Factoring Trinomials By GroupingDate: 11/16/2004 at 16:29:55 From: Laura Subject: Trick for factoring quadratics. Dear Dr. Math, I remember seeing a grouping trick for solving quadratic equations by factoring the trinomial. Somehow the middle term is broken up into two parts. Then the first two parts are factored and the second two parts are factored. After that somehow the factored parts match up and you can pull out the common factor and then you have a factored quadratic equation. For example, 6x^2 + 7x + 2 should factor into (2x + 1)(3x + 2). The grouping method separates the middle term into 3x and 4x, giving 6x^2 + 3x + 4x + 2 Factor first and second parts: 3x(2x + 1) + 2(2x + 1) Pull out the common factor: (2x + 1)(3x + 2) What trick would I use to figure out how to separate the middle term? Also, is there a way to tell if this method will work before you try it? Date: 11/16/2004 at 16:42:18 From: Doctor Vogler Subject: Re: Trick for factoring quadratics. Hi Laura, Thanks for writing to Dr. Math. Yes, this is how I first learned to factor quadratics. You first multiply together the first and last coefficients. For example, for 6x^2 + 7x + 2 you get 12. Next, you look for two factors of that number that add up to the middle number. In this case, the two factors are 3 and 4. Then you split up the middle term into those two numbers, and I think you can do the rest. Be warned: You have to be careful of signs. For example, if the product is negative, then one of the factors will be positive and the other will be negative. If the product is positive, but the middle number is negative, then both factors must be negative. For example, in 10x^2 + 26x - 12 the product is 10(-12) = -120, and the factors that add to 26 are +30 and -4. In 8x^2 - 26x + 21 the product is 8*21 = 168, and the two factors that add to -26 are -12 and -14. Yes, there is a way to determine if the polynomial is factorable. The way to test is by computing the discriminant of the polynomial, which is b^2 - 4ac for the quadratic ax^2 + bx + c = 0. If the discriminant is a perfect square, then there are factors that add up to the middle number. If not, then there are not. If there are not factors, you can still solve the equation by using the quadratic formula or completing the square. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, 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/