The equation is in standard form, so a = 8, b = -2, and c = -15.  We are looking for two factors that multiply to ac or -120 and add to b or -2.  The two factors are -12 and 10.

So you get (2x -3)(4x +5) = 0.  Set each factor equal to zero and solve.

(x² + 2) / 2 = (x² - 6x - 1) / 5. We first cross-multiply to get rid of the denominators on both sides.

5(x² + 2) = 2(x² - 6x - 1)

5x² + 10 = 2x² - 12x - 2 (Subtract 2x², and add 12x and 2 to both sides.)

3x² + 12x + 12 = 0 (Factor out 3 from the left side of the equation.)

3(x² + 4x + 4) = 0 (Factor the equation, knowing that 2 + 2 = 4 and 2*2 = 4.)

3(x + 2)(x + 2) = 0

x + 2 = 0

x = -2

Factor the polynomial by choosing values that when FOIL'ed will add to equal the middle coefficient, 3, and multiply to equal the constant, 1.

x² – 6x + 5 = (x – 1)(x – 5)

Because only (x – 5) is one of the choices listed, we choose it.

$\displaystyle \small 7x+30=x^{2}$

$\displaystyle \small x^{2}-7x-30=0$

$\displaystyle \small (x-10)(x+3)=0$

Either:

$\displaystyle \small x-10=0$

$\displaystyle \small x=10$

or:

$\displaystyle \small x+3=0$

$\displaystyle \small x=-3$

The answer is $\displaystyle 2x^{3}y^{2}+10x^{4}y^{2}$^.

To determine the answer, $\displaystyle 2x^{2}$ must be distrbuted,

$\displaystyle (2x^{2}*xy^{2}) +(2x^{2}*5x^{2}y^{2})$. After multiplying the terms, the expression simplifies to $\displaystyle 2x^{3}y^{2}+10x^{4}y^{2}$.

Factoring leads to (b+3)(b+3)=0. Therefore, solving for b leads to -3.

First you want to factor the numerator from x² – 6x + 8 to (x – 4)(x – 2)

Input the denominator (x – 4)(x – 2)/(x – 2) = (x – 4) = 0, so x = 4.

The question asks us to find the value of $\displaystyle x$, because it is in a closed equation, we can simply put all of the whole numbers on one side of the equation, and all of the $\displaystyle x$ containing numbers on the other side.

We utilize opposite operations to both sides by adding $\displaystyle 2x$ to each side of the equation and get $\displaystyle 5x+5=6$

Next, we subtract $\displaystyle 5$ from both sides, yielding

$\displaystyle 5x=1$

Then we divide both sides by $\displaystyle 5$ to get rid of that $\displaystyle 5$ on $\displaystyle 5x$

$\displaystyle x=\frac{1}{5}$

First we factor out an x then we can factor the $\displaystyle (x-1)(x-1)$

The equation presented in the problem is:

$\displaystyle 4y^{2}=169x^{2}-81$

We know that:

 $\displaystyle 169x^{2}-81=(13x)^{2}-9^{2}=(13x+9)(13x-9)$

Therefore we can see that the answer choice $\displaystyle (2y)^{2}=(13x+9)(13x-9)$ is equivalent to $\displaystyle 4y^{2}=169x^{2}-81$.

 $\displaystyle \frac{(2y)^{2}}{13x+9}=27x-9-14x$ is equivalent to  $\displaystyle 4y^{2}=169x^{2}-81$. You can see this by first combining like terms on the right side of the equation: 

$\displaystyle \frac{(2y)^{2}}{13x+9}=13x-9$

Multiplying everything by $\displaystyle 13x+9$, we get back to:

$\displaystyle (2y)^{2}=(13x+9)(13x-9)$ 

We know from our previous work that this is equivalent to $\displaystyle 4y^{2}=169x^{2}-81$.

$\displaystyle 6y^{2}=\frac{3\times (13x+9)\times (13x-9)}{2}$ is also equivalent $\displaystyle 4y^{2}=169x^{2}-81$ since both sides were just multiplied by $\displaystyle \frac{3}{2}$. Dividing both sides by $\displaystyle \frac{3}{2}$, we also get back to:

$\displaystyle (2y)^{2}=(13x+9)(13x-9)$.

We know from our previous work that this is equivalent to $\displaystyle 4y^{2}=169x^{2}-81$.

$\displaystyle y=\sqrt{\frac{338x^{2}-162}{8}}$ is also equivalent to $\displaystyle 4y^{2}=169x^{2}-81$ since

$\displaystyle y^{2}=(\sqrt{\frac{338x^{2}-162}{8}})^{2}=\frac{338x^{2}-162}{8}$

$\displaystyle 4y^{2}=\frac{338x^{2}-162}{2}=169x^{2}-81$

Only $\displaystyle y^{2}=(\frac{13x-9}{2})^{2}$ is NOT equivalent to $\displaystyle 4y^{2}=169x^{2}-81$

because

$\displaystyle y^{2}=(\frac{13x-9}{2})^{2}=\frac{169x^{2}-234x-81}{4}$

$\displaystyle 4y^{2}=169x^{2}-81-234x \neq 4y^{2}=169x^{2}-81$