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

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

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Answer 1 · 2016-01-16T02:47:42

This is a perfect square trinomial.

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

Explanation:

Perfect square trinomials are of the form:

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2 = a^2+2ab+b^2$

In this case $t^{2}$ $t^2$ and $\frac{1}{4} = {(\frac{1}{2})}^{2}$ $1/4 = (1/2)^2$ are both perfect squares. So the only question is do you get $t$ $t$ as a middle term if you square $(t + \frac{1}{2})$ $(t+1/2)$ ?

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

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