Question #9dcdb

1 Answer
Dec 10, 2014

Let us find all zeros of #f(x)=x^3-7x+6#.

By inspection, we see that

#f(1)=(1)^3-7(1)+6=0#

So, #x=1# is one of the zeros, which means that #f(x)# has the factor #(x-1)#.

By factoring out #(x-1)#,

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

by factoring out a bit further,

#f(x)=(x-1)(x-2)(x+3)=0#

Hence, the zeros of #f# are #x=-3, 1, 2#.


I hope that this was helpful.