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

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