Factoring a polynomial of degree 4 or higher can be a difficult task. However, some polynomials of higher degree can be written in quadratic form, and the techniques used to factor quadratic functions can be utilized.

## Changing Form

Factoring a polynomial, such as *x*4 – 29*x*2 + 100 might seem intimidating. In this lesson, you will learn how to change the form of certain polynomials of higher degree so that they are much easier to factor.

Let’s first discuss changing form in the world of insects. Insects typically go through multiple life stages, and the appearance of some insects is very different from one life stage to the next. In other words, they change form. Consider the monarch butterfly: in its larval stage, it is in the form of a caterpillar. During this adolescent stage, the caterpillar is eating leaves like there is no tomorrow. At the right time, the monarch caterpillar prepares for its transformation to adulthood. It emerges from its shell to reveal a radical biological change. The caterpillar has changed form so that it can now enjoy the life and responsibilities of an adult butterfly.

In mathematics, we are usually not changing the form of living things. However, changing the forms of mathematical expressions might be effective in finding solutions to problems, like the one I mentioned at the beginning of the lesson.

## The Quadratic Form

Let’s look at the **standard form** of the quadratic function in one variable: *y* = *ax*2 + *bx* + *c*. A quadratic function in one variable has a degree of 2 because the variable of the leading term has an exponent of 2. The second term in *x* (*bx*) actually has an exponent of 1, but this exponent is generally not shown. The letters *a*, *b*, and *c* represent real numbers, except that *a* cannot equal zero.

Let’s consider the following quadratic equation: *x*2 + 4*x* – 21 = 0. We can factor this equation as follows: (*x* + 7)(*x* – 3) = 0. We can now use the zero product property to solve the equation: *x* + 7 = 0, so *x* = -7. *x* – 3 = 0, so *x* = 3. However, the remainder of this lesson will focus on the task of **factoring**.

## Changing to Quadratic Form

Now let’s look at a polynomial expression that has a degree higher than two, as follows: *x*4 + *x*2 – 12. This expression does not fit our definition of a quadratic function in one variable because it has a degree of 4. However, we can rewrite it in quadratic form. Remember that the lead term in a quadratic expression has an exponent of 2 and the other term *x* has an exponent of 1. We can first rewrite the expression as follows:

(*x*2)2 + (*x*2)1 – 12

Let’s do one more thing: let’s substitute *u* for *x*2. In other words, *u* = *x*2. We now get the following: *u*2 + *u* – 12. There we have it: a quadratic expression! It is now much easier to factor. In factored form, it is (*u* + 4)(*u* – 3). Remember that *u* = *x*2. Now change *u* back to the term that it originally replaced: (*x*2 + 4)(*x*2 – 3). Our original expression is now factored:

*x*4 + *x*2 – 12 = (*x*2 + 4)(*x*2 – 3)

You might have noticed that if we square the second term in *x*, we get a term with an exponent of 4, just like the lead term. In other words, (*x*2)2 = *x*4. This will always be the case for polynomial expressions that can be written in quadratic form.

We now need to define **quadratic form**. An expression is in quadratic form if we can rewrite in the form of *au*2 + *bu* + *c* and *u* is any expression in *x*. Please note that the variable *u* can be used to replace any expression in *x*. Therefore it will not always equal *x*2. It will replace an appropriate expression so that we can write a polynomial in quadratic form as in our next example.

## Example 1

Write the expression in quadratic form if possible:

3*x*8 – 22*x*4 – 16

We can rewrite the expression as 3(*x*4)2 – 22(*x*4)1 – 16. We can substitute *u* for *x*4 and rewrite the expression as follows: 3*u*2 – 22*u* – 16. We now have an expression in quadratic form.

## Example 2

Write the expression in quadratic form if possible:

*x*10 – 5*x*4 + 9

We can rewrite the first term as *x*10 = (*x*5)2. We can then rewrite the entire expression as (*x*5)2 – 5(*x*4)1 + 9. Notice that our terms in *x* are not the same: *x*5 does not equal *x*4. We cannot replace both of these terms with the same variable. If we substitute *u* for *x*5, we cannot substitute *u* for *x*4, and vice versa. Therefore this function cannot be written in quadratic form. Also note that the square of *x*^4 ((*x*4)2) equals *x*8, which is not the same as the lead term *x*10.

## Example 3

Factor the polynomial expression *x*4 – 29*x*2 + 100. We can rewrite the expression as (*x*2)2 – 29(*x*2)1 + 100, so we can substitute *u* for *x*2 and rewrite the expression as follows:

*u*2 – 29*u* + 100

The quadratic form of the expression can now be put in factored form as (*u* – 25)(*u* – 4). Remember that *u* = *x*2 in this problem, so change *u* back to *x*2 to get (*x*2 – 25)(*x*2 – 4). Both factors are a difference of squares and can be further factored into the following:

(*x* – 5)(*x* + 5)(*x* – 2)(*x* + 2)

## Example 4

Factor the polynomial expression *x*6 – 49. We can rewrite the expression as (*x*3)2 – 49. In this example there is not a second term in terms of *x*; therefore we only have to worry about changing the lead term. So we can substitute *u* for *x*3 and rewrite the expression as follows: *u*2 – 49. The quadratic form of the expression can now be put into factored form as (*u* – 7)(*u* + 7). Remember that *u* is equal to *x*3 in this problem, so change *u* back to *x*3 to get:

(*x*3 – 7)(*x*3 + 7)

## Lesson Summary

Factoring polynomials of a higher degree can be quite a task. However, factoring quadratic expressions is usually easier. Therefore, try to see if you can write a polynomial of higher degree in quadratic form. This will allow you to use certain factoring techniques that are used to factor quadratic expressions.

## Learning Outcome

Once you have completed this lesson, you should be able to demonstrate how to factor polynomials using quadratic form.