 and b=-1.
(b)
1 st term :
3 − 2 i
2 − i
(3 − 2 i)(2 + i)
(2 − i)(2 + i)
(conjugate)
(3 − 2 i)(2 + i)
22 − i^2
(special f ormula f or bottom)
6 + 3i − 4 i − 2 i^2
4 − (−1)
(F OIL on top)
8 − i
5
=
i
2 nd term :
1 − 2 i
4(1 + 2i)
(1 − 2 i)(1 + 2i)
(conjugate)
4(1 + 2i)
12 − (2i)^2
(special f ormula f or bottom)
4 + 8i
1 − (−4)
4 + 8i
5
=
i
So
3 − 2 i
2 − i
1 − 2 i
i) − (
i)
i −
i
i −
i)
i.
Compare to a + bi, we have a = 45 and b = − 95.
