How do you factor the trinomial $x^{2} - \frac{6}{5} x + \frac{9}{25}$ $x^2-6/5x+9/25$ ?

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How do you factor the trinomial $x^{2} - \frac{6}{5} x + \frac{9}{25}$ $x^2-6/5x+9/25$ ?

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Answer 1 · 2016-06-01T06:19:47

$x^{2} - \frac{6}{5} x + \frac{9}{25} = {(x - \frac{3}{5})}^{2}$ $x^2-6/5x+9/25=(x-3/5)^2$

Explanation:

As $\frac{9}{25}$ $9/25$ is the square of $\frac{3}{5}$ $3/5$ and middle term $\frac{6}{5}$ $6/5$ is double of this number, (compare it with ${(x - a)}^{2} = x^{2} - 2 a x + a^{2}$ $(x-a)^2=x^2-2ax+a^2$ )

$x^{2} - \frac{6}{5} x + \frac{9}{25}$ $x^2-6/5x+9/25$ is a complete square.

$x^{2} - \frac{6}{5} x + \frac{9}{25}$ $x^2-6/5x+9/25$

= $x^{2} - 2 \times \frac{3}{5} \times x + {(\frac{3}{5})}^{2}$ $x^2-2xx3/5xx x+(3/5)^2$

= ${(x - \frac{3}{5})}^{2}$ $(x-3/5)^2$

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How do you factor the trinomial $x^{2} - \frac{6}{5} x + \frac{9}{25}$ $x^2-6/5x+9/25$ ?

1 Answer

Explanation:

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