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factoring a trinomial

factoring a trinomial with a leading coefficient that is not equal to one. Why is this process more difficult than when the leading coefficient is equal to one?

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Colby W. | Math and science tutorMath and science tutor

Say we have a trinomial x^2 + bx + c. We may factor it into something of the form (x - r0)(x - r1) where r0 and r1 are the two roots of the polynomial and depend on b and c. And that's pretty much it. If the polynomial was of the form ax^2 + bx + c where a is not 1 or 0, then we can not factor it as (x - r0)(x - r1) because if we multiply it out, we get x^2 - (r0 + r1)x + r0r1 which has a leading coefficient of 1. Not what we want. So we need something of the form (ex - f)(gx - h) where e and g are not both 1. One of them could be, but not both. e and g are factors of the leading coefficient. If the leading coefficient is prime, awesome. If not, you have to factor it to get e and g. If the leading coefficient is 12, say, then (e, g) could be (1, 12) or (2, 6) or (3, 4) or (4, 3) or (6, 2) or (1, 12). The more factors the leading coefficient has, the more possibilities there are for e and g and the more trial-and-error you have to do. And no one likes that. In general, the bigger in magnitude the leading coefficient is, the possibilities you have to consider.