I need help on my homework. College algebra.

## Write p(x)=3x^2+4x-1 in p (x)=a(a-h)^2+k form using complete the square method

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# 2 Answers

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}-7/33x^2 + 4x -1 = 3(x^2 + 4x/3 - 1/3) = 3(x^2 + 4x/3 + 4/9 - 7/9) = 3(x + 2/3)^2 - 3(7/9) =

3(x + 2/3) - 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.Comment