How do you solve #y^3 - y^2 +4y -4 = 0#?

1 Answer
Feb 21, 2016

Answer:

Solution of the equation is #y=1, 2i# or #-2i#.

Explanation:

In the equation #y^3−y^2+4y−4=0#, factors of constant term #-4# are, #(1,-1,2,-2,4,-4)#. So, first identify which among these satisfies the equation. As is apparent #y=1# satisfies the equation and hence #(y-1)# is one such factor.

As such factorizing #y^3−y^2+4y−4=0# , we get

#y^2*(y-1)+4(y-1)=0#

or #(y^2+4)(y-1)=0#, i.e. either #y=1# or #y^2+4=0#.

AS the latter cannot be factorized, we get its roots by using general form of quadratic equation #ax^2+bx+c=0# which are #(-b+-sqrt(b^2-4ac))/(2a)#.

As #a=1, b=0, c=4# roots of #y^2+4=0# are #(0+-sqrt(0-4*4*1))/(2a)# or #+-sqrt(-16)/2# or #+-(4i)/2# i.e. #+-2i#

Hence solution of the equation is #y=1, 2i# or #-2i#.