Question 846843: Trying to understand how
n^2+5n+100-[(n-1)^2+5(n-1)+100]= 2n+4
Need to see steps to understand please
Many thanks to you!
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! What can you do inside parentheses (or other groupings) first?
n^2+5n+100-[(n-1)^2+5(n-1)+100] ----------- See distributive property possible?
n^2+5n+100-((n-1)^2+5n-5+100)
n^2+5n+100-((n-1)^2+5n+95)
Anything more inside parentheses?
n^2+5n+100-(n^2-2n+1+5n+95)
and still working inside parentheses,
n^2+5n+100-(n^2+3n+96)
Next, understand you are subtracting EVERYTHING inside the parenthesized expression:
n^2+5n+100-n^2-3n-96
Obviously, combine like-terms:
n^2-n^2+5n-3n+100-96
2n+4
|
|
|
