To factor this, we need two numbers that multiply to $\displaystyle \dpi{100} \small 4\times -15=60$ and sum to $\displaystyle \dpi{100} \small 4$. The numbers $\displaystyle \dpi{100} \small -6$ and $\displaystyle \dpi{100} \small 10$ work.

$\displaystyle 4y^{2}+4y-15 = 4y^{2} - 6y + 10y - 15 = 2y(2y-3) + 5(2y-3) = (2y-3)(2y+5)$

$\displaystyle {x^{2}-y^{2}}$ is a difference of squares. The difference of squares formula is

$\displaystyle {a^{2}-b^{2}} = (a-b)(a+b)$

So $\displaystyle {x^{2}-y^{2}} = (x-y)(x+y)$.

Then, $\displaystyle \frac{y-x}{x^{2}-y^{2}} = \frac{y-x}{(x+y)(x-y)} = \frac{-(x-y)}{(x+y)(x-y)} = \frac{-1}{x+y}$.

Let's factor the expression: $\displaystyle x^{2}-6x+5 = (x-1)(x-5)$.

We need to look at the behavior of the function to the left and right of 1 and 5.  To the left of $\displaystyle x=1$,

$\displaystyle x^{2}-6x+5>0$

You can check this by plugging in any value smaller than 1. For example, if $\displaystyle x=0$,

$\displaystyle (x-1)(x-5)=5$,

which is greater than 0.

When $\displaystyle x$ takes values in between 1 and 5, $\displaystyle x^{2}-6x+5 < 0$.  Again we can check this by plugging in a number between 1 and 5. 

$\displaystyle x=2,(x-1)(x-5)=-3$, which is less than 0, so no numbers between 1 and 5 satisfy the inequality.

When $\displaystyle x$ takes values greater than 5, $\displaystyle x^{2}-6x+5>0$. 

To check, let's try $\displaystyle x=6$.  Then:

$\displaystyle (x-1)(x-5)=5$

so numbers greater than 5 also satisfy the inequality.

Therefore $\displaystyle x>5\ or\ x< 1$.

First let's factor: $\displaystyle x^{2}+7x-8=(x+8)(x-1)$

x < -8: Let's try -10.  (-10 + 8)(-10 - 1) = 22, so values less than -8 don't satisfy the inequality.

-8 < x < 1: Let's try 0.  (0 + 8)(0 - 1) = -8, so values in between -8 and 1 satisfy the inequality.

x > 1: Let's try 2.  (2 + 8)(2 - 1) = 10, so values greater than 1 don't satisfy the inequality.

Therefore the answer is -8 < x < 1.

This expression can be rewritten:

$\displaystyle 6,561x^{8}-256$

$\displaystyle (81x^{4})^{2}- 16^{2}$

As the difference of squares, this can be factored as follows: 

$\displaystyle (81x^{4}+ 16)(81x^{4}- 16)$

As the sum of squares with relatively prime terms, the first factor is a prime polynomial. The second factor can be rewritten as the difference of two squares and factored:

$\displaystyle (81x^{4}+ 16)\left [(\left (9x^{2} \right )^{2}- 4^{2}) \right ]$

$\displaystyle (81x^{4}+ 16)\left (9x^{2} + 4) (9x^{2} - 4)$

Similarly, the middle polynomial is prime; the third factor can be rewritten as the difference of two squares and factored:

$\displaystyle (81x^{4}+ 16)\left (9x^{2} + 4) \left [(3x^{2} )^{2}- 2^{2} \right ]$

$\displaystyle (81x^{4}+ 16)\left (9x^{2} + 4)(3x + 2)(3x - 2)$

This is as far was we can factor, so this is the complete factorization.

Factor the equation and set it equal to zero.  $\displaystyle (x-6)*(x-3)=0$. So the funtion will cross the $\displaystyle x$-axis when

$\displaystyle x=3\ or\ x=6$

This questions tests the formula: $\displaystyle a^2-b^2=(a+b)*(a-b)$.

Therefore, we have $\displaystyle 15=5(a-b)$. So

$\displaystyle a-b=\frac{15}{5}=3$

$\displaystyle 64 - 4x^{2} + 20xy - 25y^{2}$ can be grouped as follows:

$\displaystyle 64 - \left ( 4x^{2} - 20xy + 25y^{2} \right )$

$\displaystyle 4x^{2} - 20xy + 25y^{2}$ is a perfect square trinomial, since 

$\displaystyle 2\cdot \sqrt{4x^{2}} \cdot \sqrt{25y^{2}} = 2 \cdot 2x \cdot 5y = 20xy$

$\displaystyle 64 - \left ( 4x^{2} - 20xy + 25y^{2} \right )$

$\displaystyle = 8^{2} - \left ( 2x-5y \right )^{2}$

Now use the difference of squares pattern:

$\displaystyle = 8^{2} - \left ( 2x-5y \right )^{2}$

$\displaystyle =\left [ 8+ \left ( 2x-5y \right ) \right ]\left [ 8- \left ( 2x-5y \right ) \right ]$

$\displaystyle = (8+ 2x-5y) (8- 2x+5y)$

Group the first three terms and the last three terms, then factor out a GCF from each grouping:

$\displaystyle x^{5} + 3x^{4} + 2x^{3}+ 5x ^{2} + 15x + 10$

$\displaystyle =\left ( x^{5} + 3x^{4} + 2x^{3} \right )+\left ( 5x ^{2} + 15x + 10 \right )$

$\displaystyle =x^{3} \left ( x^{2} + 3x + 2 \right )+5 \left ( x^{2} + 3x + 2 \right )$

$\displaystyle =\left (x^{3}+5 \right )\left ( x^{2} + 3x + 2 \right )$

We try to factor $\displaystyle x^{3}+5$ as a sum of cubes; however, 5 is not a perfect cube, so the binomial is a prime.

To factor out $\displaystyle x^{2} + 3x + 2$, we try to factor it into $\displaystyle (x + ?)(x + ?)$, replacing the question marks with two integers whose product is 2 and whose sum is 3. These integers are 1 and 2, so 

$\displaystyle x^{2} + 3x + 2 = \left ( x+1\right ) \left ( x+2\right )$

The original polynomial has $\displaystyle \left (x^{3}+5 \right ) \left ( x+1\right ) \left ( x+2\right )$ as its factorization.

Group the first three terms and the last three terms, then factor out a GCF from each grouping:

$\displaystyle x^{5} - 3x^{4} + x^{3}+ 8x ^{2} - 24x + 8$

$\displaystyle =( x^{5} - 3x^{4} + x^{3} ) +( 8x ^{2} - 24x + 8 )$

$\displaystyle =x^{3} ( x^{2} - 3x + 1 ) +8 ( x^{2} - 3x + 1 )$

$\displaystyle = \left ( x^{3} +8 \right )( x^{2} - 3x +1)$

$\displaystyle x^{3} +8$ is the sum of cubes and can be factored using this pattern: 

$\displaystyle x^{3} +8$

$\displaystyle = x^{3} +2^{3}$

$\displaystyle = \left ( x+2\right ) \left ( x^{2} - 2 \cdot x + 2^{2} \right )$

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

We try to factor out the quadratic trinomial as $\displaystyle (x + ?)(x + ?)$, replacing the question marks with integers whose product is 1 and whose sum is $\displaystyle -3$. These integers do not exist, so the trinomial is prime.

The factorization is therefore

$\displaystyle = \left ( x+2\right ) \left ( x^{2} - 2 x + 4 \right )( x^{2} - 3x +1)$