# How do you write 2x^3-5x^2-x+6 in standard form and as a product of linear factors?

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Good afternoon,

First, it is written in standard form already. You can tell by looking at the exponents, the exponents on the variables in each term. The terms should be arranged in order of their exponent, from highest to lowest.

2x^3, exponent of 3

-5x^2, exponent of 2

-x, can be written with an exponent of 1, -x^1

6, can be written with an exponent of 0, 6x^0

As for the linear terms, that is more difficult. There are various ways. I like synthetic division. First, take the factors from the first and last coefficients, as I show on the attachment. Make all the ratios you can from those factors, taking last factor/first factor. You have to include all possible negatives, also. So, that gave us +-6, 3, 2, 1, 3/2.

So, if the polynomial can be broken up into linear factors, factors that read like "ax+b", one of these would have to be "b". So, then, we literally start trying each one until we find one that works. So, we try +2. We set it up as I showed in the attachment in the red. We carry the first number down automatically, so 2 comes down. Then, we multiply the 2 by the number we picked, +2, giving us +4. We put that under the -5 and add the numbers, -5 + 4, giving us -1. Repeat the pattern, multiplying -1 by 2, giving us -2, putting that under the -1. Add for -3. Multiply by -2, giving -6, put that under the upper 6. Add for 0. If we get 0, then we know we have a factor. If not, you have to work everything again, choosing a different number. If you don't get a 0 for the end ever, then there was either an error, or the polynomial can't be factored.

We got the 0. So, it can be factored. We just found one linear factor. What is it? What is the number we used? +2. So, the linear factor would be "x - 2" (you can't forget the subtraction tweek here; if a negative number, we would have x + 2). The numbers that are left, the 2, -1, -3, that becomes the quadratic x+2 would be multiplied by, 2x^2 - 1x - 3. So, we can factor that like a normal quadratic, "(2x-3)(x+1)"

So, 2x^3-5x^2-x+6 factored into linear factors would be (2x-3)(x+1)(x-2).