Algebra II For Dummies. Sterling Mary Jane

+ bd:

      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

,

, or

. The last pair is your choice, because 4 + 5 = 9. Arranging the factors and x’s into two binomials, you get x² + 9x + 20 = (x + 4)(x + 5).

Factoring four or more terms by grouping

      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).

      

Factoring by grouping only works if a new common factor appears – the exact same one in each term.

      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. In this chapter, you find out how to deal with linear equations, what to do with the answers in linear inequalities, and how to rewrite linear absolute value equations and inequalities so that you can solve them.

Linear Equations: Handling the First Degree

      Linear equations feature variables that reach only the first degree, meaning that the highest power of any variable you solve for is one. The general form of a linear equation with one variable is

      In this equation, the one variable is the x. The a, b, and c are coefficients and constants. (If you go to Chapter 12, you can see linear equations with two or three variables.) But, no matter how many variables you see, the common theme to linear equations is that each variable has only one solution or value that works in the equation.

      The graph of the single solution of a linear equation, if you really want to graph it, is one point on the number line – the answer to the equation. When you up the ante to two variables in a linear equation, the graph of all the solutions (there are infinitely many) is a straight line. Any point on the line is a solution. Three variable solutions means you have a plane – a flat surface.

      

Generally, algebra uses the letters at the end of the alphabet for variables; the letters at the beginning of the alphabet are reserved for coefficients and constants.

Tackling basic linear equations

      

To solve a linear equation, you isolate the variable on one side of the equation. You do so by adding the same number to both sides – or you can subtract, multiply, or divide the same number on both sides.

      For example, you solve the equation 4x – 7 = 21 by adding 7 to each side of the equation, to isolate the variable and the multiplier, and then dividing each side by 4, to leave the variable on its own:

      4x –7 + 7 = 21 + 7 becomes 4x = 28

      giving you x = 7.

      When a linear equation has grouping symbols such as parentheses, brackets, or braces, you deal with any distributing across and simplifying within the grouping symbols before you isolate the variable. For instance, to solve the equation 5x – [3(x + 2) – 4(5 – 2x) + 6] = 20, you first distribute the 3 and –4 inside the brackets:

      You then combine the terms that combine and distribute the negative sign (–) in front of the bracket; it’s like multiplying through by –1:

      Simplify again, and you can solve for x:

      

When distributing a number or negative sign over terms within a grouping symbol, make sure you multiply every term by that value or sign. If you don’t multiply each and every term, the new expression won’t be equivalent to the original.

      To check your answer from the previous example problem, replace every x in the original equation with –2. If you do so, you get a true statement. In this case, you get 20 = 20. The solution –2 is the only answer that works – focusing your work on just one answer is what’s nice