facotr

• Feb 21st 2008, 04:41 AM
Graffitixgirl
facotr
what do you do with these? these are the only ones i dont get...
(p-5)^2
(x-y)^2
(d-2e)^2
(3p-5q)^2
• Feb 21st 2008, 04:46 AM
Charbel
Quote:

Originally Posted by Graffitixgirl
what do you do with these? these are the only ones i dont get...
(p-5)^2
(x-y)^2
(d-2e)^2
(3p-5q)^2

Um ok .. well it's the same as...

(n)^2 = n x n

sooo

(n+1)^2 = (n+1)(n+1)

glade to give the answers but try them yourself first=]

(p-5)^2 = p^2 -10p + 25
(x-y)^2 = x^2 - 2xy + y^2
(d-2e)^2 = d^2 - 4de + 4e^2
(3p-5q)^2 = 9p^2 - 30pq + 25q^2
• Feb 21st 2008, 04:47 AM
earboth
Quote:

Originally Posted by Graffitixgirl
what do you do with these? these are the only ones i dont get...
(p-5)^2
(x-y)^2
(d-2e)^2
(3p-5q)^2

I assume that you should expand the bracket.

If so there are 2 different ways to do it:

1. Re-write the square as a product and multiply using the distribution rule

\$\displaystyle (p-5)^2 = (p-5)(p-5)\$

2. All 4 terms belong to one binomial formula:

\$\displaystyle (a-b)^2 = a^2 - 2ab + b^2\$

Plug in your terms and values into this formula and you'll get the result immediately.