How do you find all zeros of f(x)=1/3x^2+1/3x-2/3?

Jun 1, 2017

Zeros of $f \left(x\right)$ are $x = 1$ and $x = - 2$.

Explanation:

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

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

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

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

Hence, zeros of $f \left(x\right)$ are $x = 1$ and $x = - 2$.