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:

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

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: $\displaystyle \small x^2y^3z(z+x^2)$. 

First pull out 3u from both terms.

3u⁴ – 24uv³ = 3u(u³ – 8v³) = 3u[u³ – (2v)³]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember, but pat yourself on the back for getting to such hard questions! The difference of cubes formula is a³ – b³ = (a – b)(a² + ab + b²). In our problem, a = u and b = 2v:

3u⁴ – 24uv³ = 3u(u³ – 8v³) = 3u[u³ – (2v)³]

                = 3u(u – 2v)(u² + 2uv + 4v²)

$\displaystyle 2x^{2}-7x+6$

$\displaystyle =2x^{2}-3x-4x+6$

$\displaystyle =2x^{2}-3x-(4x-6)$

$\displaystyle =x(2x-3)-2(2x-3)$

$\displaystyle =(x-2)(2x-3)$

$\displaystyle x^2 -10x - 24$

To factor, we are looking for two terms that multiply to give $\displaystyle -24$ and add together to get $\displaystyle -10$.

Possible factors of $\displaystyle -24$:

$\displaystyle (-1,24),\ (1,-24),\ (-2,12),\ (2,-12),\ (-3,8),\ (3,-8),\ (-4,6),\ (4,-6)$

Based on these options, it is clear our factors are $\displaystyle 2$ and $\displaystyle -12$.

$\displaystyle (2)(-12)=-24\ \text{and}\ 2+(-12)=-10$

Our final answer will be:

$\displaystyle (x-12)(x+2)$

$\displaystyle x^2 + 20x + 96$

To factor, we are looking for two terms that multiply to give $\displaystyle 96$ and add together to get $\displaystyle 20$. There are numerous factors of $\displaystyle 96$, so we will only list a few.

Possible factors of $\displaystyle 96$:

$\displaystyle (3,32),\ (-3,-32),\ (4,24),\ (-4,-24),\ (6,16),\ (8,12)$

Based on these options, it is clear our factors are $\displaystyle 8$ and $\displaystyle 12$.

$\displaystyle (8)(12)=96\ \text{and}\ 8+12=20$

Our final answer will be:

$\displaystyle (x + 12)(x+8)$

We can factor this trinomial using the FOIL method backwards. This method allows us to immediately infer that our answer will be two binomials, one of which begins with $\displaystyle 3x$ and the other of which begins with $\displaystyle x$. This is the only way the binomials will multiply to give us $\displaystyle 3x^{2}$.

The next part, however, is slightly more difficult. The last part of the trinomial is $\displaystyle -2$, which could only happen through the multiplication of 1 and 2; since the 2 is negative, the binomials must also have opposite signs.

Finally, we look at the trinomial's middle term. For the final product to be $\displaystyle -x$, the 1 must be multiplied with the $\displaystyle 3x$ and be negative, and the 2 must be multiplied with the $\displaystyle x$ and be positive. This would give us $\displaystyle \left ( -3x\right )+\left ( 2x\right )$, or the $\displaystyle -x$ that we are looking for.

In other words, our answer must be 

$\displaystyle \left ( 3x+2\right )\left ( x-1\right )$ 

to properly multiply out to the trinomial given in this question.

$\displaystyle 8x^2+32x+30$

Factor out the largest quantity common to all terms:

$\displaystyle 2(4x^2+16x+15)$

Factor the simplified quadratic:

$\displaystyle 2(2x+3)(2x+5)$