1. ## Question on algebra

There are many solution but i find the one which i could understand was:

Question: Factorise 121 - (p+2)Square

Originally Posted by skeeter
$11^2 - (p+2)^2 = [11 - (p+2)][11 + (p+2)] = (9-p)(13+p)$
But, what i don't understand is this solution:

Question: Factorise 121 - (p+2)Square

Answer/Solution: 121 - (p)Square + 2(p)(2) + 2Square
= 121 - (p)Square + 4p + 4
= 117 - (p)Square + 4p
= (9-p)(13+p)
I mean, how do you know if it's "(9-p)(13+p)" ?

Thank you so much for your time

2. You can expand as I showed you earlier. Expand both expressions and you'll get the same final expression.

3. Originally Posted by Unknown008
You can expand as I showed you earlier. Expand both expressions and you'll get the same final expression.
Hi, but how can you expand from "117 - (p)Square + 4p" onwards?
Can you show it to me?

4. $\displaystyle = -(p^2 - 4p - 117)$.

Now what numbers multiply to give $\displaystyle -117$ and add to give $\displaystyle -4$?

5. Originally Posted by Prove It
$\displaystyle = -(p^2 - 4p - 117)$.

Now what numbers multiply to give $\displaystyle -117$ and add to give $\displaystyle -4$?
OOOOOO, so thats how you do it, i was searching for this.
So,
-13 x 9 = -117
9 - 13 = -4

Thanks so much !

6. Well, what I initially meant is:

11^2 - (p+2)^2 = 121 - p^2 - 4p - 4 = 117 - p^2 - 4p

And taking the other,

(9-p)(13+p) = 117 - p^2 - 13p + 9p = 117 - p^2 - 4p

Which is the same as before. Hence, both are equivalent, meaning that the answer (9-p)(13+p) is correct.