The inequality in the question possesses an absolute value; therefore, we most solve for the variable being less than positive 6 and greater than negative 6. Let's start with the positive solution.

\(\displaystyle (2x - 4) < 6\) 

Add 4 to both sides of the inequality.

\(\displaystyle 2x-4+4< 6+4\)

\(\displaystyle 2x< 10\)

Divide both sides of the inequality by 2.

\(\displaystyle \frac{2x}{2}< \frac{10}{2}\)

\(\displaystyle x < 5\)

Now, let's solve for the negative solution

\(\displaystyle (2x - 4) > -6\)

Add 4 to both sides of the inequality.

\(\displaystyle 2x - 4+4> -6+4\)

\(\displaystyle 2x>-2\)

Divide both sides of the inequality by 2.

\(\displaystyle \frac{2x}{2}>\frac{-2}{2}\)

\(\displaystyle x > -1\)

Using these solutions we can write the following statement: 

\(\displaystyle -1< x< 5\)

The smallest integer that satisfies this equation is 0, and the largest is 4. Their product is 0. 

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 \(\displaystyle (x-1)(x-1)\)