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

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

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Answer 1 · 2016-05-14T11:53:02

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

Explanation:

This is a perfect square trinomial:

Note that $m^{2}$ $m^2$ and $\frac{1}{9} = {(\frac{1}{3})}^{2}$ $1/9 = (1/3)^2$ are both perfect squares. So the remaining question is whether the middle term matches when you square $(m + \frac{1}{3})$ $(m+1/3)$ ...

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

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

1 Answer

Explanation:

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