2.1: 2.1 Factoring Review

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?

Factoring Functions

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 \(a, b, c \ldots\) in the following general polynomial expression stand for regular numbers like \(0,5,-\frac{1}{4}, \sqrt{2}\) and the \(x\) represents the variable.

\(a x^{n}+b x^{n-1}+\ldots+f x^{2}+g x+h\)

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.

Greatest Common Factor Method

The first method you should always try is to factor out the greatest common factor (GCF) of the expression.

To factor the following expression, first apply the GCF method:

\(-\frac{1}{2} x^{4}+\frac{7}{2} x^{2}-6\)

To find the GCF, it is common to try to factor out the \(a\) value. In this case, try factoring out \(-\frac{1}{2}\)

\(-\frac{1}{2} x^{4}+\frac{7}{2} x^{2}-6=-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\)

In order to check to see that this is an equivalent expression, you need to distribute the \(-\frac{1}{2}\). When you distribute, the first coefficient matches because it just gets multiplied by 1 , the second term becomes \(\frac{7}{2}\) and the third term becomes - \(6 .\) Note that this expression is not completely factored yet but it is simplified as much as it can be with just the GCF method.

Factoring Into Binomials Method

The second method you should try is to see if you can factor the expression into the product of two binomials.

To continue factoring the expression from the Greatest Common Factor section, factor the following expression into the product of two binomials and a constant:

\(-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\)

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

Start by rewriting the problem: \(-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\)

Then choose a temporary substitution: Let \(u=x^{2}\).

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

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

This type of temporary substitution that enables you to see the underlying structure of an expression is very common in calculus. The expression is still not completely factored and since there are no more trinomials, you must apply the last method.

Difference of Squares Method

The third method of basic factoring is the difference of squares. It is recognizable as one square monomial being subtracted from another square monomial.

To finish factoring the resulting expression from the Factoring Into Binomials section, factor the expression into four linear factors and a constant:

\(-\frac{1}{2}\left(x^{2}-3\right)\left(x^{2}-4\right)\)

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

\(3=(\sqrt{3})^{2}\)

So the fully factored expression would be:

\(-\frac{1}{2}(x-\sqrt{3})(x+\sqrt{3})(x-2)(x+2)\)

Examples

Example 1

Earlier, you were asked how you use the difference of perfect squares factoring technique on polynomials that don't contain perfect squares and why it would be useful. One reason why it might be useful to completely factor an expression like \(-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\) into linear factor is if you wanted to find the roots of the function \(f(x)=-\frac{1}{2}\left(x^{4}-7 x^{2}+12\right)\). The roots are \(x=\pm \sqrt{3},\pm 2\)

You should recognize that \(x^{2}-3\) can still be thought of as the difference of perfect squares because the number 3 can be expressed as \((\sqrt{3})^{2}\). 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.

Example 2

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

\(\begin{aligned} \frac{x^{5}}{3}-\frac{11 x^{3}}{3}+6 x &=\frac{1}{3} x\left(x^{4}-11 x^{2}+18\right) \\ &=\frac{1}{3} x\left(x^{2}-2\right)\left(x^{2}-9\right) \\ &=\frac{1}{3} x(x+\sqrt{2})(x-\sqrt{2})(x+3)(x-3) \end{aligned}\)

Example 3

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

\(-\frac{2}{7} x^{4}+\frac{74}{63} x^{2}-\frac{8}{63}\)

For \(-\frac{2}{7} x^{4}+\frac{74}{63} x^{2}-\frac{8}{63},\) let \(u=x^{2}\).

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

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

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

Example 4

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

\(x^{4}+x^{2}-72\)

\(x^{4}+x^{2}-72=\left(x^{2}-8\right)\left(x^{2}+9\right)\)

\(x^{4}+x^{2}-72=\left(x^{2}-8\right)\left(x^{2}+9\right)\)

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

\(x^{4}+x^{2}-72=(x-2 \sqrt{2})(x+2 \sqrt{2})\left(x^{2}+9\right)\)

Review

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

1. \(x^{2}+5 x+6\)

2. \(x^{4}+5 x^{2}+6\)

3. \(x^{4}-16\)

4. \(2 x^{2}-20\)

5. \(3 x^{2}+9 x+6\)

6. \(\frac{x^{4}}{2}-5 x^{2}+\frac{9}{2}\)

7. \(\frac{2 x^{4}}{3}-\frac{34 x^{2}}{3}+\frac{32}{3}\)

8. \(x^{2}-\frac{1}{4}\)

9. \(x^{4}-\frac{37 x^{2}}{4}+\frac{9}{4}\)

10. \(\frac{3}{4} x^{4}-\frac{87}{4} x^{2}+75\)

11. \(\frac{1}{2} x^{4}-\frac{29}{2} x^{2}+50\)

12. \(\frac{x^{4}}{2}-\frac{5 x^{2}}{9}+\frac{1}{18}\)

13. \(x^{4}-\frac{13}{36} x^{2}+\frac{1}{36}\)

14. How does the degree of a polynomial relate to the number of linear factors?

15. If a polynomial does not have strictly linear factors, what does this imply about the type of roots that the polynomial has?