# How do you solve the polynomial x^4-4x^3+5x^2-4x+4=0?

Dec 6, 2015

Use the rational root theorem to help find one of the roots. Separate out the corresponding factor, then factor by grouping to find roots $2$, $2$, $i$ and $- i$.

#### Explanation:

Let $f \left(x\right) = {x}^{4} - 4 {x}^{3} + 5 {x}^{2} - 4 x + 4$

By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be expressible in lowest terms as $\frac{p}{q}$ for some integers $p$, $q$, where $p$ is a divisor of the constant term $4$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots are:

$\pm 1$, $\pm 2$, $\pm 4$

Also since the signs of $f \left(x\right)$ alternate for odd and even powers of $x$, $f \left(x\right) = 0$ has no negative roots. So that only leaves the following possible rational roots:

$1$, $2$, $4$

Let's try each in turn:

$f \left(1\right) = 1 - 4 + 5 - 4 + 4 = 2$
$f \left(2\right) = 16 - 32 + 20 - 8 + 4 = 0$

So $x = 2$ is a root and $\left(x - 2\right)$ a factor.

${x}^{4} - 4 {x}^{3} + 5 {x}^{2} - 4 x + 4 = \left(x - 2\right) \left({x}^{3} - 2 {x}^{2} + x - 2\right)$

The remaining cubic can be factored by grouping:

${x}^{3} - 2 {x}^{2} + x - 2 = \left({x}^{2} - 2 {x}^{2}\right) + \left(x - 2\right) = {x}^{2} \left(x - 2\right) + \left(x - 2\right) = \left({x}^{2} + 1\right) \left(x - 2\right)$

Putting it all together:

$f \left(x\right) = {x}^{4} - 4 {x}^{3} + 5 {x}^{2} - 4 x + 4 = {\left(x - 2\right)}^{2} \left({x}^{2} + 1\right)$

$= {\left(x - 2\right)}^{2} \left(x - i\right) \left(x + i\right)$

So the $4$ roots are $2$, $2$, $i$ and $- i$

graph{x^4-4x^3+5x^2-4x+4 [-9.625, 10.375, -1.36, 8.64]}