Module 3 Study Guide The second module covers the following sections of the textbook: 5.4-5.8 and 6.1-6.5. Most people would consider this the hardest module of the semester. Really, it boils down to your ability to understand factoring -- if you know how to factor polynomials, then you will succeed with this module. Section 5.4 Really, section 5.4 would fit better in the last module, but needed to be moved due to even things out. The topic is division of polynomials. Objective A deals with dividing a polynomial by a monomial, and should be pretty easy to swallow. This example summarizes perfectly: In order for this to work, each term in the numerator must be perfectly divisible by the denominator. So, if the numerator were 16 x 5 8 x 3 +4 instead, this simplification would not be possible. Notice that objective B is NOT listed as a required objective. This is division of two polynomials. You should definitely understand the concept and look through the examples, but you will not be required to divide two general polynomials in the quiz. Finally, objective C introduces a new technique called synthetic division. Synthetic division is a very useful technique that allows you to divide polynomials by a divisor of the form x + a or x - a. The process works because of the fact that we limit the divisor to these two forms. You will be required to perform synthetic division on the quiz. Sections 5.5-5.7, Factoring The next three sections tackle factoring. You should already have a good handle on factoring from your previous algebra classes. The techniques you probably already know are the GCF method, the difference of two perfect squares method, the trinomial x 2 +b x+c method, and hopefully, the trinomial a x 2 +b x+c method. Quick examples of each of these below: GCF Method: Notice that a polynomial like 2x 2 8 xy+9 y 2 can't be factored by this method. has no common factor among all three terms, so it
Difference of Perfect Squares: a 2 b 2 =(a+b)(a b) Notice you CANNOT factor the sum of perfect squares. So 4 x 2 + 81 y 2 not by this technique. can't be factored, at least Note: I will not require you to learn "Factoring a Perfect Square Trinomial", since this can usually be accomplished by other techniques. Factoring trinomials of the form x 2 +b x+c... you need to find factors of the number "c" that add up to "b". For example, in the problem below, we need to find factors of -18 (for example, -6 and 3), that add up to 7. Our example -6 and 3 doesn't work (they add up to 3), but... Factoring trinomials of the form a x 2 +b x+c using trial factors. Similar to above, but now you need to account for the leading term ax 2 when setting up your trial. The key here is to break up your factors so the F and L in FOIL list all possibilities, and then keep trying until you get back the correct middle term. So, for example: 6x 2 x 15 Possible F (first) combinations (3x )(2x ) and (6x )(x )... don't need to worry about the order here, we'll take care of that with the L's (Lasts) Possible L (last) combinations (numbers that multiply to -15) : -1 & 15, -3 & 5, in all orders and combinations. So, in other words, we need to try -15 & 1, 3 & -5, 5 & -3...
Once you've determined all possible Firsts and Lasts, now perform the foil method on each combination until you get back the middle term (-x = -1x). Try for both (6x )(x ) and (3x )(2x ). Middle Term Equals -1x Notice (6x + 1)(x - 15) -90x + 1x no (6x - 1)(x + 15) gives +90x - 1x (6x + 3)(x - 5) -30x + 3x no (6x - 3)(x + 5) gives 30x - 3x (6x + 5)(x - 3) -18x + 5x no (6x - 5)... (6x + 15)(x - 1) -6x + 15x no... (3x + 1)(2x - 15) -45x + 2x no (3x - 1)(x + 15) gives +45x -2x (3x + 3)(2x - 5)... no (3x + 5)(2x - 3) -9x + 10x CLOSE! (3x - 5)(2x + 3) gives 9x - 10x = -1x So, the correct answer is (3x - 5)(2x + 3). Persistance is the key! Be aware that many polynomials of this form will not factor, for example 7 x 2 3 x+4 doesn't factor, though one of the potential factorizations comes "close". What about 7 x 2 3 x 4? "New" factoring methods The factoring methods that you may not be familiar with are the grouping method, and sums/differences of perfect cubes. Grouping is pretty straightforward. Starting with 4 terms, you break into two groups, factor the GCF out of each group separately, then factor out the common binomial, if it exists. Example: 3 x 2 12 xy + 5 xy 20 y 2 = 3 x 2 12 xy + 5 xy 20 y 2 GCF factor each of the groups: 3 x(x 4 y) + 5 y( x 4 y) Common binomial is (x - 4y) (x 4y) ( 3 x + 5 y ) so factor it out Answer is (x y)(3 x + 5 y) Note: if the groups are connected by a minus sign, you need to be careful factoring the second group. See this example and notice that -6ay - 3a factors to -3a(2y + 1)... the plus sign is very important. Verify this by distributing the -3a back to the 2y + 1.
Sum/Difference of Perfect Squares Remember that factoring and distributing are inverse processes. If you factor, then distribute, you get back to the original problem. For example: Factor x 2 +7 x +12 = (x+3)(x+4), then FOIL (x +3)(x+4) = x 2 +4 x+3 x+12 = x 2 +7 x +12 So, you can ALWAYS check your work when you factor. For Sum/Difference of Perfect Squares, mathematicians a long time ago derived these formulas for the sum and difference of perfect squares. The factorization of a 3 +b 3 is (a+b)(a 2 a b + b 2 ), and the factorization of a 3 b 3 is (a b)(a 2 + a b + b 2 ) Things to note: The second trinomial will not factor further under normal circumstances. Don't beat yourself up trying to figure out where this formula came from. You can PROVE this formula by distributing the right side... lots of stuff will cancel and you'll get back the left side.' At least for this class, you don't need to memorize these formulas, I'll provide them for you on the quiz. Two examples should suffice: Final notes on factoring You should be competent with each of these factoring techniques for the quiz. Additionally, you should be able to factor completely using multiple methods for the same problem. So, for: 3 x 3 6 x 2 +24 x you should be able to GCF a 3x: 3 x(x 2 2 x+8) and then factor the remaining trinomial. Final answer would be 3 x(x 2)(x 4).
Finally, you should be able to factor polynomials that are similar to the problems we have already looked at. So, for example: (x+2) 2 y 2 is a difference of perfect squares problem (where a = x+2 and b=y), so (x+2 + y)(x+2 y) x 2 + 5 x 14 is similar to x 2 + 5 xy 14 y 2 and x 4 + 5 x 2 14 (x+7)(x 2)... (x+7 y)( x 2 y)... (x 2 +7)( x 2 2) If you recognize that these polynomials look similar (they all have the same basic "form"), then factor using the appropriate method, and then fiddle with the x's and y's until things "work out". When I say "work out", I mean try to multiply your factors back to the original problem. If you can, then you've factored correctly. Solving Equations by Factoring Based on the principal that if two quantities multiply together and equal zero, then one or the other must be zero. For non-linear equations, apply this principal by setting the equation equal to 0, factoring, then set each factor equal to zero. This example should help. Notice the first step was to subtract 15 and add 7x to both sides to set the equation equal to zero.
Rational Expressions A rational expression is nothing more than a fraction of two polynomials. First and foremost, you need to know how to simplify a rational expression. The key is to cancel common FACTORS from top and x 2 16 bottom. You can't just cancel random "stuff". So, for example: x 2 10 x+16 because the x 2 and the 16 are terms, not factors! This means you need to factor polynomials before you can cancel!!!!!!! However, sometimes things won't work out: x 2 16 x 2 10 x+16 = (x +4)( x 4) ( x 2)(x 8) No common factors on the top and bottom. You're done. Factors need to be exact matches or opposites to cancel. Opposites cancel and leave a factor of -1: 3x 6 3( x 2) 3 = where (x - 2) and (2 - x) are exact opposites, so answer is 4 x 2 (2+ x)(2 x) 2+x You need a keen understanding of what a factor is: In the example to the right, you get down to 3x 2 which has a common factor of 3x 6 x Multiplying and Dividing Same rules for multiplying and dividing as with regular fractions. Multiply numerator to numerator, denominator to denominator, then reduce the fraction. It makes sense to factor before combining, otherwise it becomes nearly impossible to reduce your final answer.
Division Example - flip the second fraction, change to multiplication: At this point, reduce. Since we're dealing with monomials (only single terms), apply exponent rules: 45 70 a2 3 b 1 2 x 1 2 y 2 1 = 9 14 a 1 b 1 x 1 y 1 = 9 y 14 a b x Of course, you can use any method of cancellation you wish to simplify above, as long as it works. Division Example 2 - again, flip the second fraction, change to multiplication: Addition and Subtraction Addition and Subtraction of Fractions requires a common denominator. If you have a common denominator, just add (or subtract) the numerators, and keep the denominator the same: x 2 +3 x 5 2 x+18 x 5 = x2 +3 2x 18 x 5 Notice 2x + 18 becomes -2x - 18 because we subtracted it. Now, simplify: x 2 2 x 15 x 5 and reduce if possible. You'll get x + 3 as your final answer. Getting a common denominator If you don't have common denominators, you'll need to find one. This can be accomplished by factoring each denominator and asking the question: what is the LCM of these terms? Examples: (x + 2) and (x - 3) Nothing in common, LCM is the product (x + 2)(x - 3) 3(x+2) and (x+2)(x - 3) Common (x+2), 3 and (x-3) are not common, so LCM is 3(x - 3)(x + 2) 6 x 2 y and 4 x 5 LCM is 12 x 5 y because this is the smallest expression that each denominator divides evenly.
Building up Once we determine a LCM of denominators, the next step is to build up each term. Using examples from previous page. Make each denominator equal the LCM by multiplying: 5 x+2 x x 3 = 5 (x+2) (x 3) (x 3) x (x 3) (x+2) (x+2) Now, simplify numerators, don't start canceling again: 5 x 15 LCM + x2 2 x LCM And finally, combine: 5 x 15 + x 2 2 x LCM = x2 +3 x 15 ( x+2)(x 3) Try to factor numerator and cancel what you can. In this problem we're done. Do lots and lots and lots of examples until you can do these problems in your sleep. Complex Fractions Complex fractions will come up in numerous occasions in subsequent math courses, as well as in chemistry, physics, and business applications. A complex fraction is a fraction within a fraction. The idea is to rewrite a complex fraction as a simple fraction. The simplest way to do this is by multiplying by a form of 1, where this form of 1 is Numerical Example: 2 + 3 4 5 6 1 2 Notice there are three embedded fractions: 3/4, 5/6 and 1/2 The LCM of all the denominators is 12. So multiply top and bottom of this big fraction by 12 12 : (2 + 3 4 ) 12 2 12 + 3 4 12 24 + 9 ( 5 6 1 = 2 ) 12 5 6 12 1 = 2 12 10 6 = 33 4
The same principal holds for algebra. The LCM of all fractions below is x 2 Ratio and Proportion Finally, a topic you're probably pretty familiar with. A ratio is simply an equation of two fractions. You solve ratio problems by CROSS-MULTIPLYING. The book makes this harder than it needs to. Look: Solve x+3 3x 2 = 6 5 this is a simple ratio, so cross multiply and set these products equal: 5(x + 3) = 6(3x - 2)... now solve... 5x + 15 = 18x - 12... you'll get x = 27/13 You can only use this method if you have two fractions equal to each other. However, a lot of times you can rewrite equations to be in this form: Example 1: 3 = 2x 1 2 x is the same as 3 1 = 2x 1 2 x cross-multiply 3 to (2x) and 1 to (2x - 1), and get the equation: 6x = 2x - 1 Exmple 2: 1 3 x = 2 7 subtract 1 from both sides: 3 x = 2 7 7 7 same as 3 x = 5 7 equation is -5x = -21, answer is x = 21/5.
Proportion We, as humans, innately understand proportionality. We know when things are "in proportion" or "out of proportion". So when we say something like: "it takes 35 minutes to solve 18 problems"... we can think of this as a proportion: Amount of time (in minutes) is proportional to Number of problems solved This can be expressed as a fraction: Amount of Time (in minutes) Number of Problems Solved So, if we ask the question... "how long would it take to solve 50 problems?" we can generate the following equation: 35 minutes 18 problems =? minutes 50 problems... which algebraically looks like: 35 18 = x 50 The key is consistency. Make a decision on what will go in the numerator (in this case minutes) and what will go in the denominator (in this case problems). It will work just as well if you flip this, but you must be consistent! In other words: 18 problems 35 minutes = 50 problems? minutes will work just was well. Try to solve 18 35 = 50 x and verify it's exactly the same as 35 18 = x 50 Finally, one more point about consistency: make sure you're using the same units for your proportion. Don't mix hours and minutes in the same problem... convert to one or the other. Same problem as above, with a slight twist: "It takes 35 minutes to solve 18 problems, how many problems can we solve in one and a half hours?" Wrong: 35 minutes 18 problems = 1.5 hours? problems Correct: 35 minutes 18 problems = 90 minutes? problems because we're consistently using minutes Well, that's all I have for you. Good luck! Remember, as always, practice LOTS and LOTS of problems, not just the problems in the online homework.