muskan Posted July 11, 2017 Share Posted July 11, 2017 How can we factorize cubic polynomials? e.g. y3+ 2y2+ y+ 1 Link to comment Share on other sites More sharing options...
Country Boy Posted July 15, 2017 Share Posted July 15, 2017 In general, we can't! One way is to look for "zeroes" of the polynomial: if x= a makes the polynomial equal to 0, then (x- a) is a factor. But it is not easy to find zeroes and, generally, when we say "factor" we mean to factor with integer coefficients- and that may not be possible. The best I can do is notice that we can factor "y" out of the first three terms: [tex]y(y^2+ 2y+ 1)+ 1[/tex] and that [tex]y^2+ 2y+ 1= (y+ 1)^2 so that [tex]y^3+ 2y^2+ y+ 1= y(y+ 1)^2+ 1[/tex] but that has not "factored" the original polynomial. Link to comment Share on other sites More sharing options...
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