Level J Algebra: Sample Questions Exam
1314-MJ1E3 -V1
Level J Algebra First Two Weeks
Sample Questions Exam
1. Solve, x ( x + 9 ) = 0.
Reference:
Algebra 1, Part 2, Chapter 8.7, Activity (1a), Page 78
Solution:
x = 0 or x = −9
2. Factor, x 2 + 5 x + 2 xy + 10 y. .
Reference:
Algebra 1, Part 2, Chapter 8.6.1 Activity 1 #(d), Page 73
Solution: x 2 + 5 x + 2 xy + 10 y = x( x + 5) + 2 y ( x + 5) = ( x + 5)( x + 2 y )
3. Factor, a 2 + 4 a − 32.
Reference:
Algebra 1 Part 2, Chapter 8.4.3, Activity (a), Page 61
Solution:
a 2 + 4a − 32 = ( a + 8)( a − 4 )
4. Factor, b 2 + 14b + 21.
Reference:
Algebra 1 Part 2, Chapter 8.4.1, Activity (2a), Page 59
Solution:
b 2 + 14b + 21 = ( b + 3)( b + 7 )
5. Factor the GCMF: 24ab + 8a.
Reference:
Algebra 1 Part 2, Chapter 8.2.1, Activity (2c), Page 50
Solution:
24ab + 8a = 8a ( 3b + 1)
6. Find the GCMF of 24ab and 8a.
Reference:
Algebra 1, Part 2, Chapter 8.1.2 Activity #1(a), Page 49
Solution:
24ab = 23 × 3ab and 8a = 23 a.
GCMF (24ab,8a ) = 8a
SABIS® Proprietary
Page 1 of 7
1314-MJ1E3 -V1
7. Simplify,
Reference:
Level J Algebra First Two Weeks
−18 x3 − 63x 2
.
3x
Algebra 1 Part 2, Chapter 7.7.1, Activity (a), Page 39
Solution:
−6 x 2 − 21x
8. Divide. Give the quotient and the remainder. Check the answer.
(2x
Reference:
4
+ 5 x3 − 6 x 2 ) ÷ ( x 2 + 3x )
Algebra 1 Part 2, Chapter 7.7.2, Activity 1 (a), Page 41
Solution:
2 x2 − x − 3 x 2 + 3x 2 x 4 + 5 x3 − 6 x 2
−
2x 4 + 6 x 3
− x3 − 6 x 2
+
− x3 − 3x
2
− 3x 2
+
− 3x 2 − 9 x
9x
Quotient =2 x 2 − x − 3, Remainder = 9x
Check:
( 2x
2
− x − 3 )( x 2 + 3 x ) + 9 x = 2 x 4 + 5 x 3 − 6 x 2 .
9. Simplify, ( −3) 2 ⋅ 3−4.
Reference:
Algebra 1 Part 2, Chapter 7.6.3, Activity (1e), Page 38
Solution:
1
9
SABIS® Proprietary
Page 2 of 7
1314-MJ1E3 -V1
Level J Algebra First Two Weeks
10. Simplify. ( Assume all variables are different from zero ).
−4 y
12 x3 y
Level J Algebra First Two Weeks
Sample Questions Exam
1. Solve, x ( x + 9 ) = 0.
Reference:
Algebra 1, Part 2, Chapter 8.7, Activity (1a), Page 78
Solution:
x = 0 or x = −9
2. Factor, x 2 + 5 x + 2 xy + 10 y. .
Reference:
Algebra 1, Part 2, Chapter 8.6.1 Activity 1 #(d), Page 73
Solution: x 2 + 5 x + 2 xy + 10 y = x( x + 5) + 2 y ( x + 5) = ( x + 5)( x + 2 y )
3. Factor, a 2 + 4 a − 32.
Reference:
Algebra 1 Part 2, Chapter 8.4.3, Activity (a), Page 61
Solution:
a 2 + 4a − 32 = ( a + 8)( a − 4 )
4. Factor, b 2 + 14b + 21.
Reference:
Algebra 1 Part 2, Chapter 8.4.1, Activity (2a), Page 59
Solution:
b 2 + 14b + 21 = ( b + 3)( b + 7 )
5. Factor the GCMF: 24ab + 8a.
Reference:
Algebra 1 Part 2, Chapter 8.2.1, Activity (2c), Page 50
Solution:
24ab + 8a = 8a ( 3b + 1)
6. Find the GCMF of 24ab and 8a.
Reference:
Algebra 1, Part 2, Chapter 8.1.2 Activity #1(a), Page 49
Solution:
24ab = 23 × 3ab and 8a = 23 a.
GCMF (24ab,8a ) = 8a
SABIS® Proprietary
Page 1 of 7
1314-MJ1E3 -V1
7. Simplify,
Reference:
Level J Algebra First Two Weeks
−18 x3 − 63x 2
.
3x
Algebra 1 Part 2, Chapter 7.7.1, Activity (a), Page 39
Solution:
−6 x 2 − 21x
8. Divide. Give the quotient and the remainder. Check the answer.
(2x
Reference:
4
+ 5 x3 − 6 x 2 ) ÷ ( x 2 + 3x )
Algebra 1 Part 2, Chapter 7.7.2, Activity 1 (a), Page 41
Solution:
2 x2 − x − 3 x 2 + 3x 2 x 4 + 5 x3 − 6 x 2
−
2x 4 + 6 x 3
− x3 − 6 x 2
+
− x3 − 3x
2
− 3x 2
+
− 3x 2 − 9 x
9x
Quotient =2 x 2 − x − 3, Remainder = 9x
Check:
( 2x
2
− x − 3 )( x 2 + 3 x ) + 9 x = 2 x 4 + 5 x 3 − 6 x 2 .
9. Simplify, ( −3) 2 ⋅ 3−4.
Reference:
Algebra 1 Part 2, Chapter 7.6.3, Activity (1e), Page 38
Solution:
1
9
SABIS® Proprietary
Page 2 of 7
1314-MJ1E3 -V1
Level J Algebra First Two Weeks
10. Simplify. ( Assume all variables are different from zero ).
−4 y
12 x3 y