Factoring a Quadratic EquationDate: 05/13/98 at 16:26:27 From: Alexander Yung Subject: Algebra Factor y^2 + 2y - 35 = 0. Date: 05/20/98 at 10:09:52 From: Doctor Alice Subject: Re: Algebra Hello, Alexander! Well, your problem is a quadratic equation. And you want to solve it by factoring a quadratic trinomial into a product of two binomials. Look at the y^2 first. It breaks down into y times y: (y )(y ) = 0 Now look at the 35. List its pairs of factors. Factors are numbers that when multiplied give us our product. So what two numbers can you think of that when multiplied give you 35? What about 35 and 1? Or 7 and 5? Now, which of those pairs could give us a 2 (coefficient of the linear term, that means the number in front of the y^1 term) if we played our signs right? Of course: 7 and 5. So we have: (y 7)(y 5) = 0 Now for the signs -- sometimes the hardest part of factoring. Look at the sign of the constant term -35. When is a product of two numbers negative? When one of them is + and one of them is -. Now we have either: (y - 7)(y + 5) = 0 or (y + 7)(y - 5) = 0 Which one? Use the FOIL method to multiply the outer and inner products, because we need to get +2y. The left equation expands to contain 5y - 7y, and the right choice has -5y + 7y. We want the right one. So the factorization of your quadratic equation is: (y + 7)(y - 5) = 0 To finish it off, set each factor equal to 0 and solve. I hope I have helped. Write back if you are still confused. -Doctor Alice, The Math Forum Check out our web site! 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/