Sat 2015 Practice Test #1 Answer Explanations

Choice C is correct.
To determine which quadrant does not contain 
any solutions to the system of inequalities, graph the inequalities. 
Graph the inequality 
y
ƪ[
+ 1 by drawing a line through the 
y
LQWHUFHSWDQGWKHSRLQWDVVKRZQ7KHVROXWLRQVWR
this inequality are all points contained on and above this line. Graph 
_
1
the inequality 
y
> 2
[ơE\GUDZLQJDGDVKHGOLQHWKURXJKWKH
y
LQWHUFHSWơDQGWKHSRLQWDVVKRZQ7KHVROXWLRQVWR
this inequality are all points above this dashed line. 
x 
y 
II 
I 
IV
III 
y 
x
– 1 
y 
2
x
+ 1
> 
> 2 
1 
The solution to the system of inequalities is the intersection of 
the regions above the graphs of both lines. It can be seen that the 
solutions only include points in quadrants I, II, and III and do not 
include any points in quadrant IV. 
Choices A and B are incorrect because quadrants II and III contain 
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Choice D is incorrect because there are no solutions in quadrant IV. 
QUESTION 29 
Choice D is correct.
If the polynomial 
S
(
[
) is divided by 
[ơWKHUHVXOW
S
(
_
[
) 
= 
T
(
[
) + 
U
can be written as 
[ơ
[ơ
_ 
, where 
T
(
[
) is a polynomial and 
U
is the remainder. Since 
[ơLVDGHJUHHSRO\QRPLDOWKHUHPDLQGHU
is 
a real number. Hence, 
S
(
[
) can be written as 
S
(
[
) = (
[ơ
T
(
[
) + 
U
, 
where 
U
is a real number. It is given that 
S
ơVRLWPXVWEHWUXH
WKDWơ S
(
ơ
T
(
U 
T
(
U
= 
U
. Therefore, the remainder 
when 
S
(
[
) is divided by 
[ơLVơ
Choice A is incorrect because 
S
(
ơGRHV
not imply that 
S
&KRLFHV%DQG&DUHLQFRUUHFWEHFDXVHWKHUHPDLQGHUơRULWV
opposite, 2, need not be a root of 
S
(
[
). 
QUESTION 30