I need help on my homework. College algebra.

p(x) = 3x

^{2}+ 4x -1p(x) - a(a-h)

^{2}- k ( think you mean p(x) = a(x+d)^{2}+ eSince in our eqn the coefficient of the x

^{2}term is not 1, we need to divide out the 3 by multiplying by 3/3p(x) = (3x

^{2}+ 4x - 1)3/3p(x) = 3( x

^{2+}4x/3 - 1/3)now focusing on the 4x/3 term, if we divide the coeffieceint of that term 4/3 by 2 we get 2/3

squaring it we get (2/3)

^{2}= 4/9 which is not = to -1/3if we add and subtract the 4/9 to the our equation with in the parathesis we are keeping it the same eqn

p(x) = 3(x

^{2}+ 4x/3 +4/9 -1/3 - 4/9)p(x) = 3[(x

^{2}+4x/3 +4/9) -1/3 -4/9]not x

^{2}+4x/3 +4/9 can be written as (x+2/3)^{2}p(x) = 3[(x+2/3)

^{2}- 3/9 -4/9]p(x) = 3[(x+2/3)

^{2}- 7/9]not totally in the form yet, we need to remove the -7/9 term from the paranthesis. Do this by multiplying by the 3 using distributive property

p(x) = 3(x+2/3)

^{2}-3(7/9)p(x) = 3(x+2/3)

^{2}-7/3
## Comments

^{2}- 7/3^{2}- k, I think you meant to say p(x) = a(x-h)^{2}+ k.^{2}that would give you a^{3}+ ah^{2}which is something else entirely.