Factoring Trinomials by the Grouping MethodDate: 12/08/2005 at 12:31:59 From: Rosemarie Subject: Factoring trinomials by the grouping method I would like a proof that would explain why in general factoring trinomials by grouping works. For example, if factoring 21x^2 + 43x - 14, we need to find 2 numbers whose sum is 43 and whose product is -294. I understand that it works (49 and -6) but not the reason behind it, especially why the numbers must have the product of -294. Thank you for your help. Date: 12/08/2005 at 13:21:33 From: Doctor Peterson Subject: Re: Factoring trinomials by the grouping method Hi, Rosemarie. Let's take your example, which I'll factor by grouping: (1) 21x^2 + 43x - 14 (2) 21x^2 + 49x - 6x - 14 (3) 7x(3x + 7) - 2(3x + 7) (4) (3x + 7)(7x - 2) Working backward from the end result, look at where the numbers in line (2) come from: 3x*7x + 7*7x + 3x*-2 + 7*-2 The product ac is 3*7 * 7*-2 (the coefficients in the outer terms). The product of the middle terms is 7*3 * 7*-2, which is a product of the same four numbers, grouped differently. This will always be true. So the middle two numbers have to have the same product as ac; and they have to add up to b (43) so that line (2) is equal to line (1). That is, by choosing a pair of numbers whose product is ac and whose sum is b, we are ensuring that the factoring by grouping will work. Let's express this generally. Suppose we're factoring ax^2 + bx + c Our goal is to factor it as, say, (px + q)(rx + s) = prx^2 + psx + qrx + qs So we see that pr = a ps + qr = b qs = c So we want a pair of numbers, m=ps and n=qr, whose sum is b. Notice that ac = pr * qs = ps * qr = mn So another requirement on the two numbers m and n is that their product be equal to ac. And it turns out that this is sufficient to complete the factoring. Once we write ax^2 + bx + c = ax^2 + mx + nx + c we can factor out the GCF px from the first two terms, and factor out the GCF q from the second pair of terms: prx^2 + psx + qrx + qs = px(rx + s) + q(rx + s) = (px + q)(rx + s) I'm assuming here that p and q will not only be factors of a, m, n, and c, but GCF's; that takes a little more to prove, but is true (as long as there is no common factor of all three coefficients, which can complicate things a bit). An explanation of a related method, with more detail, can be found here: Factoring Quadratics When a Doesn't = 1 http://mathforum.org/library/drmath/view/62562.html And here is a briefer explanation: Factoring a Trinomial http://mathforum.org/library/drmath/view/63894.html If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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