Simplifying Square Roots

$\sqrt{4}\times \sqrt{25}=2\times 5=10$, and $\sqrt{4\times 25}=\sqrt{100}=10$, since $10\times 10=100$. Both routes give $\boxed{{\displaystyle 10}}$. Rooting a product and multiplying the roots always agree, and that is the rule the rest of this lesson runs on.

Since $12=4\times 3$, $$\sqrt{12}=\sqrt{4}\times \sqrt{3}=2\sqrt{3}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 2}}$. The $3$ has no perfect-square factor, so it stays under the radical. Squaring back gives $(2\sqrt{3}{)}^2=4\times 3=12$.

Since $72=36\times 2$, $$\sqrt{72}=\sqrt{36}\times \sqrt{2}=6\sqrt{2}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 6}}$. The leftover $2$ is square-free, so $6\sqrt{2}$ is simplest, and $(6\sqrt{2}{)}^2=36\times 2=72$ confirms it.

Since $48=16\times 3$, $$\sqrt{48}=\sqrt{16}\times \sqrt{3}=4\sqrt{3}\mathrm{.}$$ The number left under the radical is $\boxed{{\displaystyle 3}}$. The $4$ comes out front and the $3$ stays put, because $3$ has no perfect-square factor bigger than $1$.

Since $98=49\times 2$, $$\sqrt{98}=\sqrt{49}\times \sqrt{2}=7\sqrt{2}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 7}}$. To find that factor, run through the squares $4,9,16,25,36,49$ and keep the largest one that divides the radicand.

Factor $30=2\times 3\times 5$. Every prime shows up once, so there is no repeated prime to build a square from, and none of $4$, $9$, $16$, or $25$ divides $30$. The largest perfect-square factor is $\boxed{{\displaystyle 1}}$. Pulling out $\sqrt{1}=1$ changes nothing, so $\sqrt{30}$ is already in simplest form.

Combine the roots, $\sqrt{10}\times \sqrt{5}=\sqrt{50}$. The largest perfect square dividing $50$ is $25$, so $$\sqrt{50}=\sqrt{25}\times \sqrt{2}=5\sqrt{2}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 5}}$. Combining first often exposes a square factor that neither root showed on its own.

Since $507=169\times 3$ and $169=13^2$, $$\sqrt{507}=\sqrt{169}\times \sqrt{3}=13\sqrt{3},$$ so $a=\boxed{{\displaystyle 13}}$. Because $507$ is odd, only odd squares can divide it, which shortens the search to $9,25,49,121,169$.

Since $8=4\times 2$, $$\sqrt{8}=\sqrt{4}\times \sqrt{2}=2\sqrt{2}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 2}}$.

Since $75=25\times 3$, $$\sqrt{75}=\sqrt{25}\times \sqrt{3}=5\sqrt{3}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 5}}$.

Since $44=4\times 11$, $$\sqrt{44}=\sqrt{4}\times \sqrt{11}=2\sqrt{11}\mathrm{.}$$ The $11$ is square-free and stays under the radical, so the number left under it is $\boxed{{\displaystyle 11}}$.

Since $128=64\times 2$, $$\sqrt{128}=\sqrt{64}\times \sqrt{2}=8\sqrt{2}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 8}}$.

Join the roots, $\sqrt{3}\times \sqrt{12}=\sqrt{3\times 12}=\sqrt{36}$. Since $36$ is a perfect square, $\sqrt{36}=6$, a clean whole number. So the value is $\boxed{{\displaystyle 6}}$.

Factor $60=2\times 2\times 3\times 5$. The only repeated prime is $2$, giving the perfect square $2\times 2=4$. No larger perfect square divides $60$, since $3$ and $5$ appear only once. So the largest perfect-square factor is $\boxed{{\displaystyle 4}}$. (It would simplify $\sqrt{60}$ to $2\sqrt{15}$.)

Factor $33=3\times 11$. Each prime appears once, so there is no repeated factor to form a perfect square. The largest perfect square dividing $33$ is $\boxed{{\displaystyle 1}}$, which means $\sqrt{33}$ is already in simplest form.

Join the roots, $\sqrt{15}\times \sqrt{6}=\sqrt{15\times 6}=\sqrt{90}$. The largest perfect square dividing $90$ is $9$, and $90=9\times 10$, so $$\sqrt{90}=\sqrt{9}\times \sqrt{10}=3\sqrt{10}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 3}}$.

Since $675=225\times 3$ and $225=15^2$, $$\sqrt{675}=\sqrt{225}\times \sqrt{3}=15\sqrt{3}\mathrm{.}$$ The whole number in front is $\boxed{{\displaystyle 15}}$. Squaring back, $(15\sqrt{3}{)}^2=225\times 3=675$.