Multiply 18 and 1 which remains as 18, now you are looking for two numbers that add to 9 and when multiplied together equal 18. The two numbers are 6 and 3.
= (18q+6)(18q+3)/18
*factor what you can out of everything*
=6(3q+1) 3(6q+1) /18
The 6 and 3 that are factored out equal 18 when multiplied out and it becomes 18/18 so it cancels it self out. Your final answer is:
(3q+1)(6q+1)
Sorry if I didn't explain it that well I'm new to this, hope that helped!
Answers & Comments
18q^2+9q+1
This is how I'd factor it:
Multiply 18 and 1 which remains as 18, now you are looking for two numbers that add to 9 and when multiplied together equal 18. The two numbers are 6 and 3.
= (18q+6)(18q+3)/18
*factor what you can out of everything*
=6(3q+1) 3(6q+1) /18
The 6 and 3 that are factored out equal 18 when multiplied out and it becomes 18/18 so it cancels it self out. Your final answer is:
(3q+1)(6q+1)
Sorry if I didn't explain it that well I'm new to this, hope that helped!
It’s not a difference of squares considering the only perfect square is the 9. You factor this expression by grouping.
Given a=18, b=9, and c=1, what are two factors of a*c (or 18*1) that add up to make b or 9? That would be 6 and 3. Replace 9q with 6q+3q.
18(q^2)+6q+3q+1
Group the first two terms together, and factor them.
Group the second two terms together, and factor that if possible.
[18(q^2)+6q] + [3q+1]
6q[3q+1] + 1[3q+1]
You can then factor that expression into this:
(6q+1)(3q+1)