Luis Rojas

2022-07-03

alenahelenash

To determine the constant that should be added to the binomial ${x}^{2}+\frac{1}{6}x$ in order to make it a perfect square trinomial, we need to consider the square of a binomial pattern.
The square of the binomial $\left(a+b{\right)}^{2}$ can be expanded as follows:
$\left(a+b{\right)}^{2}={a}^{2}+2ab+{b}^{2}$
Comparing this pattern to the given binomial ${x}^{2}+\frac{1}{6}x$, we can see that ${a}^{2}$ corresponds to ${x}^{2}$, $2ab$ corresponds to $\frac{1}{6}x$, and ${b}^{2}$ corresponds to the unknown constant we need to find.
From the comparison, we can deduce that $2ab=\frac{1}{6}x$. In this case, $a=x$ and $b$ is the unknown constant we're looking for.
To find $b$, we can rewrite the equation as:
$2ab=\frac{1}{6}x$
Substituting $a=x$:
$2x·b=\frac{1}{6}x$
Now, we can solve for $b$:
$b=\frac{\frac{1}{6}x}{2x}=\frac{1}{12}$
Therefore, the constant that needs to be added to the binomial ${x}^{2}+\frac{1}{6}x$ to make it a perfect square trinomial is $\frac{1}{12}$.
Now, let's write and factor the trinomial.
The original binomial is ${x}^{2}+\frac{1}{6}x$. To make it a perfect square trinomial, we add ${\left(\frac{1}{12}\right)}^{2}$ to it:
${x}^{2}+\frac{1}{6}x+{\left(\frac{1}{12}\right)}^{2}={x}^{2}+\frac{1}{6}x+\frac{1}{144}$
Now, we can factor the trinomial:
${x}^{2}+\frac{1}{6}x+\frac{1}{144}={\left(x+\frac{1}{12}\right)}^{2}$
And that's the factored form of the trinomial.

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