# How do you find all the real and complex roots and use Descartes Rule of Signs to analyze the zeros of P(x) = x^3 - 4x^2 + x + 6?

$x = - 1 , 2 , 3$

#### Explanation:

The given cubic polynomial: $P \left(x\right) = {x}^{3} - 4 {x}^{2} + x + 6$

we find that the sum of coefficients of odd powers of $x$ is equal to sum of coefficients of even powers of variable $x$ hence $x = - 1$ is a root of cubic polynomial ${x}^{3} - 4 {x}^{2} + x + 6$ i.e. $\left(x + 1\right)$ is a factor of ${x}^{3} - 4 {x}^{2} + x + 6$ & it can be factorized as follows

${x}^{3} - 4 {x}^{2} + x + 6$

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

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

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

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

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

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

hence, the roots of given cubic polynomial $P \left(x\right)$ are given as follows

$P \left(x\right) = 0$

$\left(x + 1\right) \left(x - 2\right) \left(x - 3\right) = 0$

$x = - 1 , 2 , 3$