Mth 65 Exam 1 Practice Franz Helfenstein Name
Be prepared to neatly show your work. Plan to use a pencil, simplify your results and box in your
answers! Scratched out pen marks and other such messiness will reduce your score.
a)
T
F
True or False. Circle T if always true otherwise circle F.
(a + b) = a3 + b3
j) T F
If P(x) = 3x2 + 2x – 1 then P(-1) = 0
b)
T
F
(a – b)2 = a2 – ab + b2
k)
T
F
( 2 + x)( 2 – x) = 2 – x2
c)
T
F
(-2)5 = 32
l)
T
F
2a2b(a2 + 3ab) = 2a4b + 6a3b2
m)
T
F
x2 + 4
x =x+4
n)
T
F
-x(2y – 3x – 5) = -2xy +3x2 + 5x
3
d)
T
F
(-4)(13) + 52
= DNE
77
e)
T
F
x-3 · x7 = -x10
f)
T
F
x2 + y2 = x + y
o)
T
F
Quadratics are Second Order
g)
T
F
(3x)3 – x(5x)2 x
=4
42 – 23
p)
T
F
1
y = x2 + 4x – 5 is a parabola
h)
T
F
a(x·y) = (ax)(ay)
q)
T
F
(Quadratic) x (Cubic) is 6th Order
i)
T
F
9x2 – (3x)2
=0
6 – 2·3
r)
T
F
4
3
7
+
2 =
x
x
x3
2)
P(x) = 2x2 + 5x – 7
(a) P(t) =
(b) P(-2) =
(c) P(2) + 2x – 4 =
(d) P(x+1) =
(e) [P(x) + 1]2 =
(f) 4[P(x+2)] =
(g) x2 P(x) – 5x4 =
3)
Q(2,-3) + 3
Q(x, y) = 3x2 – 5xy + 4y2. Evaluate 2 Q(0,1) =
4)
(a) Solve for x: 3(2 – x) – x = 4
(b) Solve for H: 2(3H – 2) = 4H + 6
1 x 1
(c) Solve for x: 3 + 2 = 4
(d) Solve for x:
5)
Simplify: 6x5 – 4x3 (2x2 – 3x3) =
6) Simplify: x(6x·y + 3 x2 y) – 4(2x2 y + 3 x3 y) =
7)
Simplify: (4x5 + 3x3 – 5x2 + 3x + 2) + (9x5 – 11x4 – 7x2 – 3x + 5) =
8)
Simplify: (-7x3 – 5x2 + 13x + 2) – (-11x3 – 7x2 – 3x + 5) =
3ax + 1
= 2x + b
a
Multiply & Simplify
9) (2x + 3)(4x – 7) =
10) (5x2 – 7x)(3x3 – 5x + 2) =
11) (x – 1)(x2 + x + 1) =
12) 3x3 (3x3 – 5x + 2) – (5x2 – 7)(2x4 + 3x) =
13) (2x + 3)3 =
14)
(2x + 3)2(4x – 7) =
15) Given Q2(x) = x2 – 3x + 1 simplify in standard form: Q2(x+1) =
16) Multiply and Simplify: [2x2 y3 – 3x3 y2 ]2 =
10x5 – 4x3 (3x3 – 5x2)
=
6x3
x3 – 14x + 15
19) Divide:
=
x–3
17) Simplify:
18)
20)
6x2 – 7x – 3
=
2x – 3
8x3 + 6x2 – 11x – 3
Divide:
=
2x – 1
Divide:
21) Simplify to all positive exponents:
a) a2 b4 a5 a3 b12 20 =
b)
a-3 a5 a-4 b-3 b6 b-1 =
c)
a2 (a5)2 (b12)3 =
d)
(a3b4)2 (a3b2)4 =
e)
(a2b3)3 (ab4)2
=
(a4b2)2 (a3b2)4
f)
a2 a15 a3 b2 b b6
=
a5 a7 a b10 b2 b4
g)
a-2 b5 c4 x-3 y-3 z-1
=
a3 b7 c9 x-5 y6 z8
h)
(a-2b1)3 (2ab-3)-2
=
(2a-3b2)-2 (a3b2)4
i)
2x2y3(xy2 - x2y + 2x2) =
j)
[(x2y5z4)3]2 =
k)
(x5 - y3)3 =
l)
2x3(7x5 – 4x4 + 3x3 – 2x2 + 9x + 1) =
m)
24 x y2 (22 x2 y2)5 (23 x4 y)4 =
n)
23 x6 y-5 (2-2 x-1 y2)3 =
o)
(x4 y-2)-3 z0
=
x9 (y-2 z3)-6
p)
x12 (y-5 z11)2
=
(x9 y-3)-2 z8
q)
x7 y-2 z4 (x-2 y4)-3 =
r)
2x7 y-2 z4(3x6y-5 + 4x-1 y2) =
s)
2x3y2z4(2xy5z5 – 3x2yz4 + 7x2y3z) =
t)
(2x2y5z4 – 3x3y7z3)(3xy2y2z) =
22) Write in scientific notation: (a) 3.2 Megawatts = __watts
(b) 5 nanosecond = __ sec.
23) Give your answer in scientific notation with 2 decimal digits: (i.e. x.xx · 10±n)
356
(a) 93,000,000
(b) -0.0000021
(c) 12
(d) -342
(e) 1000 =
(f) 7 billion
(-2.28 x 10-2) (1.64 x 104)
102 + 2.5·104 + 3.6(2.4·104 + 7.1·104)
g)
=
h)
=
(-3.28 x 10-3) (7.64 x 103 )
5 – 4.2 + 2.2(3.53 + 8.1)
i)
k)
2.98 · 103 + 9.75 · 102 =
x=
b – b2 – 4ac
2a
j)
(-2.28 x 10-2) (1.64 x 104)
=
-3.28 x 103 + 9.64 x 102
Find x given: a = -1.23·101, b = -3.45·105, c = 5.67·104
24) Make a T-table then plot the points and sketch the graph for P(x) =
-x3 + 25x
10
25) The Falling Body Model, H(t) = -16 t2 + v0 t + h0 , gives the height, H (in feet), as a function of
time, t (in seconds). v0 represents the initial velocity and h0 represents the initial height.
(a)
An arrow is shot upward from ground level with initial velocity of 380 ft/sec. What is the
height after 5 seconds?
(b)
A ball is thrown upward at 60 ft/sec from the top of a 120 ft tall building. How far is the ball
from the ground after 4 seconds?
1
26) Divide to show that 1 – x = x0 + x1 + x2 + x3 + x4 + . . .
1
1
27) Using the results from problem 26 should 1 – x · 1 + x = x0 + x2 + x4 + x6 + x8 + . . . ?