How do you find the roots of x^3-5x^2+2x+8=0?

Dec 24, 2016

Roots of ${x}^{3} - 5 {x}^{2} + 2 x + 8 = 0$ are $\left\{- 1 , 2 , 4\right\}$.

Explanation:

As the coefficient of ${x}^{3}$ in ${x}^{3} - 5 {x}^{2} + 2 x + 8 = 0$ is $1$,

One of the roots must be factor of $\frac{8}{1} = 8$ i.e. it can be $\pm 1. \pm 2 , \pm 4 \mathmr{and} \pm 8$.

As sum of coefficients ($1 - 5 + 2 + 8 = 6 \ne 0$) is not zero, it is evident that $1$ is not the root of ${x}^{3} - 5 {x}^{2} + 2 x + 8 = 0$.

But $- 1$ is a root as $- 1 - 5 - 2 + 8 = 0$ and hence as per factor theorem $\left(x + 1\right)$ is a factor of ${x}^{3} - 5 {x}^{2} + 2 x + 8 = 0$

Dividing ${x}^{3} - 5 {x}^{2} + 2 x + 8$ by $\left(x + 1\right)$ we get

${x}^{3} - 5 {x}^{2} + 2 x + 8 = {x}^{2} \left(x + 1\right) - 6 x \left(x + 1\right) + 8 \left(x + 1\right) = \left(x + 1\right) \left({x}^{2} - 6 x + 8\right)$

now we can further factorize ${x}^{2} - 6 x + 8$ by splitting middle term

${x}^{2} - 6 x + 8 = {x}^{2} - 4 x - 2 x + 8 = x \left(x - 4\right) - 2 \left(x - 4\right) = \left(x - 2\right) \left(x - 4\right)$

Hence, roots of ${x}^{3} - 5 {x}^{2} + 2 x + 8 = 0$ are $\left\{- 1 , 2 , 4\right\}$.