Factoring a Trinomial
Date: 08/22/2003 at 04:09:26 From: Jen Subject: Factoring a trinomial What is factoring by grouping? When factoring a trinomial, why is it necessary to write the trinomial in four terms? 2x^3 - 5 x^2 + 3 x First, we factor out x such that 2x^3 - 5 x^2 + 3 x = x (2x^2 - 5 x + 3) We obtain the form ax^2 + b x + c where a = 2, b = -5, and c = 3 Because ac = (2) (3) = 6, and b = -5, we need two integers with a product of 6 and a sum of -5. These integers are -2 and -3. Replacing -5 x by -2 x -3 x, we obtain a four-term polynomial 2x^3 - 5 x^2 + 3 x = x (2x^2 - 5 x + 3) = x (2x^2 - 2 x -3 x + 3) = x [2x (x - 1) - 3 (x - 1)] By grouping, we have x [2x (x - 1) - 3 (x - 1)] = x (x - 1) (2x - 3)
Date: 08/22/2003 at 13:09:10 From: Doctor Peterson Subject: Re: factoring a trinomial Hi, Jen. I don't know that I would say any of this is "necessary"; there are several strategies you can use to factor, and you might just be smart enough to see the answer immediately. (I'm not!) So the question is, how does the technique you are being taught work? I explained the method here: Finding a Single Pair of Factors http://mathforum.org/library/drmath/view/53308.html and here I showed why it works (actually two variants): Factoring Quadratics When a Doesn't = 1 http://mathforum.org/library/drmath/view/62562.html Factoring Trinomials http://mathforum.org/library/drmath/view/56442.html Actually, I think we can explain the basic method as you and I present it more easily. We have ax^2 + bx + c and we find p and q so that pq = ac p + q = b Then we rewrite the quadratic as ax^2 + (p + q)x + pq/a ax^2 + px + qx + pq/a Now we factor out ax from the first pair of terms and q from the second: ax(x + p/a) + q(x + p/a) and then factor out the "x + p/a" from both terms: (ax + q)(x + p/a) or, for the sake of symmetry, a(x + q/a)(x + p/a) Since it may not be true that both p and q are divisible by a, this may not be exactly how the answer looks using integers; but we have seen that writing b as p + q does always allow us to factor by grouping (that is, to find common factors in two pairs of consecutive terms, and then factor the result). And anything that allows us to finish off the problem so easily, while perhaps not strictly "necessary," is certainly welcome! 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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