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

Sep 29, 2016

The roots are:

$x = 1 \text{ }$ with multiplicity $2$

$x = - 3 \text{ }$ (with multiplicity $1$)

#### Explanation:

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

Note that the sum of the coefficients of $f \left(x\right)$ is $0$. That is:

$1 + 1 - 5 + 3 = 0$

Hence $f \left(1\right) = 0$ and $\left(x - 1\right)$ is a factor:

${x}^{3} + {x}^{2} - 5 x + 3 = \left(x - 1\right) \left({x}^{2} + 2 x - 3\right)$
$\textcolor{w h i t e}{{x}^{3} + {x}^{2} - 5 x + 3} = \left(x - 1\right) \left(x - 1\right) \left(x + 3\right)$

Hence the roots of the given equation are:

$x = 1$ with multiplicity $2$

$x = - 3$

graph{x^3+x^2-5x+3 [-10.545, 9.455, -22.2, 27.8]}