To convert a base ten number to base five, divide the number by five, with the remainder being the digit in the units place; continue, dividing each successive quotient by five and putting the remainder in the next position to the left until the final quotient is less than five.

$\displaystyle 286\div 5 = 57 \textrm{ R }\underline{1}$ - 1 is the last digit.

$\displaystyle 57 \div 5 = 11 \textrm{ R }\underline{2}$ - 2 is the second-to-last digit.

$\displaystyle 11 \div 5 = \underline{2} \textrm{ R }\underline{1}$ - 1 is the third-to-last digit; 2 is the first digit.

286 is equal to $\displaystyle 2121_{\textrm{five}}$.

Group all like terms by their order:   

$\displaystyle 2x^2 - 3y - 4x^2 + 3y^2 - 3x$

$\displaystyle 2x^2 - 4x^2 + 3y^2 - 3x- 3y$

Simplify:

$\displaystyle -2x^2 + 3y^2 - 3x - 3y$

This can be solved using the FOIL method.  The steps are shown below.

$\displaystyle -(2-x)(6-x)$

$\displaystyle -((2)(6) + (2)(-x) + (-x)(6)+(-x)(-x))$

$\displaystyle -(12-2x-6x+x^2)$

$\displaystyle -(12-8x+x^2)$

$\displaystyle -12+8x-x^2$

Therefore, after reordering, the answer is:  $\displaystyle -x^2+8x-12$

$\displaystyle \left (2x^{-4} \right )^{-2}$

$\displaystyle =\left ( \frac{2}{x^{4}} \right )^{-2}$

$\displaystyle =\left ( \frac{x^{4}}{2} \right )^{2}$

$\displaystyle = \frac{\left (x^{4}\right )^{2}}{2^{2}}$

$\displaystyle = \frac{ x^{4 \cdot 2} }{2^{2}}$

$\displaystyle = \frac{ x^{8} }{4}$

Use the grouping technique, then distribute out the greatest common factor of each group as follows:

$\displaystyle 8t^{3} -56t^{2} +3t -21$

$\displaystyle =\left ( 8t^{3} -56t^{2} \right ) +\left (3t -21 \right )$

$\displaystyle =\left ( 8t^{2} \cdot t -8t^{2} \cdot 7 \right ) +\left (3 \cdot t -3 \cdot 7\right )$

$\displaystyle =8t^{2} \left ( t - 7 \right ) + 3\left ( t - 7 \right )$

$\displaystyle =\left (8t^{2} + 3 \right ) \left ( t - 7 \right )$

The common factor of the terms $\displaystyle 9t^{2}$ and $\displaystyle 81t$ can be found as follows:

$\displaystyle GCF (9,81) = 9$, and the lesser of the two powers of $\displaystyle t$ is $\displaystyle t$; therefore, $\displaystyle GCF (9t^{2},81t) = 9t$, their product. Distribute this out:

$\displaystyle 9t^{2} - 81t$

$\displaystyle = 9t \cdot t - 9t \cdot 9$

$\displaystyle = 9t \left (t - 9 \right )$

This is as far as we can go with the factoring.

Use the grouping technique - group the terms into pairs, then factor out the greatest common factor of each pair.

$\displaystyle 30xt + 15x -22t - 11$

$\displaystyle = \left (30xt + 15x \right ) -\left (22t + 11 \right )$

$\displaystyle = \left (15x \cdot 2t + 15x \cdot 1 \right ) -\left ( 11 \cdot 2t + 11 \cdot 1 \right )$

$\displaystyle = 15x \left ( 2t + 1 \right ) - 11 \left ( 2t + 1 \right )$

$\displaystyle = \left (15x - 11 \right ) \left ( 2t + 1 \right )$

A polynomial of the form $\displaystyle Ax^{2} + Bx + C$ can be factored by first splitting the term into two terms whose coefficients add up to $\displaystyle B$ and have product $\displaystyle AC$, then factoring out by the grouping technique.

We are looking for two integers whose sum is $\displaystyle B = 21$ and whose product is $\displaystyle AC = 2 (-50) = -100$. Through some trial and error, we can see that the integers are $\displaystyle \left \{ 25, -4 \right \}$, so:

$\displaystyle 2x^{2} + 21x - 50$

$\displaystyle =2x^{2} - 4x + 25x - 50$

$\displaystyle =\left (2x^{2} - 4x \right )+\left ( 25x - 50 \right )$

$\displaystyle =\left (2x \cdot x - 2x \cdot 2 \right )+\left ( 25 \cdot x - 25 \cdot 2 \right )$

$\displaystyle = 2x \left (x-2 \right )+ 25 \left (x-2 \right )$

$\displaystyle = \left (2x+ 25 \right ) \left (x-2 \right )$

First, factor out the greatest common factor of the terms, which is $\displaystyle 4x y$:

$\displaystyle 4x ^{3}y - 100xy$

$\displaystyle = 4x y \cdot x^{2} - 4x y \cdot 25$

$\displaystyle = 4x y \left ( x^{2} - 25 \right )$

$\displaystyle x^{2} - 25$ is the difference of squares, so we can factor further:

$\displaystyle 4x y \left ( x^{2} - 25 \right )$

$\displaystyle = 4x y \left ( x^{2} - 5 ^{2}\right )$

$\displaystyle 4x y \left ( x + 5 \right ) \left ( x - 5 \right )$

$\displaystyle 7 (2x + 8)- 5 (7x - 3)$

$\displaystyle = 7 \cdot 2x +7 \cdot 8- 5 \cdot 7x + 5 \cdot 3$

$\displaystyle = 14x +56- 35x + 15$

$\displaystyle = 14x- 35x +56+ 15$

$\displaystyle = \left (14 - 35 \right )x +\left (56+ 15 \right )$

$\displaystyle = -21x+ 71$