To factor means to write an expression as a product instead of a sum. Factoring is particularly useful when solving equations set equal to zero because then logically at least one factor must be equal to zero. In PreCalculus, you should be able to factor even when there is no obvious greatest common factor or the difference is not between two perfect squares.

How do you use the difference of perfect squares factoring technique on polynomials that don’t contain perfect squares and why would this be useful?

#### Watch This

http://www.youtube.com/watch?v=EDebmfT5Nsk James Sousa: Factoring: Greatest Common Factor

http://www.youtube.com/watch?v=0yBDsZvfT0g James Sousa: Factoring: Difference of Squares

http://www.youtube.com/watch?v=cWapvAWXdoY James Sousa: Factoring: Basic Trinomials with a=1

#### Guidance

A polynomial is a sum of a finite number of terms. Each term consists of a constant that multiplies a variable. The variable may only be raised to a non-negative exponent. The letters \begin{align*}a, b, c \ldots\end{align*} in the following general polynomial expression stand for regular numbers like \begin{align*}0, 5, -\frac{1}{4}, \sqrt{2}\end{align*} and the \begin{align*}x\end{align*} represents the variable.

\begin{align*}ax^n+bx^{n-1}+ \ldots + fx^2+gx+h\end{align*}

You have already learned many properties of polynomials. For example, you know the commutative property which states that terms of a polynomial can be rearranged to create an equivalent polynomial. When two polynomials are added, subtracted or multiplied the result is always a polynomial. This means polynomials are closed under addition, and is one of the properties that makes the factoring of polynomials possible. Polynomials are not closed under division because dividing two polynomials could result in a variable in the denominator, which is a rational expression (not a polynomial).

There are three methods for factoring that are essential to master. The first method you should always try is to factor out the greatest common factor (GCF) of the expression (see Example A). The second method you should implement after factoring out a GCF is to see if you can factor the expression into the product of two binomials (see Example B). This type of factoring is usually recognizable as a trinomial where \begin{align*}x^2\end{align*} has a coefficient of 1. The third type of basic factoring is the difference of squares. It is recognizable as one square monomial being subtracted from another square monomial.

The rigor of the following factoring examples and exercises is greater than an introductory level factoring lesson but the techniques are the same.

**Example A**

Use the GCF technique to factor the following expression. Check that the factored expression matches the original.

\begin{align*}-\frac{1}{2}x^4+\frac{7}{2}x^2-6\end{align*}

**Solution:** Many students just learning factoring may conclude that the three terms share no factors besides one. However, the name GCF is deceiving because this expression has an infinite number of equivalent expressions many of which are more useful. In order to find these alternative expressions you must strategically factor numbers that are neither the greatest factor nor common to all three terms. In this case, \begin{align*}-\frac{1}{2}\end{align*} is an excellent choice.

\begin{align*}-\frac{1}{2}x^4+\frac{7}{2}x^2-6=-\frac{1}{2}(x^4-7x^2+12)\end{align*}

In order to check to see that this is an equivalent expression, you need to distribute the \begin{align*}-\frac{1}{2}\end{align*}. When you distribute, the first coefficient matches because it just gets multiplied by 1, the second term becomes \begin{align*}\frac{7}{2}\end{align*} and the third term becomes -6.

**Example B**

Factor the expression from Example A into the product of two binomials and a constant.

\begin{align*}-\frac{1}{2}(x^4-7x^2+12)\end{align*}

**Solution: ** Many students familiar with basic factoring may be initially stuck on a problem like this. However, you should recognize that beneath the \begin{align*}4^{th}\end{align*} degree and the \begin{align*}-\frac{1}{2}\end{align*} the problem boils down to being able to factor \begin{align*}u^2-7u+12\end{align*} which is just \begin{align*}(u-3)(u-4)\end{align*}.

*Start by rewriting the problem: \begin{align*}-\frac{1}{2}(x^4-7x^2+12)\end{align*}*

*Then choose a temporary substitution:* Let *\begin{align*}u=x^2\end{align*}. *

*Then substitute and factor away. Remember to substitute back at the end.*

*\begin{align*}-\frac{1}{2}(u^2-7u+12) &=-\frac{1}{2}(u-3)(u-4)\\
&= -\frac{1}{2}(x^2-3)(x^2-4)\end{align*}*

This type of temporary substitution that enables you to see the underlying structure of an expression is very common in calculus.

**Example C**

Factor the resulting expression from Example B into four linear factors and a constant.

\begin{align*}-\frac{1}{2}(x^2-3)(x^2-4)\end{align*}

**Solution:** Many students may recognize that \begin{align*}x^2-4\end{align*} immediately factors by the difference of squares method to be \begin{align*}(x-2)(x+2)\end{align*}. This problem asks for more because sometimes the difference of squares method can be applied to expressions like \begin{align*}x^2-3\end{align*} where each term is not a perfect square. The number 3 actually is a square.

\begin{align*}3=\left(\sqrt{3}\right)^2\end{align*}

So the expression may be factored to be:

\begin{align*}-\frac{1}{2} \left(x- \sqrt{3}\right)\left(x+ \sqrt{3}\right)(x-2)(x+2)\end{align*}

**Concept Problem Revisited**

One reason why it might be useful to completely factor an expression like \begin{align*}-\frac{1}{2}(x^4-7x^2+12)\end{align*} into linear factors is if you wanted to find the roots of the function \begin{align*}f(x)=-\frac{1}{2}(x^4-7x^2+12)\end{align*}. The roots are \begin{align*}x=\pm \sqrt{3}, \pm 2\end{align*}.

You should recognize that \begin{align*}x^2-3\end{align*} can still be thought of as the difference of perfect squares because the number 3 can be expressed as \begin{align*}\left(\sqrt{3}\right)^2\end{align*}. Rewriting the number 3 to fit a factoring pattern that you already know is an example of using the basic factoring techniques at a PreCalculus level.

#### Vocabulary

A ** polynomial** is a mathematical expression that is often represented as a sum of terms or a product of factors.

** To factor** means to rewrite a polynomial expression given as a sum of terms into a product of factors.

** Linear factors** are expressions of the form \begin{align*}ax+b\end{align*} where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are real numbers.

#### Guided Practice

1. Factor the following expression into strictly linear factors if possible. If not possible, explain why.

\begin{align*}\frac{x^5}{3}-\frac{11x^3}{3}+6x\end{align*}

2. Factor the following expression into strictly linear factors if possible. If not possible, explain why.

\begin{align*}-\frac{2}{7}x^4+\frac{74}{63}x^2-\frac{8}{63}\end{align*}

3. Factor the following expression into strictly linear factors if possible. If not possible, explain why.

\begin{align*}x^4+x^2-72\end{align*}

**Answers:**

1. \begin{align*}\frac{x^5}{3}-\frac{11x^3}{3}+6x\end{align*}

\begin{align*}&= \frac{1}{3}x(x^4-11x^2+18)\\ &=\frac{1}{3}x(x^2-2)(x^2-9)\\ &=\frac{1}{3}x(x+\sqrt{2})(x-\sqrt{2})(x+3)(x-3)\end{align*}

2. \begin{align*}-\frac{2}{7}x^4+\frac{74}{63}x^2-\frac{8}{63}\end{align*}. Let \begin{align*}u=x^2\end{align*}.

\begin{align*}&= -\frac{2}{7}u^2+\frac{74}{63}u-\frac{8}{63}\\ &= -\frac{2}{7} \left(u^2-\frac{37}{9}u+\frac{4}{9}\right)\end{align*}

Factoring through fractions like this can be extremely tricky. You must recognize that \begin{align*}-\frac{1}{9}\end{align*} and -4 sum to \begin{align*}-\frac{37}{9}\end{align*} and multiply to \begin{align*}\frac{4}{9}\end{align*}.

\begin{align*}&= -\frac{2}{7} \left(u-\frac{1}{9}\right)(u-4)\\ &= -\frac{2}{7} \left(x^2-\frac{1}{9}\right)(x^2-4)\\ &= -\frac{2}{7} \left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right)(x-2)(x+2)\end{align*}

3. \begin{align*}x^4+x^2-72\end{align*}

\begin{align*}=(x^2-8)(x^2+9)\end{align*}

Notice that \begin{align*}(x^2-8)\end{align*} can be written as the difference of perfect squares because \begin{align*}8=\left(\sqrt{8}\right)^2=\left(2\sqrt{2}\right)^2\end{align*}. On the other hand, \begin{align*}x^2+9\end{align*} cannot be written as the difference between squares because the \begin{align*}x^2\end{align*} and the 9 are being added not subtracted. This polynomial cannot be factored into strictly linear factors.

\begin{align*}=(x-2 \sqrt{2})(x+2 \sqrt{2})(x^2+9)\end{align*}

#### Practice

Factor each polynomial into strictly linear factors if possible. If not possible, explain why not.

- \begin{align*}x^2+5x+6\end{align*}
- \begin{align*}x^4+5x^2+6\end{align*}
- \begin{align*}x^4-16\end{align*}
- \begin{align*}2x^2-20\end{align*}
- \begin{align*}3x^2+9x+6\end{align*}
- \begin{align*}\frac{x^4}{2}-5x^2+\frac{9}{2}\end{align*}
- \begin{align*}\frac{2x^4}{3}-\frac{34x^2}{3}+\frac{32}{3}\end{align*}
- \begin{align*}x^2-\frac{1}{4}\end{align*}
- \begin{align*}x^4-\frac{37x^2}{4}+\frac{9}{4}\end{align*}
- \begin{align*}\frac{3}{4}x^4-\frac{87}{4}x^2+75\end{align*}
- \begin{align*}\frac{1}{2}x^4-\frac{29}{2}x^2+50\end{align*}
- \begin{align*}\frac{x^4}{2}-\frac{5x^2}{9}+\frac{1}{18}\end{align*}
- \begin{align*}x^4-\frac{13}{36}x^2+\frac{1}{36}\end{align*}
- How does the degree of a polynomial relate to the number of linear factors?
- If a polynomial does not have strictly linear factors, what does this imply about the type of roots that the polynomial has?