Question

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

Answer 1

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`.