Please explain this problem

2(x+h)^2-16(x+h)+35-(2x^2-16x+35) =
4xh+2h-16h
How does one get this answer? I don't understand where the 4xh and 16h "come from."

Rule for binomial sq, (a+b) ^2 = a^2 + 2ab + b^2, so 2( x+h)^2 = 2( x^2 +2xh +h^2) or 2x^2+4xh+2h^2
then + (-16) (x+h) or -16x -16h + 35 - ( quantity) subtract quantity same as (-1) times quantity or

35 + (-1) (2x^2) + (-1) ( -16x) + (-1) (-16h)+ (-1) (35) = 35 + (-2x^2) + 16x +16h+ (-35), so all parts =

2x^2+4xh+2h^2 +(-16x)+(-16h) + 35 +( -2x^2) +16x +(-35) combine all like terms
2x^2 +(-2x^2) +4xh +2h^2 + (-16x) +(16x) +(-16h) +35 +(-35)=
4xh +2h^2 -16h
4xh +2h^2 -16 h is the answer if your answer is 4xh+2h-16h, then the answer is incorrect, it violates the binomial sq. rule

4xh comes from squaring (x + h). Here's how:
(x + h)² = x² + xh + xh + h² = x² + 2xh + h²
Multiply that by the 2 in front of it all:
2x² + 4xh + 2h²
If you combine the remaining terms, you will get:
4xh - 16h + 2h²
You may now divide by the h in the denominator and then Let h become 0.

2(x+h)^2 - 16(x+h) + 35 - (2x^2-16x+35)

2(x^2+2xh+h^2) - (16x+16h) + (35-2x^2+16x-35)

2x^2+4xh+2h^2-16x-16h+35-2x^2+16x-35

4xh+2h^2-16x-16h+35+16x-35

4xh+2h^2-16h

I'm not sure if you mis-typed the 2h, but I'm pretty positive that this is the correct answer.

2(x+h)^2-16(x+h)+35-(2x^2-16x+35)
= 2(x^2 + 2xh + h^2) - 16x - 16h + 35 - 2x^2 +16x -35
... rearranging the terms
= (2x^2 - 2x^2) + (-16x + 16x) + (35 - 35) + 4xh + 2h^2 - 16h
= 4xh -16h +2 h^2

One does not get that answer, one should get:
2h^2 + 4hx - 16h