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