How do I solve this fraction question? If $a = -1/5$, how do I calculate:
$$3 a + 2 a^2$$
I did $3\times(-1/5) + (-1/5) \times (-1/5) \times 2$, but can't figure out what the right way to solve this is.
 A: That's the correct way.
\begin{align}
\left(3 \times \frac{-1}{5}\right)+\left(\frac{-1}{5} \times 2 \times \frac{-1}{5}\right) &= \frac{-3}{5}+\frac{2}{25} \\
&=\frac{-15}{25}+\frac{2}{25} \\
&= \frac{-13}{25}\\
\end{align}
A: A helpful tip:
You can only add fractions with the same denominator (the numbers at the bottom). If the denominators are not equal, you have to multiply them by a constant number $k$  to add them.
Starting from $$\frac{3 \times -1}{5} + \frac{-1 \times -1 \times 2}{5\times5},$$
since $+a \times -b = -ab$, and $-a \times -b = ab$ for arbitrary numbers $a,b$, we have:
$$-\frac{3}{5} + \frac{2}{25}$$
Since $\frac{1}{5}$ is $5$ times that of $\frac{1}{25}$, $k=5$, and we have to multiply both numerator and denominator by $5$ to simplify:
$$-\frac{15}{25} + \frac{2}{25}$$
Now you can simplify the expression by cancelling $\frac{10}{25}$ and adding the $2$ fractions together:
$$-\frac{15}{25} + \frac{2}{25} = \frac{-15+2}{5}$$
I'll leave the last step for you to figure out.
A: It might help you to add a denominator of $1$ to integers: $$\frac{3}{1} \times \frac{-1}{5} = \frac{-3}{5}.$$
So far so good, right? Next, $$\frac{2}{1} \times \left(\frac{-1}{5}\right)^2 = \frac{2}{1} \times \frac{1}{25} = \frac{2}{25}.$$
And then $$\frac{-3}{5} + \frac{2}{25} = \frac{-15}{25} + \frac{2}{25} = ?$$
