Basic Numeracy
BNU1501
Department of Decision Sciences
Exam May/June 2019
Question 1
Simplify the following expression as far as possible:
6𝑎2 − 9𝑎𝑏 + 2𝑏 2 − 3(𝑏𝑎 − 𝑏 2 )
2 2 2 2
[1] 6𝑎 − 𝑏 + 12𝑎 𝑏
[2] 6𝑎2 − 12𝑎𝑏 + 5𝑏 2
[3] 6𝑎2 + 3𝑏 2 − 12𝑏𝑎
[4] 6𝑎2 − 12𝑎𝑏 − 𝑏 2
Answer:
6𝑎2 − 9𝑎𝑏 + 2𝑏 2 − 3(𝑏𝑎 − 𝑏 2 )
Reorder the terms:
6𝑎2 − 9𝑎𝑏 + 2𝑏 2 − 3(𝑎𝑏 − 3𝑏 2 )
Distribute 3 through the parentheses:
6𝑎2 − 9𝑎𝑏 + 2𝑏 2− − 3𝑎𝑏 + 3𝑏 2
Collect like terms:
6𝑎2 − 12𝑎𝑏 + 2𝑏 2 + 3𝑏 2
6𝑎2 − 12𝑎𝑏 + 5𝑏 2
BNU1501
Department of Decision Sciences
Exam May/June 2019
Question 1
Simplify the following expression as far as possible:
6𝑎2 − 9𝑎𝑏 + 2𝑏 2 − 3(𝑏𝑎 − 𝑏 2 )
2 2 2 2
[1] 6𝑎 − 𝑏 + 12𝑎 𝑏
[2] 6𝑎2 − 12𝑎𝑏 + 5𝑏 2
[3] 6𝑎2 + 3𝑏 2 − 12𝑏𝑎
[4] 6𝑎2 − 12𝑎𝑏 − 𝑏 2
Answer:
6𝑎2 − 9𝑎𝑏 + 2𝑏 2 − 3(𝑏𝑎 − 𝑏 2 )
Reorder the terms:
6𝑎2 − 9𝑎𝑏 + 2𝑏 2 − 3(𝑎𝑏 − 3𝑏 2 )
Distribute 3 through the parentheses:
6𝑎2 − 9𝑎𝑏 + 2𝑏 2− − 3𝑎𝑏 + 3𝑏 2
Collect like terms:
6𝑎2 − 12𝑎𝑏 + 2𝑏 2 + 3𝑏 2
6𝑎2 − 12𝑎𝑏 + 5𝑏 2