How do you find the roots of x^3-5x^2+4x+10=0?

Nov 14, 2016

The roots are: $- 1$ and $3 \pm i$

Explanation:

$f \left(x\right) = {x}^{3} - 5 {x}^{2} + 4 x + 10$

By the reational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $10$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 5 , \pm 10$

We find:

$f \left(- 1\right) = - 1 - 5 - 4 + 10 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{3} - 5 {x}^{2} + 4 x + 10 = \left(x + 1\right) \left({x}^{2} - 6 x + 10\right)$

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - 3\right)$ and $b = i$ as follows:

${x}^{2} - 6 x + 10 = {x}^{2} - 6 x + 9 + 1$

$\textcolor{w h i t e}{{x}^{2} - 6 x + 10} = {\left(x - 3\right)}^{2} - {i}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 6 x + 10} = \left(\left(x - 3\right) - i\right) \left(\left(x - 3\right) + i\right)$

$\textcolor{w h i t e}{{x}^{2} - 6 x + 10} = \left(x - 3 - i\right) \left(x - 3 + i\right)$

Hence the other two zeros are: $x = 3 \pm i$