Can anyone work these questions out for me and tell me how you work it out?

Question

Please!
1. x^2 - 6x -5 = (x - a)^2 + B
Find the values of a and B

2. Simplify 2x^2 - 2x -12 / 2x^2 - 18

Any help is greatly appreciated

Thanks everyone :)

Answer 1

1. Look at x^2 - 6x
you want to put it in (x-a)^2 form, and that is x^2 - 2ax + a^2
-2ax = -6x
a = 3
(x-3)^2, but now what else do you need?
(x-3)^2 = x^2 -6x + 9
but you want -5 at the end, so
(x-3)^2 - 14

2. I assume the top and bottom are in parentheses
2(x-3)(x+2) on top
2(x-3)(x+3) on bottom
2(x-3) cancels
(x+2)/(x+3) is your answer

Answer 2

x^2 - 6x -5 = (x - a)^2 + B
Using FOIL ( firsts,outers,inners,lasts) the RHS looks like this:
x^2 - ax - ax + a^2 +B
=x^2 - 2ax + a^2 + B
From the x-term you can work out a
Put that value in a^2 and you should be able to find B

I take it the 2nd one is
(2x^2 - 2x - 12 )/(2x^2 - 18)?
If so looking at the numerator function
2x^2 - 2x - 12 can be re-written as
2(x^2 - x - 6) = 2(x - 3)(x + 2)
and the denominator as 2(x^2 - 9) = 2(x - 3)(x + 3)
Now you are in a position to cancel?

Answer 3

1. x^2 - 6x -5 = (x - a)^2 + B
Factor as a binomial:
(x - 3)^2 + (-14)

2. 2x^2 - 2x -12 / 2x^2 - 18
I think you left out the brackets.
Extract 2 as a common factor:
2(x^2 - x - 6)/2(x^2 - 9)
Cancel out the common factor:
(x^2 -x - 6)/(x^2 - 9)
Factor as binomials:
(x - 3)(x + 2)/(x - 3)^2
Cancel out (x - 3):
(x + 2)/(x - 3)

Answer 4

1. May be a=3, B=-14.
2. (x+2)/(x+3)