SOLUTION: Hello, Im trying to help my 15yr old son with his algebra homework. I can remember the basic Factorising and simplyfying but cant understand this question : show that (n+3)2

SOLUTION: Hello, Im trying to help my 15yr old son with his algebra homework. I can remember the basic Factorising and simplyfying but cant understand this question : show that (n+3)2 - n

Question 790022: Hello, Im trying to help my 15yr old son with his algebra homework. I can remember the basic Factorising and simplyfying but cant understand this question :
show that (n+3)2 - n(n-6) = 3(4n+3)
Heres what i calculate
(n+3)2 -n(n-6)
(n+3)(n+3) -2n+6n
n2+9+3n+3n
n2+9 n2+9-2n+6n = n2+9+4n
What am I missing?
chris.levett180@virginmedia.com
Many Thanks
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

I'm only going to alter the left side and will NOT change the right side.

(n+3)^2 - n(n-6) = 3(4n+3)

(n+3)(n+3) - n(n-6) = 3(4n+3)

n(n+3)+3(n+3) - n(n-6) = 3(4n+3)

n^2+3n+3n+9 - n^2+6n = 3(4n+3)

12n+9 = 3(4n+3)

3(4n+3) = 3(4n+3)

Since the last equation is true, this means that the original equation (n+3)^2 - n(n-6) = 3(4n+3) is true for all real numbers n.

Basically this means that we've just proven that both sides are identical to each other.
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