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

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

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Answer 1 · 2016-09-02T01:33:35

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

Explanation:

Because this polynomial has a coefficient of $1$ $1$ for the squared term, $x^{2}$ $x^2$ , you can easily find out that the factored form will look some like this:
$(x +- _ )(x +- _ )$

Now you just look at the constant term, which is $4/3$ , and see what numbers you can multiply by itself to get that number.
Now since the constant is a fraction this might be a little difficult but
that number is $2/3$ .
And now this problem is being extremely nice to you because $2/3+2/3=4/3$
Because the $4/3$ is negative and the $4/9$ is positive we get:
$(x-2/3)(x-2/3)$ or $(x-2/3)^2$

Hope this helped! :3

Answer 2 · 2016-09-02T01:39:04

$(x-2/3)(x-2/3)$

Explanation:

$-(2/3)xx-(2/3) = 4/9$

$-(2/3)-(2/3) = -(4/3)$

$(x-2/3)(x-2/3) = x^2-(2/3)x -(2/3)x +(4/9)$

$= x^2-(4/3)x+(4/9)$

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

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Explanation:

Explanation:

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