p^2 + 2pr + r^2
2006-09-12 15:22:19 · 8 answers · asked by J' K '06 1 in Science & Mathematics ➔ Mathematics
use the rule that (a+b)^2 = a^2 + 2ab + b^2. Thus, you would get p^2 + 2pr + r^2 = (p+r)^2. Get it?
2006-09-12 15:57:01 · answer #1 · answered by Linda O'Chuffy 2 · 0⤊ 0⤋
simple....(p+r)(p+r) = p^2 + 2pr + r^2
2006-09-12 22:26:13 · answer #2 · answered by alrivera_1 4 · 0⤊ 0⤋
Everyone is giving you the answer but no one is showing you how to get (p+r)(p+r).
first find the multiplyers of each variable. In this case it is understood that they are each 1. 1 and 1 is the only possibility that you have so you have only one choice to put into your parenthesis as in this (p r)(p r). Now all you need to determine is the choice of sign. You do this by looking at the first sign in your equation. In this case it is a '+' sign so you know that the signs will be the same. The second sign in your equation is '+' so you know that they will both be a '+' sign. So you have (p + r)(p + r).
2006-09-13 00:22:15 · answer #3 · answered by closetcoon_fan 5 · 0⤊ 0⤋
p^2 + 2pr + r^2
= p r pr
p r pr
= p+r
p+r
=(ptr)(p+r)
2006-09-13 03:36:52 · answer #4 · answered by ayie 2 · 0⤊ 0⤋
(p+r)^2
2006-09-12 22:25:11 · answer #5 · answered by rwbblb46 4 · 0⤊ 0⤋
It is a square binomial.
First the square of the first term, then the square root of the third term with the sign of the second term. Everything elevated to the second potence.
(p+r)^2
2006-09-12 22:26:34 · answer #6 · answered by daeylcq 2 · 0⤊ 0⤋
ANS : (p + r)(p + r)
2006-09-12 22:59:54 · answer #7 · answered by Sherman81 6 · 0⤊ 0⤋
(p+r)(p+r)
2006-09-12 22:30:54 · answer #8 · answered by nondescript 7 · 0⤊ 0⤋