Completing the square with a coefficient on x²

[Math Is Hard]

Hello,
I was working on a lengthy problem and got stuck on something simple. I have a quadratic expression 4(x^2) - 4x + 3 that I should have been able to complete the square on to rewrite as (2x - 1)^2 + 2
My knowledge of applying this technique when a coefficient is attached to the leading term is rusty so I got out my old algebra books. The solution they offered was to set 4(x^2) - 4x + 3 = 0
and then divide each side by four. With the coefficient removed, then the square can be completed the usual way. But this isn't giving me what I need at all.
Any advice on this technique?

Thanks mucho!

 

[quantumdude]

When you set aside the quadratic and linear terms, you can factor by grouping as in the following example:

3x²-12x+2
(3x²-12x)+2
3(x²-4x)+2

Note that I have factored the 3 out of the first two terms only. Now, I complete the square inside the parenteses.

3(x²-4x+4[/color])+2-12[/color]

Note the parts in red[/color]. I added 4 to the expression in parentheses, but that 4 was multiplied by 3 to make 12. So, to keep the expression equal to my original expression, I subtract 12 (not 4) from it.

Finally, I factor the perfect square in parentheses and collect terms.

3(x-2)²-10

edit: typo

 

[Math Is Hard]

Thanks, Tom. Glad to see you are still on the homework help boards.