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I know how to complete the square with positive $x$ coefficients but how do you complete the square with negative $x$ coefficients?

For example: \begin{align*} f(x) & = x^2 + 6x + 11 \\ & = (x^2 + 6x) + 11 \\ & = (x^2 + 6x + \mathbf{9}) + 11 - \mathbf{9} \\ & = (x+3)^2 + 2. \end{align*} For positive you would add 9 inside the parenthesis and subtract 9 outside. However, I get the wrong answer when I do it with a negative coefficient. Do you do the same thing when there is a negative coefficient in front of $x^2$ or is it the other way around? (subtract inside parenthesis and add outside of parenthesis).

Here is an example of a negative coefficient: $f(x) = -3x^2 + 5x + 1.$ I tried to solve this and entered in my answer online but it was wrong.

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    Oops @David, you omitted an exponent of 2 in the $3x$ term.2012-09-21

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$ \begin{split} f(x) &= -3(x^2-5x/3 - 1/3)\\ &= -3( x^2 - 5x/3 + (5/6)^2 - (5/6)^2 - 1/3)\\ &= -3( (x - 5/6)^2 - 25/36 - 12/36) \\ &= -3(x - 5/6)^2 -3(- 25/36 - 12/36) \\ &= -3(x - 5/6)^2 + 37/12. \end{split}$

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    A little more explanation for this would have been nice, but thank you.2015-04-15
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$(2ax+b)^2 = 4a^2x^2 + 4abx + b^2$

So, multiplying by $4a$, or in this case $-4a$, will avoid fractions until the last step.

$\begin{align} f(x) &= -3x^2 + 5x + 1 \\ -12f(x) &= 36x^2 -60x - 12 \\ -12f(x) &= 36x^2 - 60x + 25 - 25 - 12 \\ -12f(x) &= (6x-5)^2 - 37 & \left(b = \frac{4ab}{4a} = \dfrac{-60}{12}=-5 \right)\\ f(x) &= -\dfrac{1}{12}(6x-5)^2 + \dfrac{37}{12} \end{align}$