How do you find the zeroes for $f (x) = x^{3} + 2 x^{2} - 5 x - 6$ $f(x) = x^3 + 2x^2 - 5x - 6$ ?

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Answer 1 · 2015-09-13T05:48:24

-1,2,-3

The real zeros would be out of the factors of 6, may be $\pm 1, \pm 2, \pm 3, \pm 6$ $+-1, +-2, +-3,+-6$ .

Consider -1 first, it makes f(-1)=0 hence -1 is one of the zeros. Like wise x=2 and x= -3 also makes f(2)=0 and f(-3)=0.

Thus all the three zeros are determined. These are -1,2 and -3

How do you find the zeroes for f(x)=x3+2x2−5x−6f(x) = x^3 + 2x^2 - 5x - 6?