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Can I factorize the expression: $$\sqrt {x^2} + \sqrt {x}$$ for any given $x$?

In the expression: $$\sqrt {10^2} + \sqrt {10}$$ is $\sqrt {10}$ a common factor, and the other factor is $\sqrt {10} +1$?

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    Yes.... you can take out $\sqrt {10}$ as a common factor .... I general, $\sqrt {a^2} + \sqrt{a} = a + \sqrt {a} = \sqrt {a} ( \sqrt {a} + 1)$ . Hope this makes sense :) – 2017-01-19
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    $\sqrt{a^2}=|a|=a$, because $a\ge 0$, because $\sqrt{a}$ must exist because it's in the expression $\sqrt{a^2}+\sqrt{a}$. – 2017-01-19

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Yes, you can, and yes, it is.

However, this isn't considered the simplest form.

The simplest form of $$\sqrt {10^2} + \sqrt {10}$$ is just $$10 + \sqrt {10}$$

If you're handling a bigger expression with lots of variables and factors, the factorization you're asking about may help you arrive at a simpler form. Or it may not.

In any case, it's certainly mathematically correct.

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    Thank you for the answer and the explanation that helps me to understand the general idea behind it. – 2017-01-20
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    @Ryna, you're welcome. On this site, if your question has been answered to your satisfaction, you can also "accept" an answer by clicking the green checkmark to the left of the answer. See ["What should I do when someone answers my question?"](http://math.stackexchange.com/help/someone-answers) :) – 2017-01-20