Solving Non Quadratic Polynomials - Algebra 2

This can be most easily solved by setting and, subsequently, . This changes the degree-4 polynomial in to one that is quadratic in , which can be solved as follows:

The quadratic factors do not fit any factoring pattern and are prime, so this is as far as the polynomial can be factored.

To find , we must start inwards and work our way outwards, i.e. starting with :

We can now use this value to find as follows:

Our final answer is therefore

First, we can factor a from both terms:

Now we can make a clever substitution. If we make the function now looks like:

This makes it much easier to see how we can factor (difference of squares):

The last thing we need to do is substitute back in for , but we first need to solve for by taking the square root of each side of our substitution:

Substituting back in gives us a result of:

Using the difference of cubes formula:

Find x and y:

Plug into the formula:

Which Gives:

And cannot be factored more so the above is your final answer.

Use the difference of squares to factor out the numerator.

The term is prime, but can still be factored by another difference of squares.

Replace the fraction.

Simplify the top and bottom.

The answer is:

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept .

By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

Set each factor to zero and solve:

Since no real number squared is equal to a negative number, no real solution presents itself here.

The solution set is .

The first step is to determine if all of the terms have a greatest common factor (GCF). Since a GCF does not exist, we can move onto the next step.

Create smaller groups within the expression. This is typically done by grouping the first two terms and the last two terms.

Factor out the GCF from each group:

At this point, you can see that the terms inside the parentheses are identical, which means you are on the right track!

Since there is a GCF of (5x+1), we can rewrite the expression like this:

And that is your answer! You can always check your factoring by FOILing your answer and checking it against the original expression.

This can be most easily solved by setting and, subsequently, . This changes the degree-4 polynomial in to one that is quadratic in , which can be solved as follows:

The quadratic factors do not fit any factoring pattern and are prime, so this is as far as the polynomial can be factored.

To find , we must start inwards and work our way outwards, i.e. starting with :

We can now use this value to find as follows:

Our final answer is therefore

First, we can factor a from both terms:

Now we can make a clever substitution. If we make the function now looks like:

This makes it much easier to see how we can factor (difference of squares):

The last thing we need to do is substitute back in for , but we first need to solve for by taking the square root of each side of our substitution:

Substituting back in gives us a result of:

Using the difference of cubes formula:

Find x and y:

Plug into the formula:

Which Gives:

And cannot be factored more so the above is your final answer.

Use the difference of squares to factor out the numerator.

The term is prime, but can still be factored by another difference of squares.

Replace the fraction.

Simplify the top and bottom.

The answer is:

Rewriting the equation as , we can see there are four terms we are working with, so factor by grouping is an appropriate method. Between the first two terms, the Greatest Common Factor (GCF) is and between the third and fourth terms, the GCF is 4. Thus, we obtain . Setting each factor equal to zero, and solving for , we obtain from the first factor and from the second factor. Since the square of any real number cannot be negative, we will disregard the second solution and only accept .

By substituting - and, subsequently, this can be rewritten as a quadratic equation, and solved as such:

We are looking to factor the quadratic expression as , replacing the two question marks with integers with product and sum 5; these integers are .

Substitute back:

The first factor cannot be factored further. The second factor, however, can itself be factored as the difference of squares:

Set each factor to zero and solve:

Since no real number squared is equal to a negative number, no real solution presents itself here.

The solution set is .

The first step is to determine if all of the terms have a greatest common factor (GCF). Since a GCF does not exist, we can move onto the next step.

Create smaller groups within the expression. This is typically done by grouping the first two terms and the last two terms.

Factor out the GCF from each group:

At this point, you can see that the terms inside the parentheses are identical, which means you are on the right track!

Since there is a GCF of (5x+1), we can rewrite the expression like this:

And that is your answer! You can always check your factoring by FOILing your answer and checking it against the original expression.

This can be most easily solved by setting and, subsequently, . This changes the degree-4 polynomial in to one that is quadratic in , which can be solved as follows:

The quadratic factors do not fit any factoring pattern and are prime, so this is as far as the polynomial can be factored.

To find , we must start inwards and work our way outwards, i.e. starting with :

We can now use this value to find as follows:

Our final answer is therefore

First, we can factor a from both terms:

Now we can make a clever substitution. If we make the function now looks like:

This makes it much easier to see how we can factor (difference of squares):

The last thing we need to do is substitute back in for , but we first need to solve for by taking the square root of each side of our substitution:

Substituting back in gives us a result of: