How in the hell do i calculate this...?!?

Question

2(x+3)(x-2)-(2x-1)^2

Answer 1

first expand out the 2(x+3)(x-2) then expand out (2x-1)^2 then put a minus between them and simplify.

expanding 2(x+3)(x-2)
= 2[(x+3)(x-2)]
= 2[x^2 - 2x + 3x - 6]
= 2[x^2 + x - 6]
= 2x^2 + 2x - 12

now expanding (2x-1)^2
= 4x^2 - 4x + 1

now the equation becomes:
= 2x^2 + 2x - 12 - (4x^2 - 4x + 1)
= 2x^2 + 2x - 12 - 4x^2 + 4x - 1
= -2x^2 + 6x - 13

hope this helps! =D

Answer 2

Multiply the first two (x+3) x (x-2) = x^2 + x - 6
Then multiply that by 2 = 2 x^2 + 2x - 12
Then multiply the last (2x-1) x (2x-1) = 4x^2 - 4x + 1
Now combine 2x^2 + 2x - 12 - 4x^2 + 4x - 1
= -2x^2 + 6x - 13

Hope that helps.

Answer 3

How in the hell do i calculate this...?!?
2(x+3)(x-2)-(2x-1)^2
2(x^2+x-6)-(4x^2-4x+1)
2x^2+2x-6
-4x^2+4x-1
-2x^2+6x-7

Answer 4

2(x+3)(x-2)-(2x-1)^2 = 2(x^2 + x -6) - (4x^2 - 4x + 1)
= 2x^2 + 2x - 12 - 4x^2 + 4x - 1
= -2x^2 + 6x -13

Answer 5

ok first u hav to solve

2(x+3)(x-2)

2(x^2+3x-2x-6) (open the brackets 1st)

2(x^2+x-6)

= 2x^2+2x-12 (multiplying 2) now leave this as it is

next

(2x-1)^2

(2x-1)(2x-1)

open the brakets

4x^2-2x-2x+1

4x^2-4x+1

now putting both the equations together

2(x+3)(x-2)-(2x-1)^2

we get

(2x^2+2x-12)-(4x^2-4x+1)

opening the brakets

2x^2 +2x-12-4x^2 +4x-1

put them in order

2x^2-4x^2+2x+4x-12-1

-2x^2+6x-13 is ur answer

or

-(2x^2 -3x+13)

Answer 6

= 2(x+3)(x-2) - (2x-1)^2
= 2x+6 (x-2) - (2x-1) (2x-1)
= 2x^2 - 4x + 6x -12 - (4x^2 -2x -2x +1)
= 2x^2 - 4x + 6x -12 - 4x^2 + 2x + 2x -1
= 2x^2 - 4x^2 - 4x +6x + 2x + 2x -12 -1
= -2x^2 + 6x -13

Answer 7

2(x + 3)(x - 2) - (2x - 1)²

- - - - - - - - - - - - - - - -

(x + 3)(x - 2) = x² + 3x - 2x - 6 = x² + x - 6

2(x² + x - 6) = 2x² + 2x - 12

2x² + 2x - 12

- - - - - - - - - - - - - - - - - - -

(2x - 1)( 2x - 1) = 4x² - 2x - 2x - 1 = 4x² - 4x + 1

4x² - 4x + 1

- - - - - - - - - - - - -

2x² + 2x - 12 - (4x² - 4x + 1) =

2x² + 2x - 12 - 4x² + 4x - 1= - 2x² + 6x - 13

The answer is - 2x² + 6x - 13

- - - - - - s-

Answer 8

velocity and % are an similar for all intents and applications for this. They anticipate you to apply the equation Vf^2=Vi^2 + 2*a*d Vf= very last velocity(what you're searching for) Vi= initial velocity (the speed of 75ft/s) A= acceleration (gravity, so 9.8 m/s/s or 32.1ft/s/s D = displacement (475ft) do purely the maths and look at were given the answer

Answer 9

=2(x+3)(x-2)-(2x-1)^2
=2(x+3)(x-2)-4x-1
=2x+6(x-2)-4x-1
=2x^2-2x+6x-12-4x-1
=2x^2-13

I believe that this is how you do this.

Answer 10

2(x^2+x-6)-((2x-1)(2x-1))
2x^2+2x-12-(4x^2-4x+1)
2x^2+2x-12-4x^2+4x-1
-2x^2+6x-13