Three Factoring Techniques for Quadratic PolynomialsDate: 02/21/2008 at 17:48:06 From: Sarah Subject: factor the following: 3(40x^3+31x^2+6x) How would you factor 3(40x^3+31x^2+6x)? There is an exponent to the 3rd power, and the numbers are too large and can not be reduced any more. The original question was 120x^3+93x^2+18x, but I reduced that. I also know that the parenthesis should look like this... 3(_x__)(_x__)(_x__), but what goes into the underlined areas????? Date: 02/21/2008 at 23:17:53 From: Doctor Greenie Subject: Re: factor the following: 3(40x^3+31x^2+6x) Hi, Sarah -- Let's look at the original problem: 120x^3 + 93x^2 + 18x You saw the common factor of 3; but there is also a common factor of x. So this original expression can be factored to 3x(40x^2 + 31x + 6) So if the quadratic expression factors, the final factored form will be of the form 3x(_x + _)(_x + _) Now we have to deal with that quadratic expression with the large coefficients. There are several methods that can be used to do this, and there are often many small variations of a particular method. I will demonstrate three methods that I have found easy to use (IF you understand how to apply them!!). We need to factor the expression 40x^2 + 31x + 6 ( = ax^2 + bx + c ) 1. The "ac" method One method commonly taught is to multiply the "a" and "c" coefficients together to get a "target number": ac = 40*6 = 240 Then our task is to find two numbers whose product is 240 and whose sum is the coefficient of the middle term, 31. We find the two numbers to be 15 and 16. The next step is to write the middle term of the original expression, 31x, using these two numbers--that is, we write the middle term as 15x + 16x. When we do that, we can then finish the factoring by grouping. 40x^2+31x+6 (40x^2+16x) + (15x+6) 8x(5x+2) + 3(5x+2) (8x+3)(5x+2) Note if you try using this method that, on the third line in this example, you should end up with a common binomial factor if you have done the rest of the process correctly. 2. Organized trial and error I learned factoring by this basic method; and I have just recently been shown a way to organize this method so that a lot of the guesswork is removed. We have to start with a guess of some sort. For this example, I will start by guessing that the leading terms of the binomial factors are 10x and 4x. I start the process by writing these leading coefficients 10 and 4 like this: 4 | --------- 10 | On the right side of the "|" separator, I write all the possible combinations of factors that give me the constant product of 6: 4 | 6 1 3 2 ------------- 10 | 1 6 2 3 This display shows that, IF the leading coefficients are 4 and 10, the POSSIBLE factorizations are (4x+6)(10x+1) (4x+1)(10x+6) (4x+3)(10x+2) (4x+2)(10x+3) But every one of these possible factorizations contains a factor which contains a common factor. (And note that we can see that this will be the case by looking at our array of numbers--we don't need to write out all the possible factorizations). Since the original expression does not contain a common factor, none of these can be the correct factorization of the original expression. So the leading coefficients of our factors can't be 10 and 4. So let's try leading coefficients of 8 and 5 and go through the same process. 8 | 6 1 3 2 ------------- 5 | 1 6 2 3 We can immediately reject the possibilities which pair the "6" or "2" with the leading coefficient of 8, because those would produce factors with a common factor. So we only need to look at the remaining possibilities: 8 | 1 3 ---------- 5 | 6 2 To get the middle term of the product from this array, we can perform a sort of "cross multiplication". If we choose the "1" and "6" constant terms, then the coefficient of our middle term is going to be 8(6)+5(1) = 48+5 = 53--not what we want. But when we choose the "3" and "2" constant terms, we get the desired middle term: 8(2)+5(3) = 16+15 = 31. So we find that the factorization of our expression is (8x+3)(5x+2) 3. Pure magic Well, it's not really magic--it's just hard to show why this method works; and it's hard to UNDERSTAND why it works. But in the few cases where I have seen it used, students who understand it find it quite easy to use.... We start like the ac method described above--we multiply the 40*6 to get 240. But this time we make a new equation with a leading coefficient of 1 and a constant term of 240: x^2 + 31x + 240 Next we factor this by usual methods. Since the leading coefficient is 1, it is relatively easy to find the numbers whose product is 240 and whose sum is 31; those numbers are again 15 and 16. So we factor this new expression as (x+15)(x+16) Now comes the "magic" part. We replace each "x" in this factored form with "40x"--using the leading coefficient of our original expression: (40x+15)(40x+16) Finally, we remove the common factor from each of these factors: (8x+3)(5x+2) Abracadabra! Poof! Our "magic" method has given us the correct factorization. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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