Question
Write the following squares of binomials as trinomials:
$\big(9\text{a}+\frac{1}{6}\big)^2$
$\big(9\text{a}+\frac{1}{6}\big)^2$
✓
Answer
We will use the identities $(a+b)^2=a^2+2 a b+b^2$ and $(a-b)^2=a^2-2 a b+b^2$ to convert the squares of binomials as trinomials.
$\big(9\text{a}+\frac{1}{6}\big)^2$
$=(9\text{a})^2+2(9\text{a})\big(\frac{1}{6}\big)+\big(\frac{1}{6}\big)^2$
$=81\text{a}^2+3\text{a}+\frac{1}{36}$
$\big(9\text{a}+\frac{1}{6}\big)^2$
$=(9\text{a})^2+2(9\text{a})\big(\frac{1}{6}\big)+\big(\frac{1}{6}\big)^2$
$=81\text{a}^2+3\text{a}+\frac{1}{36}$
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