FactoringDate: 12/03/2001 at 21:42:09 From: Sommer Pruitt Subject: Factoring Hi Dr. Math, This problem is giving me a hard time: 6x squared + 31x + 5 I don't understand it. Thanks for your time, Sommer Date: 12/04/2001 at 02:44:56 From: Doctor Jeremiah Subject: Re: Factoring Hi Sommer, When you factor this you will end up with (6x+a)(x+b), because when you multiply those two terms together you get: (6x+a)(x+b) 6x^2 + ax + 6bx + ab 6x^2 + (a + 6b)x + ab And the 6x^2 (6x squared) is then accounted for. Now we have two things to figure out: a and b. We know that 31 = (a + 6b), and we know that 5 = ab, because those are the values that were in the original equation, so we have two equations and two unknowns: 31 = a + 6b 5 = ab All we need to do is to come up with integers that work in these equations. Look at the second equation first. There are only really two possible answers for a and b. Either a = 1 and b = 5, or a = 5 and b = 1. Now look at the first equation, (31 = a + 6b). The only way to get 31 is for b to equal 5 and a to equal 1, so: a = 1, b = 5 Our original factors were (6x+a)(x+b), and now we know a and b so we can fill them in: (6x+a)(x+b) <== a = 1, b = 5 (6x+1)(x+5) Now all we need to do is check our answer. We can do that by multiplying the factors together: (6x+1)(x+5) 6x(x+5) + 1(x+5) 6x(x) + 6x(5) + x + 5 6x^2 + 30x + x + 5 6x^2 + 31x + 5 And since that is what we wanted, the factors really are (6x+1)(x+5). - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/