By the end of this section, you should be able to:

1. Factor polynomials

This is yet another computational lesson, so be sure to try some of the practice flashcards.

In this lesson, we'll take a look at factoring polynomials. To factor something means to write it as a product of two or more divisors, without leaving a remainder. As we learned in "Operations With Polynomials and Radicals", a polynomial is an expression that combines multiple (poly-) terms separated by addition, subtraction, multiplication, and/or division.

For example, \(\color{red}{a}^2+\color{red}{a}b+\color{red}{a}c\) can be factored into \(\color{red}{a}(a+b+c)\). There are four main ways of factoring which we will discuss in this lesson:

**Common Factoring**

Common factoring is as we did in the example; it means to isolate a term that is common throughout the polynomial. For example, we can factor \(9x^2 + 27x\), which can be rewritten as \((\color{red}{x})(\color{red}{9})(x) + (\color{red}{9})(3)(\color{red}{x})\) into \(\color{Red}{9x}(x+3)\), by common factoring the 9 and \(x\) terms.

Another example. Factor \(4x^4 + x^3+ 28x^2+7x \).

This one looks tricky, but let's try to expand our terms first: \(4(x)(x)(x)(x) + (x)(x)(x) + 7(4)(x)(x) + 7(x)\). We find that we have two terms from which we can factor out \(x\) and 7, and another two terms from which we can factor out \(x^2\) and 4.

Factoring these out, we get two factored terms: \(4x^3(4x+1) + 7x(4x+1)\). Notice something? We can common factor out these common factors! It's common factor-ception!

Simplifying this expression, we get \((4x+1)(4x^3+7x)\). We can then factor this further by common factoring an \(x\) from the second term, giving \(x(4x+1)(4x^2+7)\).

WOW! Just how much further deeper can this go? Not much, unfortunately; we've fully factored this polynomial. Onwards!

**Factoring a Difference of Squares**

In the lesson "Operations With Polynomials and Radicals", we were (re-)introduced to perfect squares; integers that are the square of another integer. We can use this knowledge to our advantage to factor out a special type of polynomial; called a difference of squares.

The general form for factoring a difference of squares is \(a^2 - b^2 = (a+b)(a-b)\). Let's try some examples:

Factor \(16x^2 - 4\).

Well, we can first start with what we already know: common factoring. Factoring out the number 4, we get \(4(x^2 -1)\). Here, we have a difference of squares; \(x^2\) is the square of \(x\) and 1 is the square of... well, 1. Following our formula, we get \(4(x+1)(x-1)\). With that, the expression is factored!

Another one. Factor \((4x+9)^2-9\).

This one is somewhat in disguise as the square is the sum of **two** terms, rather than only just one term. Letting \(a = 4x+3\) and \(b=3\), we have a difference of squares. Using our formula, we get \(((4x+9)+3)((4x+9)-3)\). Simplifying, we get \((4x+12)(4x+6)\). We can further simplify this expression by common factoring 4 from the first term and 2 from the second, giving: \((4)(x+3)(2)(x+3)\). WOW! This polynomial just keeps on giving. Expressing the \((x+3)\) terms as a square and simplifying, we get \(8(x+3)^2\)... and voilà! The expression is fully factored.

**Factoring Polynomials Where **\(\boldsymbol{a=1}\)

As you might know, we can express a polynomial as \(ax^2 + bx + c\). In this case, we will look at polynomials where the first term, which precedes the square, does not have a coefficient; that is, it's coefficient is 1.

For example, factor \(x^2 + 20x + 75\).

In this case, we are looking for two numbers which multiply to 75 and add to 20. Writing out the factors of 75 in a table, we can find which products we are looking for:

Aha! 5 and 15 are the numbers we need. We can now factor the expression, giving us \((x+5)(x+15)\).

Factor \(x^2 -8x + 7\).

Now, we are looking for two numbers that add to -8 and multiply to positive 7. The only way two numbers can be negative when added and positive when multiplied is if they were both negative. Luckily, we don't have to rewrite the factors of 7 as it is prime; its only factors are 1 and 7; in this case, -1 and -7. Factoring the expression, we get \((x-1)(x-7)\).

We can check our math by expanding the polynomial. As you might remember, this is done through FOIL, an acronym for the order in which we multiply binomial (two-numbered) terms—first, outer, inner, last. This is illustrated in two different ways in the diagram below:

Applying this technique to our function, we get \((x)(x) + 7(x) + (x) + 7\) = \(x^2 + 8x + 7\); right on the money.

Another example: factor \(x^4+22x^2+96\).

You might notice that the first two terms have been raised to the fourth and second power. No need to worry; we can use variables to eliminate this problem.

Let \(u = x^2\). Substituting into our original equation, we get \(u^2 + 22u + 96\). We need two numbers that multiply to 96 and add to 22; 16 and 6 come to mind (through practice, you will gain spidey-senses enabling you to find factors like nobody's business).

Factoring, we get \((u+16)(u+6)\). Resubstituting \(u = x^2\), we get \((x^2 + 16)(x^2+6)\). You might be tempted to treat \(x^2 + 16\), because it has two squares, as a difference of squares. However, it's important to remember that the terms must be separated by a subtraction symbol in order for that technique to apply; not the case here!

**Factoring Polynomials Where **\(\boldsymbol{a\neq1}\)

Factor \(9x^2 + 30x + 25\).

As we can see here, \(a=9\). In order to factor this type of expression we will need two terms: \((ax+b)(cx+d)\) where \(a \times c = 9\), \(b \times d = 25\), and \(ad + bc = 30\).

We know this because the only way we can get an \(x^2\) term is by multiplying \(ax \times cx\), giving \(acx^2\). Because we know that our only \(x^2\) term has coefficient 9, \(a \times c = 9\). Our constant term, 25, can only come from multiplying our constants, \(b \times d\). Therefore, \(b \times d = 25\). Finally, our terms with x must arise from multiplying a constant and an x-term. This can only occur in two ways: \(a \times d\), or \(b \times c\). We add these two possibilities to get the coefficient of our x-term, 30. Therefore, \(ad + bc = 30\).

Here's a diagram to clarify:

The only way to solve this problem (aside from the blessed quadratic equation \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)) is by trial and error. Luckily, on the computer, I am able to harness the powers of Bill Gates and use his crown jewel, Microsoft Excel. Below is a table which illustrates how we find the coefficients for factoring:

From the table, the only row where all three criteria (\(a \times c = 9\), \(b \times d = 25\), and \(ad + bc = 30\)) are met is the second. Therefore, we know that \(a = 3\), \( b= 5\), \( c= 3\), and \(d= 5\). This gives \((3x+5)(3x+5)\), or, more simply \((3x+5)^2\).

Unfortunately, the only trick to these types of problems is practice, practice, practice. Speaking of practice, have you tried those flashcards yet?

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