This is a difference of perfect cubes so it factors to \(\displaystyle (x-1)(x^2+x+1)\). Only the first expression will yield an answer when set equal to 0, which is 1. The second expression will never cross the \(\displaystyle x\)-axis. Therefore, your answer is only 1.

Factor the equation to \(\displaystyle 2x(x^4-4x^2+4)\). Set \(\displaystyle 2x=0\) and get one of your \(\displaystyle x\)'s to be \(\displaystyle 0\). Then factor the second expression to \(\displaystyle (x^2-2) (x^2-2)\). Set them equal to zero and you get \(\displaystyle \pm \sqrt{2}\). 

First, begin by factoring out a common term, in this case \(\displaystyle 16x\):

\(\displaystyle 16x^{3} -48x^{2} + 32x = 16x(x^{2}-3x+2)\)

Then, factor the terms in parentheses by finding two integers that sum to \(\displaystyle -3\) and multiply to \(\displaystyle 2\):

\(\displaystyle 16x(x-2)(x-1)\)

Here you have an expression with three variables. To factor, you will need to pull out the greatest common factor that each term has in common.

Only the last two terms have \(\displaystyle c\) so it will not be factored out. Each term has at least \(\displaystyle a^2\) and \(\displaystyle b\) so both of those can be factored out, outside of the parentheses. You'll fill in each term inside the parentheses with what the greatest common factor needs to be multiplied by to get the original term from the original polynomial:

\(\displaystyle a^3b^2 + a^4b + a^2bc^3 - a^3b^3c^2 = a^2b(ab + a^2 + c^3 - ab^2c^2)\)

For the trinomial \(\displaystyle \small \small \small x^{2}+Bx+36\) to be factorable, we would have to be able to find two integers with product 36 and sum \(\displaystyle B\); that is, \(\displaystyle B\) would have to be the sum of two integers whose product is 36.

Below are the five factor pairs of 36, with their sum listed next to them. \(\displaystyle B\) must be one of those five sums to make the trinomial factorable.

1, 36: 37

2, 18: 20

3, 12: 15

4, 9: 13

6, 6: 12

Of the five choices, only 16 is not listed, so if \(\displaystyle B = 16\), then the polynomial is prime.

To factor trinomials like this one, we need to do a reverse FOIL. In other words, we need to find two binomials that multiply together to yield \(\displaystyle 7x^{2} - 11x - 6\).

Finding the "first" terms is relatively easy; they need to multiply together to give us \(\displaystyle 7x^{2}\), and since \(\displaystyle 7\) only has two factors, we know the terms must be \(\displaystyle 7x\) and \(\displaystyle x\). We now have \(\displaystyle (7x\)    \(\displaystyle )(x\)    \(\displaystyle )\), and this is where it gets tricky.

The second terms must multiply together to give us \(\displaystyle -6\), and they must also multiply with the first terms to give us a total result of \(\displaystyle -11x\). Many terms fit the first criterion. \(\displaystyle (-2)(3)\), \(\displaystyle (-3)(2)\), \(\displaystyle (-6)(1)\) and \(\displaystyle (-1)(6)\) all multiply to yield \(\displaystyle -6\). But the only way to also get the "\(\displaystyle x\)" terms to sum to \(\displaystyle -11x\) is to use \(\displaystyle (7x + 3)(x - 2)\). It's just like a puzzle!

To find the greatest common factor, we must break each term into its prime factors:

\(\displaystyle \small x^2y^3z^2 = x \cdot x \cdot y \cdot y \cdot y \cdot z \cdot z\)

\(\displaystyle \small x^4y^3z= x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot z\)

The terms have \(\displaystyle \small x^2\), \(\displaystyle \small y^3\), and \(\displaystyle \small z\) in common; thus, the GCF is \(\displaystyle \small x^2y^3z\).

Pull this out of the expression to find the answer: \(\displaystyle \small x^2y^3z(z+x^2)\). 

Use the \(\displaystyle ac\)-method to split the middle term into the sum of two terms whose coefficients have sum \(\displaystyle -31\) and product \(\displaystyle 6*35 = 210\). These two numbers can be found, using trial and error, to be \(\displaystyle -21\) and \(\displaystyle -10\).

\(\displaystyle -21*-10=210\) and \(\displaystyle -21+(-10)=-31\)

Now we know that \(\displaystyle 6x^{2}-31x+35\) is equal to \(\displaystyle 6x^{2}-10x-21x+35\).

Factor by grouping.

\(\displaystyle (6x^{2}-10x)-(21x-35)\)

\(\displaystyle 2x(3x-5)-7(3x-5)\)

\(\displaystyle (2x-7)(3x-5)\)

First, we note that the coefficients have an LCD of 3, so we can factor that out:

\(\displaystyle 3(x^{2} - 15x - 32)\)

We try to factor further by factoring quadratic trinomial \(\displaystyle x^{2} - 15x - 32\). We are looking to factor it into two factors\(\displaystyle (x+ ? )(x+?)\), where the question marks are to be replaced by two integers whose product is \(\displaystyle -32\) and whose sum is \(\displaystyle -15\). 

We need to look at the factor pairs of \(\displaystyle -32\) in which the negative number has the greater absolute value, and see which one has sum \(\displaystyle -15\):

\(\displaystyle \begin{matrix} \textrm{Pair} & \textrm{Sum} \\ 1,-32& -31\\ 2,-16&-14 \\ 4,-8& -4 \end{matrix}\)

None of these pairs have the desired sum, so \(\displaystyle x^{2} - 15x - 32\) is prime. \(\displaystyle 3(x^{2} - 15x - 32)\) is the complete factorization.

When working with a rational expression, you want to first put your monomials in standard format.

Re-order the bottom expression, so it is now reads \(\displaystyle -4x+2\). 

Then factor a \(\displaystyle -1\) out of the expression, giving you \(\displaystyle -1(4x-2)\).

 The new fraction is  \(\displaystyle \frac{4x-2}{-1(4x-2)}\).

Divide out the like term, \(\displaystyle (4x-2)\), leaving \(\displaystyle \frac{1}{-1}\), or \(\displaystyle -1\).