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

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

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