How do you factor $\frac{1}{4} + x + x^{2}$ $1/4+x+x^2$ ?

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How do you factor $\frac{1}{4} + x + x^{2}$ $1/4+x+x^2$ ?

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Answer 1 · 2016-05-05T18:59:15

$\frac{1}{4} + x + x^{2} = {(\frac{1}{2} + x)}^{2}$ $1/4+x+x^2 = (1/2+x)^2$

Explanation:

This is a perfect square trinomial.

Both $\frac{1}{4} = {(\frac{1}{2})}^{2}$ $1/4 = (1/2)^2$ and $x^{2}$ $x^2$ are perfect squares, so the only remaining question is whether the middle term matches when you square $(\frac{1}{2} + x)$ $(1/2+x)$ :

${(\frac{1}{2} + x)}^{2} = {(\frac{1}{2})}^{2} + 2 (\frac{1}{2}) x + x^{2} = \frac{1}{4} + x + x^{2}$ $(1/2+x)^2 = (1/2)^2+2(1/2)x + x^2 = 1/4+x+x^2$

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How do you factor $\frac{1}{4} + x + x^{2}$ $1/4+x+x^2$ ?

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