Intermediate Single-Variable Algebra - Math

We are asked to solve the equation .

Often, when solving equations involving rational expressions, it helps to elminate fractions by multiplying both sides of the equation by the denominators of each term intervolved.

In the context of this problem, we can first multiply both sides of the equation by x+2 to eliminate the denominator of the first term.

Be sure to distribute the x+2 to each term on the left side of the equation.

Next, multiply both sides of the equation by x to elminate the term with an x in the denominator.

Then, multiply both sides of the equation by 2.

In order to solve this polynomial equation, we can use the Rational Root Theorem. According to this theorem, if there is a rational root to a polynomial equation, then that root must be in the form p/q, where p is a factor of the constant, and q is the factor of the coefficient of the highest term.

In the context of this problem, p will be a factor of 4 (which is the constant), and q will be a factor of 2 (which is the coefficient of the highest term).

The factors of 4 are . The factors of 2 are .

This means that if there is a rational root for the equation, then the numerator must be either , and the denominator must be one of . In other words, the possible rational roots are , because these are all the possible ratios we can make with a numerator of and a denominator of one of .

At this point, we have narrowed down the rational roots of the equation to the following eight choices: .

Notice, that if we were to let , then the polynomial would evaluate to 4. Also, if we were to let , then the polynomial would evaluate to -3. Because the value of the polynomial switches from a postive value of 4 when x = 0 to a negative value of -3 when x = -1, there must be a root somewhere between and .

Let's see what happens when , because -1/2 is between 0 and -1.

We will put -1/2 into the polynomial to see if it evaluates to zero.

Thus is indeed a root to the equation.

However, the question asks us to find the largest real root; this means that there could be other roots larger than . We will need to continue solving the equation to see what other roots are possible.

We can make use of the Factor Theorem of polynomials, which states that if is a root of a polynomial equation, where a is constant, then x-a must be a factor of the polynomial.

This means that for our polynomial, is a factor. We can divide the polynomial by using long division.

This means that

In order to find the remaining roots of the equation, we now need to solve the quadratic equation .

First, divide both sides of the equation by two.

Perhaps the most straightforward way to solve this is to use the quadratic formula:

When we evaluate this, we will be forced to take the square root of . Because this will result in an imaginary number, has no real solutions.

Thus, returning to our original polynomial equation , it turns out that -1/2 is indeed the only root of the equation.

The question asks us to find the largest real value of x that solves the equation. Because -1/2 is the only real value that solves the equation, the answer must be -1/2.

You can see that each term in the equation has an "x", therefore by factoring "x" from each term you can get that the equation equals .

Use the FOIL method (First, Outer, Inner, Last):

Put all of these terms together:

Combine like terms:

Use the quadratic formula to solve:

This is a difference of perfect cubes so it factors to . Only the first expression will yield an answer when set equal to 0, which is 1. The second expression will never cross the -axis. Therefore, your answer is only 1.

Factor the equation to . Set and get one of your 's to be . Then factor the second expression to . Set them equal to zero and you get .

Use the difference of perfect cubes equation:

In ,

and

First consider all the factors of 12:

1 and 12

2 and 6

3 and 4

Then consider which of these pairs adds up to 7. This pair is 3 and 4.

Therefore the answer is .

Begin by extracting from the polynomial:

Now, factor the remainder of the polynomial as a difference of cubes:

Begin by rearranging like terms:

Now, factor out like terms:

Rearrange the polynomial:

Begin by rearranging like terms:

Now, factor out like terms:

Rearrange the polynomial:

Factor:

Begin by separating into like terms. You do this by multiplying and , then finding factors which sum to

Now, extract like terms:

Simplify:

To begin, distribute the squares:

Now, combine like terms:

Begin by extracting from the polynomial:

Now, distribute the cubic polynomial:

Begin by extracting like terms:

Now, rearrange the right side of the polynomial by reversing the signs:

Combine like terms:

Factor the square and cubic polynomial:

Begin by rearranging the terms to group together the quadratic:

Now, convert the quadratic into a square:

Finally, distribute the :

Begin by extracting from the polynomial:

Now, rearrange to combine like terms:

Extract the like terms and factor the cubic:

Simplify by combining like terms:

Begin by extracting from the polynomial:

Now, rearrange to combine like terms:

Extract the like terms and factor the cubic:

Simplify by combining like terms:

Put the negative exponent on the bottom so that you have which simplifies further to .

First, multiply out the second expression so that you get .

Then, multiply your like terms, taking care to remember that when multiplying terms that have the same base, you add the exponents. Thus, you get .