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ok so my equation should look like \[1/4(4x^2+3(4)x+9)\]

and 3(4) can be written as 12 of course

you can also see if you need to factor anymore

right, so now my equation should be \[x^2+12+9\]

well you are missing 1/4( )
and a 4 in front of x^2
and a x next to the 12

\[\frac{1}{4}(4x^2+12x+9)\]

anyways let's look at the thing in ( )

can you find two numbers that multiply to be 4(9)=36
and add up to be 12

something that are multiples to 36 and add to 12?

yep

6 and 6

yes yes! :)

(6x+9) will equal 3x(2x+3)

we will bring down 1/4 later
\[2x(2x+3)+3(2x+3)\]
now one last step

oh ok, that makes sense!

I'm not sure how to do the last step

so you see (2x+3) in both terms
the terms being 2x(2x+3) and 3(2x+3)

what is left of the first term when factoring out the (2x+3)
is 2x
and of the second term is +3

2x(2x+3)+3(2x+3)
(2x+3)(2x+3)

don't forget to bring down the 1/4

so its just 1/4(2x+3)(2x+3)

or 1/4(2x+3)^2

or...

you could separate the fraction inside the ( )^2

but anyways yeah you are done

so its just \[(x+\frac{ 3 }{ 2 })^2\]

yep

Thank you so much, I really appreciate your help!

\[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]

oh yes! That's very true, we missed that. Oh well :)

no problem :)