# How do you solve x^3+3x^2-x-3=0?

Mar 18, 2016

$x = - 1 , + 1 , - 3$

#### Explanation:

You can factor it but here is a shorter trick

${x}^{3} + 3 {x}^{2} - x - 3 = 0$

${x}^{3} + 3 {x}^{2} = x + 3$

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

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

${x}^{2} = 1$

$x = \pm 1$

Thats 2 values lets go for the third one

{XD,we actually need to factor it to get the third value}
${x}^{3} + 3 {x}^{2} - x - 3 = 0$

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

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

$\implies x + 3 = 0 \implies x = - 3$

$\textcolor{b l u e}{\text{Now we have got our 3 values and we can be confident we have solved our cubic}}$