What are all the possible rational zeros for #f(x)=x^3+2x^2-5x-6# and how do you find all zeros?

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Answer 1 · 2016-08-22T22:00:23

Zeros are #{-1,2,-3}#

As coefficient of highest power of #x# is #1#, the product of all the zeros will be #6#.

A trial indicates that #1# is not a zero, but #-1# is as #(-1)^3+2×(-1)^2-5(-1)-6=-1+2+5-6=0#. Hence one factor of #x^3+2x^2-5x-6# is #(x+1)#.

Dividing former by latter

#x^3+2x^2-5x-6#

= #x^2(x+1)+x(x+1)-6(x+1)#

= #(x+1)(x^2+x-6)#

We can further factorize second by splitting middle term as follows

#(x+1)(x^2+x-6)#

= #(x+1)(x^2+3x-2x-6)#

= #(x+1)(x(x+3)-2(x+3))#

= #(x+1)(x-2)(x+3)#

Hence, Zeros are #{-1,2,-3}#