Completing the square? to get the basic shape of 1+5g+4g^2

[kmce]

I need to complete the square to get the basic shape of a function.

The Answer i am meant to come out with is

4(g+5/8)² -9/16

However I cannot get to this answer, as close I can get it is

4(g+5/8)² -9/64

So the question is, complete the square to get the basic shape of 1+5g+4g²


The way i have been doing it is

4[g²+5/4g+1/4]
4[(g+5/8)² + 1/4]
4[(g+5/8)² + 1/4 - ((5/8)²]
Therefore
4(g+5/8)² -9/64

Can anyone help with where i am going wrong here, and why i cant get the 9/16. If i dont half the 5/4 again i get the 9/16, but i obviously wouldnt get the 5/8 then

 

[MarkFL]

I would begin by stating:

\(\displaystyle f(g)=4g^2+5g+1\)

Now, factor out the coefficient of the squared term from the first two terms only:

\(\displaystyle \displaystyle f(g)=4\left(g^2+\frac{5}{4}g\right)+1\)

Within the brackets add the square of one-half the coefficient of the linear term, and then on the outside, subtract this same amount so that the net result is to add zero:

\(\displaystyle \displaystyle f(g)=4\left(g^2+\frac{5}{4}g+ \left(\frac{1}{2}\cdot\frac{5}{4}\right)^2\right)+ 1-4\left(\frac{1}{2}\cdot\frac{5}{4}\right)^2\)

\(\displaystyle \displaystyle f(g)=4\left(g^2+\frac{5}{4}g+ \left(\frac{5}{8}\right)^2\right)+1-4\cdot\frac{25}{64}\)

\(\displaystyle \displaystyle f(g)=4\left(g+\frac{5}{8}\right)^2+1-\frac{25}{16}\)

\(\displaystyle \displaystyle f(g)=4\left(g+\frac{5}{8}\right)^2-\frac{9}{16}\)

 

[HallsofIvy]

You have not distributed the leading "4" correctly. You have, correctly, \(\displaystyle 4[(g+ 5/8)^2+ 1/4- (5/8)^2]\). That is \(\displaystyle 4[(g+ 5/8)^2- 9/64]\) but now, distribute the 4: \(\displaystyle 4(g+ 5/8)^2- 4(9/64)= 4(g+ 5/8)^2- 9/16\).

 

[kmce]

Ah ok, so you need to multiply the outside again by the the common factor.

Thanks a lot