How do you factor the expression $x^{2} + \frac{2}{3} x + \frac{1}{9}$ $x^2 + 2/3x + 1/9$ ?

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How do you factor the expression $x^{2} + \frac{2}{3} x + \frac{1}{9}$ $x^2 + 2/3x + 1/9$ ?

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Answer 1 · 2016-03-08T15:52:12

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

Explanation:

This is a perfect square trinomial, which take the form:

${(x + a)}^{2} = a^{2} + 2 a x + a^{2}$ $(x+a)^2=a^2+2ax+a^2$

In

$x^{2} + \frac{2}{3} x + \frac{1}{9}$ $x^2+2/3x+1/9$

We see that $a = \frac{1}{3}$ $a=1/3$ , since $2 a = \frac{2}{3}$ $2a=2/3$ and $a^{2} = \frac{1}{9}$ $a^2=1/9$ .

${(x + \frac{1}{3})}^{2} = x^{2} + \frac{1}{3} x + \frac{1}{3} x + \frac{1}{9} = x^{2} + \frac{2}{3} x + \frac{1}{9}$ $(x+1/3)^2=x^2+1/3x+1/3x+1/9=x^2+2/3x+1/9$

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How do you factor the expression $x^{2} + \frac{2}{3} x + \frac{1}{9}$ $x^2 + 2/3x + 1/9$ ?

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