Factoring Polynomials

Transcription

1 Factoring Polynomials What is Factoring? You learned how to expand polynomials in earlier units. For example, when distributing: a(b + c) = ab + ac, you are expanding. Doing the reverse is factoring. In another example, FOILing is expanding, but reverse foiling is factoring. Recall back in middle school math, when you factored a number, such as 15, would be writing it as the multiple of two other numbers: 15 = (5)(3), where both 5 and 3 are factors of 15. The expressions you deal with to factor are usually more complex than the above example, but the same general procedure is followed in each case: rewriting an expression as a product of two or more expressions. The only difference now is that we are factoring polynomials. But recall from an earlier unit that polynomials are simply expressions involving variables (x or any other variable) with exponents that are just added or multiplied. Each term in a polynomial can be written as ax n where a is a real number and n is a non-negative integer. Common Factors One of the most basic ways to factor an expression is to take out a common factor". If every term in an expression has several factors, and if every term has at least one factor that is the same, then that factor is called a common factor. If this is the case, then the common factor can be "taken out" of every term and multiplied by the expression that remains (kind of like un-distributing something). Here s an example: 2x 2 + 8x. The first term has factors of 2, x, x and the second term has factors of 2, 4, x. Clearly, the 2 and x are both common factors, so we call 2x the common factor. In this case, 2x 2 + 8x = 2x(x + 4). Decomposition Method Some people learn how to factor by using trial and error. If this works for you, go for it! Unfortunately, unless you have the knack of it or have done tons of practice, it s not the best way. The decomposition method is a systematic method used for factoring quadratic equations. In this case, it is good to look at what happens when expanding an expression so that you can clearly see what needs to be done in reverse to factor it: Expanding: (x + 5)(x + 1) = x(x + 1) + 5(x + 1) = x 2 + x + 5x + 5 = x 2 + 6x + 5 Factoring, starting from x 2 + x + 5x + 5: = x(x + 1) + 5(x + 1) = (x + 1)(x + 5) In order to do this method, you must have the x term (middle term) split into two separate terms. While there are many ways to do this, only one ways works to factor. In the example above, 6x was broken into 5x + 1x. Why this? Here are the rules for splitting the middle term:

2 Given a general form of a quadratic expression, ax 2 + bx + c, where a is the coefficient of x 2, b is the coefficient of x and c is the constant: 1) Multiply a and c (remember the sign) 2) Write down b (remember the sign) 3) Write down all the pairs of factors of the product of ac and see which pair adds up to give you b (signs are important) 4) Rewrite the x term as the sum of the two terms with these numbers as coefficients. Once you have split the x term, you continue factoring by taking out a common factor from the first two terms, and then a common factor from the last two terms. Your expression should now look like: x(x + n) + m(x + p) where n, m, and p are constants, and p will either be equal to n or m. Refer to the example above (specifically the second line in the factoring part), You then find the common binomial and factor it out, ending with the form: (x + n)(x + m). Her is another example: 2x x + 36 Note: 2 times 36 = 72. Factors of 72 that add to 13 are 9 and 8. Therefore, we rewrite the expression as 2x 2 + (8 + 9)x + 36 This becomes 2x 2 + 8x + 9x + 36, which equals 2x(x + 4) + 9(x + 4) The common factor here is (x + 4), which can be factored out, yielding: (x + 4)(2x + 9). Note that this method is very similar to the box method that is often used to factor quadratic trinomials. Perfect Squares Any expression of the form: x 2 + 2ax + a 2 is called a perfect square because x 2 + 2ax + a 2 = (x + a) 2. Note that you can always check to see if this applies by simply expanding (x + a) 2. To recognize an expression as a perfect square or not, you should first see if the constant term is a square number (i.e. can you take the square root of it and get an integer for an answer?). If so, see if the square root of it, multiplied by 2 (only if the coefficient of x 2 is 1) gives you the coefficient of the linear term (the x term). If it does, the original expression may be factored into a perfect square. In other words, the constant term is some perfect square (e.g. 4, 9, 16, 25, 100, etc.) and the middle term is the square root of this constant term times two (6 for 9, 8 for 16, 10 for 25, 20 for 100, and so on). Note that if you do not recognize it as a perfect square, it may still be factored by the decomposition method to get the same answer.

3 Difference of Squares A difference of squares expression is one of the form: (something) 2 - (something) 2. Mathematically, anything that looks like (a 2 x 2 - b 2 ) is a difference of squares and can be factored into (ax + b)(ax - b). Notice that the factors are identical except for the sign. If you multiply out (ax + b)(ax - b), you get (ax)(ax) - abx + abx - b 2. The two middle terms ( -abx and + abx) cancel each other and you are left with (a 2 x 2 - b 2 ). Example: x 2 49 = (x 7)(x + 7) Note that not every expression is factorable using these above procedures. Some expressions are simply not factorable. You do know, however, that if a certain value, call it n, for x (or any variable that is in question) makes the whole expression zero, then (x - n) is a factor. For a quadratic expression, there are other ways to find the factors. You can either Complete the Square, use the Quadratic Formula, or use Synthetic Division to find which values make the polynomial zero. From that, you know the factors. Difference of Cubes A difference of cubes is an expression of the form: a 3 - b 3. It can be factored into (a - b)(a 2 + ab + b 2 ). To verify this, all you need to do is expand these factors: (a - b)(a 2 + ab + b 2 ) = a 3 + a 2 b + ab 2 - a 2 b - ab 2 - b 3 = a 3 - b 2 It is often difficult to memorize this factored expression so all you really need to remember is that if you ever see an expression like (a 3 - b 3 ), then (a - b) is a factor. Once you remember that, you can use synthetic or long division to find the remaining factor(s). Sum of Cubes A sum of cubes is an expression of the form: a 3 + b 3. It can be factored into (a + b)(a 2 - ab + b 2 ). To verify this, all you need to do is expand these factors: (a + b)(a 2 - ab + b 2 ) = a 3 - a 2 b + ab 2 + a 2 b - ab 2 + b 3 = a 3 + b 2 Just like the above, if you prefer not memorizing this, just remember that if you ever see an expression like (a 3 + b 3 ), then (a + b) is a factor and then use long or synthetic division to find the remaining factor(s).

4 Polynomials of Higher Degree When you first see a polynomial of degree greater than 2, that is not a difference or sum of cubes, there is no need to panic. Often, there will be a common factor in the expression which when factored out, will make the remaining expression a quadratic. For example: (9x 4-2x x 2 ), x 2 is a common factor and thus 9x 4-2x x 2 = x 2 (9x 2-2x + 10). The expression inside the brackets can now be factored using the decomposition method. In some cases, you will be given one factor of a large expression and you will be need to find the remaining ones. A good way to do this is by using synthetic or long division. Click here to see some examples. If you are not given one of the factors, there is still hope in factoring it. By now you should realize that if (x - k) is a factor [where k is a constant] of some polynomial, then k must divide exactly into the constant term of the polynomial. If the highest order term in a polynomial is not 1, then the factor is of the form (hx - k), where h must divide exactly into the coefficient of the highest order term, and k must divide exactly into the constant term of the polynomial. The possible values for k are the factors of the constant term, and the possible values of h are the factors of the leading coefficient (coefficient of the highest order term). You then must use long division with these possible factors to see if they are indeed factors, and thus find the other factors. Click here to see some examples. This method can be very time consuming and tedious if there is a lot of possible factors, so there is a shorter way to check if a possible factor is indeed a factor. Once you find the possible k and h values, and put them together in a possible factor form, you check which of them are factors by the following rules (note that some people including the UCSMP text call this the P/Q method): 1) If (x - k) is a factor of the polynomial of interest, then substituting x = k into the polynomial gives an answer of zero. 2) If (x - k) is not a factor of the polynomial of interest, then substituting x = k into the polynomial does not give an answer of zero. 3) If (hx - k) is a factor of the polynomial of interest, then substituting x = k/h into the polynomial gives an answer of zero. 4) If (hx - k) is not a factor of the polynomial of interest, then substituting x = k/h into the polynomial does not give an answer of zero. You can check all possible factors this way but it is quite possible that out of many of the possibilities, only one of them is correct, and the other factor that you need to find may be of a different form, such as (dx 2 + ex - f). That means that you may waste time substituting without finding all the factors and that you will still need to perform long division. Once you have found one factor using these rules, it is a common practice to use long division to find all remaining factors. Note that you can also use graphing to help zero in on possible factors, remembering

5 that the roots are just the opposite sign of the factors (e.g. if 3 is a root, then (x 3) is a factor. Click here to see an example. Summary of Methods for Factoring 1) Take out any common factors. 2) Recognize if a polynomial is (or isn't) a perfect square, a difference of squares, a difference of cubes or a sum of cubes. 3) If a quadratic, try the decomposition method. 4) If a polynomial is higher than degree 2, find factors of the form (hx - k) or (x - k) by substituting x = k/h and or x = k into the polynomial and then using long division. Continue the process until the polynomial is fully factored. 5) Check the factors by multiplying them together - you should get the original polynomial if the factors are correct.

6 Factoring Always look for a Greatest Common Factor FIRST!!! 2 TERMS 3 TERMS 4 TERMS (Must be in one of the following forms to factor with two terms) (X or Pattern) (Grouping) Difference of Two Perfect Squares a 2 b 2 = (a + b)( a - b) OR Box Method and Rules for Signs Last sign positive both get middle sign. Last sign negative check the O.I. in F.O.I.L Group first two and last two terms and see if each pair has a G.C.F. (May need to change order of the terms) 2x 3 8x 2 + 3x - 12 THEN Sum Of Cubes - SOAP a 3 + b 3 = (a+b)(a 2 -ab + b 2 ) OR Difference of Cubes - SOAP a 3 - b 3 = (a-b)(a 2 + ab + b 2 ) If the G.C.F. of each pair results in a common binomial, factor out the binomial.. 2x 2 (x 4) + 3(x 4) THEN Write the binomial times the binomial created by the terms left when GCF. binomial was pulled out. (x 4 )(2x 2 + 3) 1. If nothing can be done to the original expression, then it is PRIME 2. Check to see if any of your final answers will factor further. 3. When it is written as a product of primes, the polynomial is COMPLETELY FACTORED.

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