Download 1 я -8 2 я 4

8 ÿ -6 = -48
Chapter 6, Math 065
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topic. We do not promise that the problems below are problems that will appear on your exam in Math 065.
These problems have appeared on old exams.
-48
ÿ
-48
ÿ
-24
1
2
ÿ
ÿ
ÿ
3
4
6
-16
-12
-8
8 x2 - 13 x - 6
Factor out the common factor: 36 x5 y7 - 48 x8 y3
1.
36 x5 y7 - 48 x8 y3 = 12 x5 y3 I3 y4 - 4 x3 M
8 x2 + 3 x - 16 x - 6
I8 x2 + 3 xM + H-16 x - 6L
xH8 x + 3L -2 H8 x + 3L
H8 x + 3L Hx -2L
Factor completely: 25 x3 - 10 x2 - 35 x + 14
2.
25 x3 - 10 x2 - 35 x + 14
I25 x3 - 10 x2 M + H-35 x + 14L
5 x I5 x - 2M -7 H5 x - 2L
2
H5 x - 2L I5 x2 -7M
This
is
25 x2 - 16 = H5 xL2 - 42
= H5 x - 4L H5 x + 4L
a
difference
Factor: x2 - 7 x - 8
3.
ÿ
ÿ
This
-8
4
is
a
Factor completely: 3 x2 - 9 x - 12
3 Hx + 1L Hx -4L
3AIx3 - 2 x2 M + H-4 x + 8LE
3 Ix3 - 2 x2 - 4 x + 8M
3Ax2 Ix - 2M -4 Hx - 2LE
3 Hx - 2L Ix2 -4M
3 Hx - 2L Hx + 2L Hx - 2L
-24
-12
-8
-6
Solve: 3 x2 - 7 x = -4
10.
First, we get it equal to zero:
None of the pairs above add to -7, so the polynomial is prime.
3 x2 - 7 x = -4
+4 + 4
Factor: 8 x - 13 x - 6
2
3 x2 - 7 x +4 = 0
8 ÿ -6 = -48
1
2
3
4
6
-48
ÿ
-48
ÿ
-24
ÿ
ÿ
ÿ
-16
-12
-8
cubes:
3 x3 - 6 x2 - 12 x + 24
Factor: 4 x - 7 x + 6
6.
of
3 x2 - 9 x - 12
4 ÿ 6 = 24
-1
-2
-3
-4
sum
Factor completely: 3 x3 - 6 x2 - 12 x + 24
9.
2
24
ÿ
ÿ
ÿ
ÿ
27 x3 + 125 = H3 xL3 + 53
= H3 x + 5L I9 x2 - 15 x + 25M
3 Ix2 - 3 x - 4M
5.
squares:
= H3 x + 5L IH3 xL2 - H3 xL H5L + 52 M
x2 - 7 x - 8
Hx - 8L Hx + 1L
4.
of
Factor: 27 x3 + 125
8.
8
1
2
Factor: 25 x2 - 16
7.
Now we factor:
3 ÿ 4 = 12
-1
-2
-3
12
ÿ -12
ÿ -6
ÿ
-4
3 ÿ 4 = 12
12
ÿ -12
ÿ -6
-1
-2
-3
ÿ
3 x2 - 5 x - 2
3 x2 -6 x + 1 x - 2
I3 x2 - 6 xM + H1 x - 2L
-4
3 xHx - 2L + 1 Hx - 2L
Hx - 2L H3 x +1L
3 x2 - 7 x + 4 = 0
HeL
3 x2 - 4 x - 3 x + 4 = 0
I3 x2 - 4 xM + H-3 x + 4L = 0
x H3 x - 4L -1 H3 x - 4L = 0
H3 x - 4L Hx -1L = 0
3x-4=0
x-1=0
+4 + 4
+1 + 1
3x=4
x=1
3x
3
=
4
3
x=
11.
14 x3 y - 4 x2 y2 + 6 xy4
14 x3 y - 4 x2 y2 + 6 xy4
HbL
HcL
-1
-2
is
HfL
a
2 xyI7 x2 - 2 xy + 3 y3 M
1
2
3
4
36
ÿ
ÿ
ÿ
ÿ
36
18
12
9
6
ÿ
6
4 x2 + 6 x + 6 x + 9
t - 8 t - 48
I4 x2 + 6 xM + H6 x + 9L
2 xH2 x + 3L +3 H2 x + 3L
H2 x + 3L H2 x +3L
t2 - 8 t - 48
Ht - 12L Ht + 4L
x2 + x - 10
HgL
H2 x + 3L2
xy2 - 7 y2 - 4 x + 28
xy2 - 7 y2 - 4 x + 28
Ixy2 - 7 y2 M + H-4 x + 28L
y2 Ix - 7M -4 Hx - 7L
3 x2 - 5 x - 2
Hx - 7L Iy2 -4M
Hx - 7L Hy + 2L Hy - 2L
3 ÿ -2 = -6
-6
ÿ
ÿ
squares:
4 x2 + 12 x + 9
This polynomial is prime because none of the pairs add to get 1.
1
2
of
4 x2 + 12 x + 9
2
-10
ÿ 10
ÿ
5
HdL
9 y2 - 49 = H3 yL2 - 72
= H3 y - 7L H3 y + 7L
difference
4 ÿ 9 = 36
4
3
Factor each polynomial completely:
HaL
This
9 y2 - 49
-6
-3
12.
3 x2 - 5 x - 2
3 x2 -6 x + 1 x - 2
I3 x - 6 xM + H1 x - 2L
2
3 xHx - 2L + 1 Hx - 2L
Hx - 2L H3 x +1L
Solve each equation:
HaL
x2 + 5 x = 14
x2 + 5 x = 14
-14 - 14
x2 + 5 x - 14 = 0
Hx + 7L Hx - 2L = 0
x+7=0 x-2=0
-7 - 7
+2 + 2
x = -7
x=2
x2 + 5 x = 14
-14 - 14
HbL
x2 + 5 x - 14 = 0
Hx + 7L Hx - 2L = 0
x+7=0 x-2=0
-7 - 7
+2 + 2
x = -7
x=2
5 t3 - 45 t = 0
5 t3 - 45 t = 0
5 tIt2 - 9M = 0
5 tHt + 3L Ht - 3L = 0
5t =0
5t
5
=
0
5
t=0
13.
t+3=0 t-3=0
-3 - 3
+3 + 3
t = -3
t=3
The sum of two numbers is 17 and the sum of their squares is 145. Find the numbers.
x= first number
y= second number
We get the system of equations:
x + y = 17
x2 + y2 = 145
Using the method of substitution, we will solve for x in the first equation:
x + y = 17
-y
-y
x = 17 -y
And substitute it into the second equation and solve:
H17 - yL2 + y2 = 145
289 -17 y -17 y +y2 + y2 = 145
2 y2 - 34 y + 289 = 145
-145 - 145
2 y2 -34 y + 144 = 0
2 Iy2 - 17 y + 72M = 0
y-8=0
+8 + 8
y=8
2 Hy - 8L Hy - 9L = 0
y-9=0
+9 + 9
y=9
We check to see that 8 + 9 = 17 and 82 + 92 = 64 + 81 = 145