how do you do completing the square x^2-6x-7

Question

fwizbang · Answer

Start by looking at the term with one power of x.
Find its coefficient(-6)
Divide this number by 2(-3)
Square this(9)
Add and subtract the last answer to the polynomial(x^2-6x+9 -9 -7)
Since (x-3)^=x^2-6x+9, this last is just  (x-3)^2 -16 and you're done

anonymous · Answer

thank you so much for responding, but i lost you after (x^2-6x+9 -9 -7) if you can break it down a little maore that would be better

fwizbang · Answer

Once you've added and subtracted the 9, you have "completed" the square of x-3:
(x-3)*(x-3) = x*x -3*x - 3*x +3*3 = x^2-6x +9.

So,  x^2 -6x +9 -9 -7 = (x^2-6x+9) - (9+7) = (x-3)^2 - 16.

anonymous · Answer

okay now what if you have an integer in the A spot, for instance 3x^2=12x-15

fwizbang · Answer

Factor a three out of everything first, then the rest of the procedure is the same as before.

anonymous · Answer

okay thank you SO much