# Given that the polynomial function (below) has the given zero, find the other zeros? f(x) = x^4 - 5x^3 + 7x^2 - 5x + 6; -i

## I know the answers are $i , 2 , 3$ but I don't know how to find them.

Nov 16, 2017

The general factored form of a quartic equation is:

$y = k \left(x - {r}_{1}\right) \left(x - {r}_{2}\right) \left(x - {r}_{3}\right) \left(x - {r}_{4}\right) \text{ [1]}$

Where ${r}_{1} , {r}_{2} , {r}_{3} , \mathmr{and} {r}_{4}$ are the roots of the equation.

#### Explanation:

Given: f(x) = x^4 - 5x^3 + 7x^2 - 5x + 6; -i

We can substitute $- i$ for ${r}_{1}$ into equation [1]:

$y = k \left(x + i\right) \left(x - {r}_{2}\right) \left(x - {r}_{3}\right) \left(x - {r}_{4}\right) \text{ [1.1]}$

Because all of the coefficients are real, we know that imaginary roots must exist in conjugate pairs, therefore, $i$ must, also, be a root. Substitute $i$ for ${r}_{2}$ in equation [1.1]:

$y = k \left(x + i\right) \left(x - i\right) \left(x - {r}_{3}\right) \left(x - {r}_{4}\right) \text{ [1.2]}$

Because the leading coefficient of the given equation is 1, we know that $k = 1$, therefore, remove k from equation [1.2]:

$y = \left(x + i\right) \left(x - i\right) \left(x - {r}_{3}\right) \left(x - {r}_{4}\right) \text{ [1.3]}$

We know that $\left(x - i\right) \left(x + i\right) = {x}^{2} + 1$, therefore, the original polynomial must be evenly divisible by ${x}^{2} + 1$:

$\frac{{x}^{4} - 5 {x}^{3} + 7 {x}^{2} - 5 x + 6}{{x}^{2} + 1} = {x}^{2} - 5 x + 6 \leftarrow$ the factors of this quadratic must contain the remaining roots:

${x}^{2} - 5 x + 6 = \left(x - 2\right) \left(x - 3\right)$

$y = \left(x + i\right) \left(x - i\right) \left(x - 2\right) \left(x - 3\right) \text{ [1.4]}$

The roots are: $- i , i , 2 , \mathmr{and} 3$