p2+10d+7
p2 + 3p = p (p + 3)
p2 + 2pq + q2 = 1q2 + 2pq + (p2 - 1) = 0q = 1/2 [ -2p plus or minus sqrt( 4p2 - 4p2 + 4 ) ]q = -1 - pq = 1 - p
p2
P2 + 13p - 30 = 0 Answer: p= -15, p = 2
p2+10d+7
p2+2pq+q2=1
p2 + 3p = p (p + 3)
p2 + 2pq + q2 = 1q2 + 2pq + (p2 - 1) = 0q = 1/2 [ -2p plus or minus sqrt( 4p2 - 4p2 + 4 ) ]q = -1 - pq = 1 - p
rutherford
p2
Even
Assignment. Eg: void *p1, *p2; p2= p1;
p2 + p2 + p3 = 2p2 + p3 because you can add the two variables that match, while you leave the different variable alone.
p2 - 2p + 2 can be factored as (p - 1)(p - 1)which can be written as ( p - 1)2.
(p=-15)(p=1)
P2 + 13p - 30 = 0 Answer: p= -15, p = 2