# How do you find all the zeros of  f(x)=x^3+x^2-5x+3?

May 19, 2016

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

#### Explanation:

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

First note that the sum of the coefficients is $0$. That is:

$1 + 1 - 5 + 3 = 0$

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

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

Again note that the sum of the coefficients of ${x}^{2} + 2 x - 3$ is $0$, so there is another factor $\left(x - 1\right)$

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

Putting it all together:

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