Try using the box method. When I do it, it makes multiplying trinomials and binomials like these easier.
Here is how I would set up the equation:
as you can see in the picture below I multiplied x by 3x^2, -4x, and positive six which gives me a product of 3x^3 (when multiplying always add the exponents) -4x^2 and 6x. Afterwards I did the same thing by multiplying one to the trinomials (3x^2, -4x and 6). Now to simplify the answer you will add by like terms. So -4x^2 plus 3x^2= -x^2
6x+(-4x)= 2x. Since the only two numbers left (6 and 3x^3 aren’t like terms, you leave it like that. Your answer will be 3x^3 + (-x^2)+2x+6.
Solution :
We have to prove that (De-Morgan's law)
Let then
and
and so and
.
Thus, and so
Hence, .........(1)
Now we will show that
Let ⇒
Thus x is present neither in the set A nor in the set B, so by definition of the union of the sets, by definition of the complement.
and
Therefore, and we have
.............(2)
From (1) and (2),
Hence proved.