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

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Answer 1 · 2017-06-01T16:55:04

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

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

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

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

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

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

How do you find all zeros of f(x)=1/3x^2+1/3x-2/3f(x)=13x2+13x−23f(x)=1/3x^2+1/3x-2/3?