Start by looking at your last term. Since this term is negative, you will need to have a positive group and a negative group:

Now, since the middle term is positive, you can guess that the positive group will contain the larger number. Likewise, since the coefficient is only , you can guess that the factors will be close.  Two such factors of  are  and .

Therefore, your groups will be:

Begin by looking at the last element. Since it is positive, you know that your groups will contain either two additions or two subtractions. Since the middle term is negative (), your groups will be two subtractions:

Now, the factors of  are  and ,  and , and  and .

Clearly, the last is the one that works, for when you FOIL , you get your original equation!

We can factor out a , leaving .

From there we can factor again to 

.

The original statement is equivalent to a statement that is identically true regardless of the value of ; therefore, so is the original statement itself. The solution set is the set of all real numbers.

To solve this equation, simply plug 12 in for  in the equation. 

The question is equivalent to asking the following:

For what value of  does the equation 

have more than one solution?

The equation simplifies as follows:

If the absolute values of two expressions equal, then either the expressions themselves are equal or they are each other's opposite. 

Taking the latter case:

Regardless of the value of , exactly one solution is yielded this way.

The question becomes as follows: for which value of  does the other way yield a solution?

Set:

If this is a false statement, then this yields no solutions.

If this is a true statement, then this automatically yields the set of all real numbers as the solution set. We solve for :

As a result, the statement

has infinitely many solutions, and the other three statements have exactly one.

First, since the equation is quadratic, put it in standard form 

as follows:

To determine the nature of the solution set, evaluate discriminant  for :

The discriminant is positive and a perfect square, so the solution set comprises two rational numbers.

First, since the equation is quadratic, put it in standard form 

as follows:

To determine the nature of the solution set, evaluate discriminant  for :

The discriminant is positive, but not a perfect square, so the solution set comprises two irrational numbers.

Factor the polynomial as the difference of squares:

We set each binomial equal to 0 and apply the Square Root Property:

This yields two imaginary solutions.

This yields two irrational solutions.

The correct response is that the solution set comprises two irrational numbers and two imaginary numbers.

The cubic binomial in the equation can be factored to yield an equivalent equation as follows::

One of these factors must be equal to 0.

If  then , so one solution is rational.

If , we can find out about the nature of the remaining solutions using discriminant , setting :

The discriminant is negative, so the two solutions of the equation  are imaginary. These are also the two remaining solutions of .

The correct response is that the equation has one rational solution and two imaginary solutions.