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**Hint:**A square root of a number $'x'$ is number $'y'$ that ${{y}^{2}}=x$ in other words, a number $'y'$ whose square is $X-$ For example $4$ and $-4$ are square roots of $16$ because ${{4}^{2}}={{\left( -4 \right)}^{2}}=16$

Factorisation is defined as breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity or factors, which when multiplied together given the original number or matrix, etc.

Example: $24=4\times 6$

$4$ and $6$ are the factors of $24.$

**Complete step by step solution:**

As per the problem square root of $10$ times square root of $6.$

Here, square root of $10$ is represented by $\sqrt{10}$ and square root of $6$ is represented by $\sqrt{6}$ and times is represented by $'X'$ multiplication.

Therefore the representation of statement is,

$\Rightarrow \sqrt{10}\times \sqrt{6}$

Here, you can multiply $10$ by $6.$

$=\sqrt{60}$

To factorize the $60$. This method can be adopted.

Factors of $60$ are $2\times 2\times \,3\times 5.$

Therefore you can write in this form.

$=\sqrt{2\times 2\times 3\times 5}$

$2\times 2=4$ or ${{2}^{2}}=4$ as you can write ${{2}^{2}}$ on the place of $2\times 2$

$=\sqrt{{{2}^{2}}\times 3\times 5}$

Here, apply rule of $\sqrt{{{a}^{2}}}=a$ means you can write $\sqrt{{{2}^{2}}}=2$

Therefore the modified equation will be,

$=2\sqrt{3\times 5}$

$=2\sqrt{15}$

**Hence the simplification of square root of $10$ times square root of $6$ is $2\sqrt{15}$**

**Additional Information:**

When you have a root square root for example in the denominator of a friction you can ‘remove’ it multiplying and dividing the fraction for the same quantity. The idea is to avoid the rational number in the denominator.

Consider $\dfrac{3}{\sqrt{2}}$

You can remove the square root by multiplying and dividing by $\sqrt{2}$

$\dfrac{3}{\sqrt{2}}\dfrac{\sqrt{2}}{\sqrt{2}}$

The operation does not change the value of you fraction because $\dfrac{\sqrt{2}}{\sqrt{2}}=1$ anyway and you fraction does not change by multiplying $1$ to it.

Now you can multiply in the numerator and denominator.

$\dfrac{3}{\sqrt{2}}.\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{3\sqrt{2}}{\left( \sqrt{2} \right).\left( \sqrt{2} \right)}=\dfrac{3\sqrt{2}}{2}$

**Note:**Multiplication and division radicals

$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$

$\sqrt{a}\times \sqrt{b}=\sqrt{a\times b}$

Or $\sqrt{a}\sqrt{b}=\sqrt{a\times b}$

Remember that you can multiply the two separate square root terms. Factorization can be done by using LCM method also.

Write $2\times 2={{2}^{2}}$

Apply rule $\sqrt{{{a}^{2}}}=a$

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