Download Exam II Practice II Winter 2017

Math 71 Exam II Practice II
x3  8
x3 1

2 x 2  x  1 4 x 2  16
4
3
2
4
3
8
8
Factor a) 125 x  8 xy b) 3x  3 y c) 12 x  5 x  2 x
1) Multiply
2)
nE
1 1 1
 
for n. b) Solve
R  nL
a b c
x3
x 1
 2
Subtract 2
x  1 x  3x  2
Solve for x: t
x  2 1  x  3
3) Solve a)
4)
5)
I
3
6) Rationalize the denominator for a)
4
7) Simplify
b)
3 xy 3
e) factor
x 4  5x 2  4
4 x 3  4 x 2  36 x  36
f)
for a
5 2
3 22 3
if x is a nonnegative real number.
3x 3 8 x 2  3 x 5 y 3
8) A) Simplify by first writing the expression in the rational form. Assume x represents a nonnegative
4
real number. i)
3
x
2
x ii)
6
x
iii)
3
x
x
5
b) Simplify
9) Solve for x by first finding the impossible points:
1 i
2i
4
3
(2 x  x  4 x  1)  ( x 2  2)
10) Simplify a)
11) A) Divide
i 31
12) a) Simplify i)
b)
27
2
3
c) Multiply
5
27 x 13 y 21  5 27 x 21 y 14
20
20

1
x x9
i 60 c)
x2
ii)
3
x3
iii)
4
x4
if x is any real number. b) Simplify i)
x2
ii)
3
x3
iii)
4
x4
if x is a
nonnegative real number
13) Solve
2 x 2  10 x  1 by completing the square.
3x 2  x  2  0
14) Solve
15) A rectangular lawn measures 60 ft by 80 ft. Part of the lawn is torn up to install a sidewalk of uniform
width around it. The area of the new lawn is 2400 square feet. How wide is the sidewalk?
f ( x)  2  3x
1
1
1
 2
3
2
x y
4 x  4 b) xy
1
1
1
 2
1
2
2x  2
y
x
16) Find the domain of
17) A) Simplify
18) Compute the discriminant and find the number and type of solutions to 4 x  2 x  1  0
19) Squares with sides 2 in. are cut from the four corners of a sheet of metal, and the flops are folded
2
upward to form an open box. If the volume of the box is 32
the piece of metal?
20) Find the domain of
f ( x) 
3
4 x  11x  3
2
in 2
, what were the original dimensions of
Answer Key
( x  2)( x 2  2 x  4) ( x  1)( x 2  x  1) ( x 2  2 x  4)( x 2  x  1)
2) a)


(2 x  1)( x  1)
4( x  2)( x  2)
4(2 x  1)( x  2)
x(5x  2 y)( 25 x 2  10 xy  4 y 2 ) b) 3( x 4  y 4 )( x 2  y 2 )( x  y)( x  y) c) x 2 (4 x  1)(3x  2) e) Let u  x 2 ,
x 4  5x 2  4  u 2  5u  4  (u  1)(u  4)  ( x 2  1)( x 2  4)  ( x  1)( x  1)( x  2)( x  2) f) Since this is
1)
a 4-nomial, use grouping. Make sure to factor out the GCF first.
4 x 3  4 x 2  36 x  36  4( x 3  x 2  9 x  9)  4( x 2 ( x  1)  9( x  1)) 
4( x  1)( x 2  9)  4( x  1)( x  3)( x  3) 3)a) Multiply both sides by the LCD R  nL to get
I ( R  nL)  nE . Distribute I and move the terms containing n to one side to get IR  nE  nLI Factor
IR
out n from the right side and divide by the coefficient to get n 
b) Multiply both sides by the
E  LI
1
1
1
abc  abc  abc After clearing the denominators we obtain bc  ac  ab Isolate the
LCD abc:
a
b
c
terms containing a to get bc  ab  ac , Factor out a on the right side and divide both sides by b-c, we
bc
get bc  a (b  c )  a 
4) LCD=(x-1)(x+1)(X+2)
bc
x3
x  2 x2  x  6
x 1
x  1 x 2  2x  1

,

( x  1)( x  1) x  2
LCD
( x  1)( x  2) x  1
LCD
( x 2  x  6)  ( x 2  2 x  1)
x7

( x  1)( x  1)( x  2)
( x  1)( x  1)( x  2)
5) Square both sides (FOIL) to get
x  3  2 x  2  x  3 . Isolate the radical and divide both sides by -2 to get x  2  3 . Square both
2
2
sides again to get ( x  2)  3  x  2  9  x  7 x = 7. Check:
7  2  1  7  3 , yes. x=3.
4
3
6) a)
4
3xy
3 4
27 x 3 y
3

27 x y
4
27 x 3 y
xy
5 2
b)

3 2 2 3
3 2 2 3 3 2 2 3
3x 3 8 x 2  3 x 5 y 3  3x(2)3 x 2  xy3 x 2  (6 x  xy)3 x 2
7)
2
exponent form.
1
a) i)
3
x x  x 3 x 2  x 6  6 x 7  x 6 x _ ii)
5
1
2
30  10 6 15  5 6

6
3
8)Write each expression in the fractional
7
1

4
x
6
x
1 1

6
 x4
1
 x 12  12 x
iii)
2
(27) 3  (33 ) 3  3 2  9 c) 5 36 x 34 y 35  3x 6 y 7  5 3x 4 9)Domain: all real
numbers except 0 and -9. Multiply both sides by LCD  x( x  9) to get 20( x  9)  20 x  x( x  9) . This
x 5  (( x 5 ) 3 ) 2  x 6  6 x 5
3
b)
x 2  31x  180  0 . Factoring this, ( x  5)( x  36)  0 , we get x = -5, 36. 10) a)  i
1 3
12 x  9
 i 11) a) 2 x 2  x  4  2
5 5
x 2
12) a) i) x ii) x iii) x b) i) x ii) x iii) x 13)
simplifies as
2
b)
1
2
5
3 3
1
5
27
1 5
5
2 x  10 x  1  x  5 x   x 2  5x         ( x  ) 2 
 x 

2
2
2
2
4
2 2
2
2
2
1  (1)2  4(3)(2) 1  23 1  23i
5 3 3
2
x 
14) 3x  x  2  0  a  3, b  1, c  2  x 


2
2
2(3)
6
6
c)
15) Since you are removing 2x from the width and length, we get
move 2400 to the other side, we get
(60  2 x)(80  2 x)  2400 .
FOIL and
4 x 2  2800 x  4800  0 Divide each term by 4 to get
x 2  70 x  600  0 Factoring this will give x = 10 and x = 60.
But x =60 is impossible. X = 10 is the
answer. 10feet. 16) All real numbers less than or equal to 2/3. 17) Multiply each term LCD of 4(x-1) and
1 (4 x  1)
12( x  1)  1 12 x  11
4( x  1) 1
cancel out the denominator to get


1 4( x  1)
4( x  1)  2
4x  2
1(4 x  1) 
2( x  1) 1
3(4 x  1) 
b) The LCD is
x2 y2
18) The discriminant is
1 x2 y2
1 x2 y2

x y
x y
1
xy 2 1
x2 y 1
 2


2 2
2 2
2
( x  y )( x  y ) x  y
1 x y
1 x y
x y
 2
2
1
1
y
x
(2) 2  4(4)(1)  20 , which is a positive and non-square.
Two irrationals.
19)
After cutting off the corners, the length =width=x-4
Thus
Volume  2( x  4)( x  4)  32  ( x  4)( x  4)  16  x 2  8x  0
Solve this equation by factoring.
x  8x  0  x( x  8)  0  x  0, x  8
2
X=0 is impossible.
8m by 8m.
3
: set the denominator =0 to find all the impossible points:
4 x  11x  3
1
4 x 2  11  3  (4 x  1)( x  3)  0  x   , x  3 . All real numbers except -1/4, 3.
4
20)
f ( x) 
2