Outer, Inner, Last); unFOIL helps you factor the product of those binomials.
Finding perfect square trinomials
To factor x2 – 20x + 100, for example, you should first recognize that 20x is twice the product of the root of x2 and the root of 100; therefore, the factorization is (x – 10)2. An expression that isn’t quite as obvious is 25y2 + 30y + 9. You can see that the first and last terms are perfect squares. The root of 25y2 is 5y, and the root of 9 is 3. The middle term, 30y, is twice the product of 5y and 3, so you have a perfect square trinomial that factors into (5y + 3)2.
Resorting to unFOIL
When you factor a trinomial that results from multiplying two binomials, you have to play detective and piece together the parts of the puzzle. Look at the following generalized product of binomials and the pattern that appears:
So, where does FOIL come in? You need to FOIL before you can unFOIL, don’t ya think?
The F in FOIL stands for “First.” In the previous problem, the First terms are the ax and cx. You multiply these terms together to get acx2. The Outer terms are ax and d. Yes, you already used the ax, but each of the terms will have two different names. The Inner terms are b and cx; the Outer and Inner products are, respectively, adx and bcx. You add these two values. (Don’t worry; when you’re working with numbers, they combine nicely.) The Last terms, b and d, have a product of bd. Here’s an actual example that uses FOIL to multiply – working with numbers for the coefficients rather than letters:
Now, think of every quadratic trinomial as being of the form acx2 + (ad + bc)x + bd. The coefficient of the x2 term, ac, is the product of the coefficients of the two x terms in the parentheses; the last term, bd, is the product of the two second terms in the parentheses; and the coefficient of the middle term is the sum of the outer and inner products. To factor these trinomials into the product of two binomials, you can use the opposite of the FOIL, which I call unFOIL.
1. Determine all the ways you can multiply two numbers to get ac, the coefficient of the squared term.
2. Determine all the ways you can multiply two numbers to get bd, the constant term.
3. If the last term is positive, find the combination of factors from Steps 1 and 2 whose sum is that middle term; if the last term is negative, you want the combination of factors to be a difference.
4. Arrange your choices as binomials so that the factors line up correctly.
5. Insert the + and – signs to finish off the factoring and make the sign of the middle term come out right.
To factor x2 + 9x + 20, for example, you need to find two terms whose product is 20 and whose sum is 9. The coefficient of the squared term is 1, so you don’t have to take any other factors into consideration. You can produce the number 20 with
When four or more terms come together to form an expression, you have bigger challenges in the factoring. You see factoring by grouping in the previous section as a method for factoring trinomials; the grouping is pretty obvious in this case. But what about when you’re starting from scratch? What happens with exponents greater than two? As with an expression with fewer terms, you always look for a greatest common factor first. If you can’t find a factor common to all the terms at the same time, your other option is grouping. To group, you take the terms two at a time and look for common factors for each of the pairs on an individual basis. After factoring, you see if the new groupings have a common factor. The best way to explain this is to demonstrate the factoring by grouping on x3 – 4x2 + 3x – 12 and then on xy2 – 2y2 – 5xy + 10y – 6x + 12.
The four terms x3 – 4x2 + 3x – 12 don’t have any common factor. However, the first two terms have a common factor of x2, and the last two terms have a common factor of 3:
Notice that you now have two terms, not four, and they both have the factor (x – 4). Now, factoring (x – 4) out of each term, you have (x – 4)(x2 + 3).
The six terms xy2 – 2y2 – 5xy + 10y – 6x + 12 don’t have a common factor, but, taking them two at a time, you can pull out the factors y2, –5y, and –6. Factoring by grouping, you get the following:
The three new terms have a common factor of (x – 2), so the factorization becomes (x – 2)(y2 – 5y – 6). The trinomial that you create also factors (see the previous section):
Factored, and ready to go!
Chapter 2
Toeing the Straight Line: Linear Equations
In This Chapter
▶ Isolating values of x in linear equations
▶ Comparing variable values by using inequalities
▶ Assessing absolute value in equations and inequalities
The term linear has the word line buried in it, and the obvious connection is that you can graph many linear equations as lines. But linear expressions can come in many types of packages, not just equations of lines. Add an interesting operation or two, put several first-degree terms together, throw in a funny connective, and you can construct all sorts of creative mathematical challenges.
