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

Feb 21, 2016

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

#### Explanation:

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

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

${y}^{2} \cdot \left(y - 1\right) + 4 \left(y - 1\right) = 0$

or $\left({y}^{2} + 4\right) \left(y - 1\right) = 0$, 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 $a {x}^{2} + b x + c = 0$ which are $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

As $a = 1 , b = 0 , c = 4$ roots of ${y}^{2} + 4 = 0$ are $\frac{0 \pm \sqrt{0 - 4 \cdot 4 \cdot 1}}{2 a}$ or $\pm \frac{\sqrt{- 16}}{2}$ or $\pm \frac{4 i}{2}$ i.e. $\pm 2 i$

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