ALGEBRA I- MATH 006
Practice for FINAL EXAM
SOLUTIONS
1. Factor completely.
(a) 2x3 + 3x2 − 2x − 3
(b) 2x3 + 18x2 + 28x
Solution:
When in doubt, try grouping!
(a) 2x3 + 3x2 − 2x − 3 = x2 (2x + 3) − (2x + 3)
= (x2 − 1)(2x + 3)
= (x − 1)(x + 1)(2x + 3)
(b) 2x3 + 18x2 + 28x = 2x(x2 + 9x + 14)
= 2x(x + 7)(x + 2)
2. Perform the indicated operation and simplify. Leave
your answer in factored form.
x+4 x−3
−
x
x
− x2 −2x−24
(b) x−2x+1x+1
(a) x2 −7x+6
Solution:
x
x
(a) (x−1)(x−6)
− (x+4)(x−6)
(x−1)
(x+4)
x
(x+4) · (x−1)(x−6) − (x−1)
x2 +4x−(x2 −x)
= (x−1)(x−6)(x+4)
5x
= (x−1)(x−6)(x+4)
x+4 x−3
(x−2)(x+1) x−2 − x+1
·
(x−2)(x+1)
x+1
(x+4)(x+1)−(x−3)(x−2)
=
(x+1)2 (x−2)
10x−2
= (x+1)2 (x−2)
2(5x−1)
= (x+1)
2 (x−2)
=
(b)
1
·
x
(x+4)(x−6)
3. Given that x and y are positive, simplify
1
(243x5 y) 4
1
(3xy) 4
Write your answer so that each variable appears only
once.
Solution:
1
1 5 1
4 4 4
(243x5 y) 4
= 2431 x1 y1
1
(3xy) 4
34 x4 y4
1
4
=( 243
31 ) ·
5
1
1
1
x( 4 − 4 ) y ( 4 − 4 )
=81 4 x
=3x
4. A total of $10,000 is to be shared between Ben and
Jerry. Jerry is to receive 3 times as much as Ben.
How much will each receive?
Solution:
Let B=Ben and J=Jerry. Then
B + J = 10000 and J = 3B. Hence
B + 3B = 10000 and B = 2500
J = 3(2500) and J = 7500
5. Find all real solutions, if any, of each equation.
(a) 4x2 − 6x − 9 = 0
√
(b) x = 2 x − 1
(c) 2x3 + 5x2 − 8x − 20 = 0
(d) 5x+3 =
2
1
5
x
(e) 4x = 2
(f) e5x = e3−x
Solution:
2
(a) 4x2 − 6x
√ − 9 = 0, Use the quadratic formula.
6±
(−6)2 −4(4)(−9)
x= √
8
6± 180
x + 8√
x = 3±34 5
√
(b) x = 2 √
x−1
2
x = (2 x − 1)2
x2 = 4(x − 1)
x2 − 4x + 4 = 0
(x − 2)2 = 0
x=2
(c) 2x3 + 5x2 − 8x − 20 = 0, When in doubt, ...
(2x3 + 5x2 ) − (8x + 20) = 0
x2 (2x + 5) − 4(2x + 5) = 0
(x2 − 4)(2x + 5) = 0
(x − 2)(x + 2)(2x + 5) = 0
x = 2, x = −2, x = −5
2
(d) 5x+3 = 15 , Use Ax = Ay −→ x = y
5x+3 = 5−1 −→ x + 3 = −1
x = −4
2
(e) 4x = 2x
2
(2 · 2)x = 2x
2
2
2x 2x = 2x
2
22x = 2x −→ 2x2 = x
2x2 − x = 0
x(2x − 1) = 0
x = 0, x = 12
(f) e5x = e3−x −→ 5x = 3 − x
6x = 3
x = 12
3
6. Solve each inequality, and give the solution in interval notation. Also, graph the solution set (depending
on your printer, you may not notice a difference between the closed and open dots).
(a) x(9x − 5) ≤ (3x − 1)2
(b) |2 − 3x| > 1
(c) |4x − 3| ≤ 1
(d) 4x2 < 13x − 3
Solution:
(a) x(9x − 5) ≤ (3x − 1)2
9x2 − 5x ≤ 9x2 − 6x + 1
−5x ≤ −6x + 1
x ≤ 1 or (−∞, 1]
−1
0
1
Figure 1: Problem 6a: Solution Set
(b) |2 − 3x| > 1
2 − 3x < −1 or 2 − 3x > 1
−3x < −3
−3x > −1
x>1
x < 13
or (−∞, 13 ) ∪ (1, ∞)
−1
0
1
3
1
Figure 2: Problem 6b: Solution Set
4
(c) |4x − 3| ≤ 1
−1 ≤ 4x − 3 ≤ 1
2 ≤ 4x ≤ 4 h i
1
1
2 ≤ x ≤ 1 or 2 , 1
−1
0
1
1
2
Figure 3: Problem 6c: Solution Set
(d) 4x2 < 13x − 3, nonlinear inequality
4x2 − 13x + 3 < 0
(4x − 1)(x − 3) < 0
Zeros: x = 41 , x = 3
Let f (x) = (4x − 1)(x − 3)
Region Test Point(t) f (t) f (t) < 0?
(−∞, 14 ) 0
3
No
1
( 4 , 3)
1
−6 Yes
(3, ∞)
4
15
No
1
1
Solution: 4 < x < 3 or ( 4 , 3)
−1
0
1
4
1
3
Figure 4: Problem 6d: Solution Set
7. Find the distance between the points (3, −4) and
(5, 4).
Solution:
q
d = q(x2 − x1 )2 + (y2 − y1 )2
d = √(5 − 3)2 + (4 + 4))2
d = √4 + 64 √
d = 68 = 2 17
5
8. Verify that the triangle with vertices A = (−5, 3),
B = (6, 0), and C = (5, 5) is a right triangle.
Solution:
2
Method 1: qUse Distance Formula √
and c2 = a2 +
√b
d(A, B) = q(6 + 5)2 + (0 − 3)2 = 121 + 9 = 130
√
√
d(A, C) = q(5 + 5)2 + (5 − 3)2 = 100 + 4 = 104
√
√
d(B, C) = (5 − 6)2 + (5 − 0)2 = 1 + 25 = 26
d(A, B)2 = 130 = 104 + 26 = d(A, C)2 + d(B, C)2
Method 2: Use Perpendicular Lines
−3
mAB = 0−3
6+5 = 11
1
mAC = 5−3
5+5 = 5
mBC = 5−0
5−6 = −5
AC and BC are perpendicular. It is a right triangle.
9. Find an equation for the line with slope − 23 and containing the point (1, −1).
Solution:
y − y1 = m(x − x1 )
y + 1 = − 32 (x − 1)
y = − 23 x − 13
6
10. A line has equation 7x + 2y = 21.
(a) Find the slope and y-intercept of the line.
(b) Graph the line.
Solution:
2y = −7x + 21
y = − 72 x + 21
2
7
m = − 2 and y-intercept=
y
21
2
12
11
10
9
8
7
6
5
4
3
2
1
0
−3 −2 −1
−1 0
1
2
3
4
x
−2
−3
Figure 5: y = − 27 x +
7
21
2
11. A circle has equation x2 + y 2 − 2x − 4y − 4 = 0.
(a) Find the center and radius of the circle.
(b) Graph the circle.
Solution:
(x2 − 2x) + (y 2 − 4y) = 4
(x2 − 2x + 1) + (y 2 − 4y + 4) = 4 + 1 + 4
(x − 1)2 + (y − 2)2 = 9
Center= (1, 2), r = 3
y
6
5
(1,5)
4
3
(-2,2)
2
(4,2)
(1,2)
1
−3
−2
O
−1
−1
1
2
3
(1,-1)
−2
Figure 6: (x − 1)2 + (y − 2)2 = 32
8
4
5
x
12. Consider the function:


f (x) = 
x+3
if x < −2
−2x − 3 if x ≥ −2
(a) Graph the function.
(b) Find the range of the function.
Solution:
y
f (x) = x + 3
−6
−5
−4
2
1
−3
−2
O
−1
−1
1
−2
−3
−4
−5
−6
Figure 7: (b) Range: (−∞, 1]
9
2
3
4
x
f (x) = −2x − 3
13. Sketch the graph of the function f (x) = x2 − 2x and
find its range.
Solution:
−b
Vertex= ( −b
2a , f ( 2a )) = (1, f (1)) = (1, −1)
x-intercepts: x2 − 2x = 0 or x(x − 2) = 0; x = 0 and
x=2
y-intercepts: y = 0
y
10
9
8
7
6
5
4
3
2
1
(0,0)
(2,0)
0
−3 −2 −1
−1 0
−2
1
2
3
4
5
(1,-1) x
−3
Figure 8: f (x) = x2 − 2x, Range: [−1, ∞)
14. Let f (x) =
√
x + 1 and g(x) = 3x.
(a) Find (f ◦ g)(1).
(b) Find (g ◦ f )(1).
Solution:
q
√
(a) (f ◦ g)(x) = q
f (g(x)) = g(x) + 1 = 3x + 1.
√
(f ◦ g)(1) = 3(1) + 1 = 4 = 2
√
(b) (g ◦ f )(x) = g(f
(x))
=
3f
(x)
=
3
x + 1.
√
√
(g ◦ f )(1) = 3 1 + 1 = 3 2
10
15. Find the amount that results from each investment.
(a) $500 invested at 8% compounded monthly after
a period of 4 years.
(b) $500 invested at 8% compounded continuously
after a period of 4 years.
Solution:
(a) Use A = P ·(1+ nr )nt where A =amount, P =principal,
r =rate, n =number of compoundings, and t =number
of years.
(12)(4)
A = 500 · (1 + .08
= $687.83
12 )
(b) Use A = P ert where e ≈ 2.71828 is the natural
exponential base.
A = P ert = 500e(.08)(4) = $688.56
11