Factoring Trinomials in Quadratic EquationsDate: 02/15/2004 at 01:06:16 From: Shoaib Subject: Solving quadratic equations by factoring I'm stuck on this question: 15x^2 - 16x - 7 = 0 My book teaches me two methods to factor quadratic equations: guess and check, and grouping. I can't decide which method to use. Date: 02/15/2004 at 14:53:29 From: Doctor Greenie Subject: Re: Solving quadratic equations by factoring Hello, Shoaib - Factoring trinomials is a very interesting process because there are many different ways that people approach it and think about it. Many students find a method that works for them and stick with it, but finding that method may involve trying out lots of different ways. I'm going to show you a couple of ways you may not have seen before, and hopefully one of them will help you. One way that ALWAYS works is to use the quadratic formula, as explained very nicely on the following page in the Dr. Math archives: Quadratic Trinomials http://mathforum.org/library/drmath/view/52902.html This page also shows an alternative method for performing the factorization. Using the method involving the quadratic formula on your example, we have 15x^2 - 16x - 7 = 0 x = [16 + or - sqrt(16^2 - 4(15)(-7))]/30 x = [16 + or - sqrt(676)]/30 x = [16 + or - 26]/30 x = 42/30 = 7/5 or x = -10/30 = -1/3 Then we reconstruct the quadratic equation from which we obtained these roots: 15(x - (7/5))(x + (1/3)) = 0 5(x - (7/5)) * 3(x + (1/3)) = 0 (5x - 7)(3x + 1) = 0 So the factorization of the original quadratic expression is 15x^2 - 16x - 7 = (5x - 7)(3x + 1) Below are explanations of a couple of other ways to factor quadratic expressions. One of them may be the "grouping" method you say you are familiar with; the other one I suspect you haven't seen before. A Grouping Method... We have the quadratic expression 15x^2 - 16x - 7 (1) We first multiply the leading coefficient and the constant to get a "target number": (15)(-7) = -105 (2) Next we find two numbers a and b whose product is this "target number" and whose sum is the coefficient of the middle term: a*b = -105 a+b = -16 The numbers we want are -21 and 5. (3) Now we "group" the middle term using these two numbers we have found: 15x^2 - 16x - 7 = 15x^2 - 21x + 5x - 7 (4) And now we factor by grouping: 15x^2 - 21x + 5x - 7 = (15x^2 - 21x) + (5x - 7) = 3x(5x - 7) + 1(5x - 7) = (3x + 1)(5x - 7) Here's another closely related method. We start with the same quadratic expression 15x^2 - 16x - 7 (1) First, we arbitrarily multiply the constant term by the leading coefficient and change the leading coefficient to 1; this gives us in this case x^2 - 16x - 105 (So we multiplied the -7 by 15 and changed the "15" to "1"....) (2) Next we factor in this form: x^2 - 16x - 105 = (x - 21)(x + 5) (3) Now we multiply BOTH the leading coefficients by the original leading coefficient; this gives us the "factorization" (15x - 21)(15x + 5) (4) Finally, remove the common factor from each term to get (5x - 7)(3x + 1) Try any (or all) of these methods on a few more examples you have and find the one which you understand the best.... I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/