How do you find all zeros of the function #f(x)= x^3 - x^2 - 4x -6# given 3 as a zero?

1 Answer
Feb 24, 2016

Answer:

Other two zeros of f(x) are #(-1+i)# and #(-1-i)#.

Explanation:

Zeros of the function #f(x)=x^3−x^2−4x−6# are those values of #x# for which #f(x)=0#. As the term independent of #x# is #-6#, factors of #-6# i.e. #{1, -1, 2, -2, 3, -3, 6, -6}# could be among zeros of #f(x)#.

It is seen that putting #x=3# makes #f(x)=0# and hence #(x-3)# isne such factor. Dividing #f(x)# by #(x-3)# we get

#x^3−x^2−4x−6#

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

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

As the determinant (#b^2-4ac# in the function #ax^2+bx+c#) is equai to #2^2-4*1*2=-4#, is a negative number, no further real zeros are there.

Other complex zeros are given by #(-b+-sqrt(b^2-4ac))/(2a)#

i.e. #(-2+-sqrt(2^2-4*1*2))/(2*1)# or #(-2+-sqrt(-4))/2#

which simplifies to #(-1+i)# and #(-1-i)#, which are other two zeros of #f(x)#.