Math 147 - Assignment 1 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
1. Factoring by Grouping
The following expression can be completely factored as follows
x4 + 3x3 + 8x + 24 = x3 (x + 3) + 8(x + 3)
= (x3 + 8)(x + 3)
= (x3 + 23 )(x + 3)
= (x + 2)(x2 − 2x + 4)(x + 3)
Note this used the sum of cubes formula, A3 + B 3 = (A + B)(A2 − AB + B 2 ).
2. Rational Expressions
Preform the indicated operations and simplify the following expressions:
(a)
1
2
3
1
2
3
−
+ 2
=
−
+
2
2
x + 1 (x + 1)
x −1
(x + 1) (x + 1)
(x + 1)(x − 1)
(x + 1)(x − 1)
2(x − 1)
3(x + 1)
=
−
+
2
2
(x + 1) (x − 1) (x + 1) (x − 1) (x + 1)2 (x − 1)
(x2 − 1) − (2x − 2) + (3x + 3)
=
(x + 1)2 (x − 1)
x2 + x + 4
=
(x + 1)2 (x − 1)
(b)
1
(x+h)2
h
−
1
x2
=
=
=
=
=
=
x2
x2 (x+h)2
−
(x+h)2
x2 (x+h)2
h
x2 − (x + h)2 1
·
x2 (x + h)2
h
2
2
x − (x + 2xh + h2 )
hx2 (x + h)2
−2xh − h2
hx2 (x + h)2
h(−2x − h)
hx2 (x + h)2
−2x − h
x2 (x + h)2
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Math 147 - Assignment 1 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
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3. Domain of an Algebraic Expression
Find the domain of each of the following algebraic expressions.
(a)
1
1
−
1+x 2−x
Because we cannot divide by zero the domain is all real numbers such that
1 + x 6= 0
and
2 − x 6= 0
Thus we have that x 6= −1 and x 6= 2. Thus the domain is
(−∞, −1) ∪ (−1, 2) ∪ (2, ∞)
(b)
√
8 − 2x
Because square roots of negative numbers are not real, we want to ensure that the number under the root is not
negative. Thus we want
8 − 2x ≥ 0
8 ≥ 2x
4≥x
x≤4
Thus the domain is all numbers less than or equal to 4, or
(−∞, 4]
4. Equations
Solve the following equations for x.
(a)
x4 = 16x2 − 48
x4 − 16x2 + 48 = 0
(x2 )2 − 16(x2 ) + 48 = 0
(x2 − 4)(x2 − 12) = 0
x2 − 4 = 0
√
x2 = 4
x = ± 4 = ±2
or x2 − 12 = 0
or x2 = 12
√
√
orx = ± 12 = ±2 3
√
√
Thus the four solutions are x = 2, x = −2, x = 2 3 and x = −2 3
Math 147 - Assignment 1 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
(b)
√
5−x+1=x−2
√
5−x=x−3
√
( 5 − x)2 = (x − 3)2
5 − x = x2 − 6x + 9
0 = x2 − 5x + 4
0 = (x − 1)(x − 4)
x−1=0
x=1
or x − 4 = 0
or x = 4
√
√
If x = 1 then 5 − 1 + 1 = 4 + 1 = 3 while 1 − 2 = −1. So x = 1 is an extraneous solution.
√
√
If x = 4 then 5 − 4 + 1 = 1 + 1 = 2 while 4 − 2 = 2.
Thus x = 4 is the only solution.
(c)
x+1
x
−
=1
2x + 7 x + 3
x
x+1
(2x + 7)(x + 3)
−
= 1(2x + 7)(x + 3)
2x + 7 x + 3
x(x + 3) − (2x + 7)(x + 1) = (2x + 7)(x + 3)
(x2 + 3x) − (2x2 + 9x + 7) = (2x2 + 13x + 21)
−x2 − 6x − 7 = 2x2 + 13x + 21
0 = 3x2 + 19x + 28
0 = (x + 4)(3x + 7)
x+4=0
x = −4
or
3x + 7 = 0
−7
or x =
3
(d)
ax
−d=0
bx + c
ax
=d
bx + c
ax = d(bx + c)
ax = bdx + cd
ax − bdx = cd
(a − bd)x = cd
x=
cd
a − bd
Now this only works provided a − bd 6= 0 or a 6= bd (because of division by zero).
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Math 147 - Assignment 1 Solutions - Fall 2012 - BSU - Jaimos F Skriletz
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5. Dimensions of a Poster
A poster has a rectangular printed area 100 cm by 140 cm
and a blank strip of uniform width x around the edges as
shown.
The perimeter of the printed area is
P1 = 2(100) + 2(140) = 480
The perimeter of the full poster is
P2 = 2(100 + 2x) + 2(140 + 2x) = 480 + 8x
If the full perimeter is 1.5 times that of the printed area then
P2 = 1.5P1
480 + 8x = 1.5(480) = 720
8x = 240
x = 30
Thus the width of the blank strip is 30 cm.
6. Distance, Rate, Time
Lee lives 260 miles from a popular mountain retreat. On his way to do some mountain biking, Lee’s car had engine
trouble – forcing him to bike the rest of the way.
Let T1 and D1 be the time and distance Lee drove.
Let T2 and D2 be the time and distance Lee peddled his bike.
If Lee drove for 2 hours longer than he biked then T1 = T2 + 2.
If Lee drives at 60mph then D1 = 60T1 = 60(T2 + 2).
If Lee bikes at 10mph then D2 = 10T2 .
Thus since the total distance is 260 miles we have
D1 + D2 = 260
60(T2 + 2) + 10T2 = 260
60T2 + 120 + 10T2 = 260
70T2 = 140
140
=2
T2 =
70
Thus Lee spent T2 = 2 hours peddling to the mountain retreat.