danmac0710
  • danmac0710
Okay so I know how to factor a quadratic by splitting the x term, but how do I use similar method for a cubic? Please explain or point me to a good explanation somewhere.
Mathematics
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schrodinger
  • schrodinger
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mathmate
  • mathmate
For a cubic it is a little more complex because we can have factors like (ax+b) and (ax^2+bx+c). However, for a cubic, even if there is a factor of the latter kind, then there is a factor of the first. So we can concentrate on factors like (ax+b). Take an example such as x^3+4x^2+4x+3 First check if the coefficients add up to 0, if they don't, then (x-1) is not a factor. Next change the signs of odd powers (x^3, x, ...) and check again. -1+4-4+3\(\ne\)0, so (x-1) is not a factor. Since +3 has factor of 1 and 3, and 1 is not a factor, we try next -3 Group the expression as: x^2(x+3)+ x(x+3)+ (x+3) = x^3+4x^2+4x+3 ....bingo, so we have the factor. Divide the expression by (x+3) to get (x^2+x+1) as the other factor.
danmac0710
  • danmac0710
Okay thanks very much. So there is no equivalently simple method as there is for quadratics then it would seem. I just need to go through the bag of tricks until I find a factor and then group it to find the other factors from that? Thanks for your help.
mathmate
  • mathmate
It's almost that. Sometimes you can try grouping first and get it by trial and error, even for polynomials for the fourth degree. It all depends on how you are used to one technique or the other. I always try grouping first, because if I get it, it would save a lot of time.

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