$\displaystyle 4x^{2}-12xy +9y^{2} - 64x^{2}y^{2}$ can be grouped as follows:

$\displaystyle \left (4x^{2}-12xy +9y^{2} \right ) - 64x^{2}y^{2}$

The first three terms form a perfect square trinomial, since $\displaystyle 2\cdot \sqrt{4x^{2} } \cdot \sqrt { 9y^{2}} = 2 \cdot 2x \cdot 3y = 12xy$

$\displaystyle 4x^{2}-12xy +9y^{2} = \left ( 2x - 3y\right )^{2}$, so

$\displaystyle \left (4x^{2}-12xy +9y^{2} \right ) - 64x^{2}y^{2}$

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

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

Now use the dfference of squares pattern:

$\displaystyle = \left [ \left ( 2x - 3y \right ) + 8 xy \right ] \left [ \left ( 2x - 3y \right ) - 8 xy \right ]$

$\displaystyle = \left ( 2x - 3y + 8 xy \right ) \left ( 2x - 3y - 8 xy \right )$

$\displaystyle f(x)=x^2+14x+49$

$\displaystyle 0=x^2+14x+49$

$\displaystyle 0=(x+7)(x+7)$

$\displaystyle x+7=0$

$\displaystyle x=-7$

$\displaystyle f(x)=x^2+x-2$

$\displaystyle 0=x^2+x-2$

$\displaystyle 0=(x+2)(x-1)$

$\displaystyle x+2=0$ and $\displaystyle x-1=0$

$\displaystyle x=-2$ and $\displaystyle x=1$

$\displaystyle x^2-8x-9=0$

$\displaystyle (x-9)(x+1)=0$

$\displaystyle x-9=0$ and $\displaystyle x+1=0$

$\displaystyle x=9$ and $\displaystyle x=-1$

The expression is a perfect square trinomial, as the three terms have the following relationship:

$\displaystyle 625x^{4} =\left ( 25x ^{2} \right ) ^{2}$

$\displaystyle 81 = 9 ^{2}$

$\displaystyle 450x^{2} = 2 \cdot 25x^{2} \cdot 9$

$\displaystyle 625x^{4} - 450x^{2} + 81= \left ( 25x ^{2} \right ) ^{2} - 2 \cdot 25x^{2} \cdot 9 + 9 ^{2}$

We can factor this expression by substituting $\displaystyle A = 25x ^{2}, B = 9$ into the following pattern:

$\displaystyle A^{2} - 2AB + B^{2} = (A-B)^{2}$

$\displaystyle \left ( 25x ^{2} \right ) ^{2} - 2 \cdot 25x^{2} \cdot 9 + 9 ^{2} =(25x ^{2}-9)^{2}$

We can factor further by noting that $\displaystyle 25x ^{2}-9 = (5x)^{2} - 3^{2}$, the difference of squares, and subsequently, factoring this as the product of a sum and a difference.

$\displaystyle 625x^{4} - 450x^{2} + 81=(25x ^{2}-9)^{2} = \left [ (5x+3)(5x-3) \right ]^{2}$

or 

$\displaystyle 625x^{4} - 450x^{2} + 81= (5x+3)^{2}(5x-3)^{2}$

Note that $\displaystyle a^{3}+3a^{2}b+3ab^{2}+b^{3} = (a+b)^{3}$

Therefore $\displaystyle 5(a^{3}+3a^{2}b+3ab^{2}+b^{3})=40$ is equivalent to:

 $\displaystyle 5(a+b)^{3}=40$ $\displaystyle \Leftrightarrow (a+b)^{3}=40/5 = 8 \Leftrightarrow a+b = 2$

The correct answer is $\displaystyle -13/2$. Our work proceeds as follows:

$\displaystyle \frac{x^3-x}{x^3+6x^2+5x}=5$

$\displaystyle \frac{(x^2-1)(x)}{(x^2+6x+5)(x)} =5$   (Factor an $\displaystyle x$ out of the numerator and denominator)

$\displaystyle \frac{(x+1)(x-1)(x)}{(x+1)(x+5)(x)} = 5$ (Factor the quadratic polynomials)

$\displaystyle \frac{x-1}{x+5} =5$ (Cancel common terms)

$\displaystyle x-1 = 5(x+5)$ (Multiply by $\displaystyle x+5$ to both sides)

$\displaystyle x-1 = 5x +25$   (Distribute the $\displaystyle 5$)

$\displaystyle -26 = 4x$ (Simplify and solve)

$\displaystyle \frac{-13}{2} = x$

In order to solve for $\displaystyle x$ we must first factor: 

$\displaystyle 3x^2-74x-25=0$

$\displaystyle (3x+1)(x-25)=0$

The zero product property states that if $\displaystyle xy=0$ then $\displaystyle x=0$ or $\displaystyle y=0$ (or both).

Our two equations are then:

 $\displaystyle (3x+1)=0, (x-25)=0$

Solving for $\displaystyle x$ in each leaves us with:

$\displaystyle 3x+1=0$

$\displaystyle 3x=-1$

$\displaystyle x=-\frac{1}{3}$

and 

$\displaystyle x-25=0$

$\displaystyle x=25$

The roots of a function are the points at which it crosses the x axis, so at these points the value of y, or f(x), is 0. This gives us:

$\displaystyle x^2-3x-18=0$

So we will have to factor the polynomial in order to solve for the x values at which the function is equal to 0. We need two factors whose product is -18 and whose sum is -3. If we think about our options, 2 and 9 have a product of -18 if one is negative, but there's no way of making these two numbers add up to -3. Next we consider 3 and 6. These numbers have a product of -18 if one is negative, and their sum can also be -3 if the 3 is positive and the 6 is negative. This allows us to write out the following factorization:

$\displaystyle x^2-3x-18=0$

$\displaystyle (x+3)(x-6)=0$

$\displaystyle x=-3,x=6$

We could solve this question a variety of ways. The simplest would be graphing with a calculator, but we will use factoring. 

To begin, set our function equal to $\displaystyle 0$. We want to find where this function crosses the $\displaystyle x$-axis—in other words, where $\displaystyle y=0$.

$\displaystyle 0=x^2-6x+8$

Next, we need to factor the function into two binomial terms. Remember FOIL/box method? We are essentially doing the reverse here. We are looking for something in the form of $\displaystyle (x+a)(x+b)=0$.

Recalling a few details will make this easier.

1) $\displaystyle a*b$ must equal positive $\displaystyle 8$

2) $\displaystyle a$ and $\displaystyle b$ must both be negative, because we get positive $\displaystyle 8$ when we multiply them and $\displaystyle -6$ when we add them.

3) $\displaystyle a$ and $\displaystyle b$ must be factors of $\displaystyle 8$ that add up to $\displaystyle -6$. List factors of $\displaystyle 8$: $\displaystyle 1, 2, 4, 8$. The only pair of those that will add up to $\displaystyle 6$ are $\displaystyle 2$ and $\displaystyle 4$, so our factored form looks like this:

$\displaystyle 0=(x-2)(x-4)$

Then, due to the zero product property, we know that if $\displaystyle x=2$ or $\displaystyle x=4$ one side of the equation will equal $\displaystyle 0$, and therefore our answers are positive $\displaystyle 2$ and positive $\displaystyle 4$.