Factoring Math; Please help Me?

2(x-4)(x-5)

2x^2 - 18x + 40
= 2(x^2 - 9x + 20)
= 2(x^2 - 4x - 5x + 20)
= 2[(x^2 - 4x) - (5x - 20)]
= 2[x(x - 4) - 5(x - 4)]
= 2(x - 4)(x - 5)

2x^2 - 18x + 40
Factor 2 out
= 2(x^2 - 9x + 20)
= 2(x^2 - 4x - 5x - 20)
= 2[x(x-4) - 5(x-4)]
= 2(x-4)(x-5)

factor out a 2 by dividing each term by 2

2(x^2 - 9x + 20)

Now break up the x^2 to x * x
Now break up the 20 to have 5 * 4

2(x - 5)(x - 4)

I hope this helps
:o)
dr

2(x-4)(x-5)

Steps

2x^2 - 18x + 40
= 2(x^2 - 9x + 20)
= 2(x^2 - 4x - 5x - 20)
= 2[x(x-4) - 5(x-4)]
= 2(x-4)(x-5)

2[(x-5)(x-4)]

2(x^2 - 9x + 20)=2(x-4)(x-5)

you can write this equation by taking 2 common

2 ( x^2 - 9x + 20)

2( x^2 - 5x - 4x + 20 )

2 ( x(x- 5) - 4( x-5) )

2 ( x- 5) (x - 4)

uh..... I don't really know, I think 72?

the easiest way to do this is to break the middle term(b) into two terms that can be added to get the middle number and also multiplied to get the first term (a) times the last term(c)
so the two numbers must add to -18 but multiply to get 80 (2*40)
it ends up that -10 and -8 do the trick.
put these numbers in so you get:
2x^2 - 10x - 8x + 40

from here you break it into two separate parts
2x^2 - 10x and -8x + 40
then factor each one so you are left with the same answer

2x(x-5) and -8(x-5)

then just combine the terms and you have your answer of:
(2x-8)(x-5)