Question

1. assuming x and y are both positive, write the following expression in simplest radical form. $7xy^2\sqrt{36x^3y^2}$

Explanation:

Step1: Simplify the square root

First, simplify $\sqrt{36x^{3}y^{2}}$. We know that $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ (for $a,b\geq0$) and $\sqrt{a^{2}} = a$ (for $a\geq0$). So, $\sqrt{36x^{3}y^{2}}=\sqrt{36}\cdot\sqrt{x^{3}}\cdot\sqrt{y^{2}}$. Since $\sqrt{36} = 6$, $\sqrt{y^{2}}=y$ (because $y>0$), and $\sqrt{x^{3}}=\sqrt{x^{2}\cdot x}=x\sqrt{x}$ (because $x>0$). So $\sqrt{36x^{3}y^{2}} = 6\cdot x\sqrt{x}\cdot y=6xy\sqrt{x}$.

Step2: Multiply with the outside term

Now, multiply this result with $7xy^{2}$. So we have $7xy^{2}\cdot(6xy\sqrt{x})$. Multiply the coefficients and the like - terms: $7\times6 = 42$, $x\times x=x^{2}$, $y^{2}\times y = y^{3}$. So the product is $42x^{2}y^{3}\sqrt{x}$.

Answer:

$42x^{2}y^{3}\sqrt{x}$