How do you factor $- \frac{4}{3} x + \frac{4}{9} = - x^{2}$ $-4/3x+4/9=-x^2$ ?

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How do you factor $- \frac{4}{3} x + \frac{4}{9} = - x^{2}$ $-4/3x+4/9=-x^2$ ?

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Answer 1 · 2017-08-15T23:45:19

$x^{2} - \frac{4}{3} x + \frac{4}{9} = {(x - \frac{2}{3})}^{2}$ $x^2-4/3x+4/9=color(blue)((x-2/3)^2$

Refer to the explanation for the process.

Explanation:

Factor:

$- \frac{4}{3} x + \frac{4}{9} = - x^{2}$ $-4/3x+4/9=-x^2$

Add $x^{2}$ $x^2$ to both sides.

$x^{2} - \frac{4}{3} x + \frac{4}{9} = 0$ $x^2-4/3x+4/9=0$

Since the desired answer is the factors of the equation, the $0$ $0$ will not be included in the expression.

Find two numbers that when added equal $- \frac{4}{3}$ $-4/3$ and when multiplied equal $\frac{4}{9}$ $4/9$ . Both requirements are met by $- \frac{2}{3}$ $-2/3$ .

$x^{2} - \frac{4}{3} x + \frac{4}{9} = (x - \frac{2}{3}) (x - \frac{2}{3})$ $x^2-4/3x+4/9=(x-2/3)(x-2/3)$

$x^{2} - \frac{4}{3} x + \frac{4}{9} = {(x - \frac{2}{3})}^{2}$ $x^2-4/3x+4/9=(x-2/3)^2$

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Explanation:

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