Question

multiply. assume q, p, and r are greater than or equal to zero, and write your answer in simplest form. $sqrt{33q^{5}} cdot 6sqrt{98p^{2}q^{4}r^{3}}$

Explanation:

Step1: Combine the square roots

$\sqrt{33q^5} \cdot 6\sqrt{98p^2q^4r^3} = 6\sqrt{33q^5 \cdot 98p^2q^4r^3}$

Step2: Multiply terms inside root

Calculate $33 \cdot 98 = 3234$, and combine like variables: $q^5 \cdot q^4 = q^{9}$, so:
$6\sqrt{3234p^2q^9r^3}$

Step3: Simplify the radical

Factor perfect squares: $3234 = 49 \cdot 66$, $q^9 = q^8 \cdot q$, $r^3 = r^2 \cdot r$. Extract perfect squares:
$6 \cdot \sqrt{49} \cdot \sqrt{p^2} \cdot \sqrt{q^8} \cdot \sqrt{r^2} \cdot \sqrt{66qr}$
$= 6 \cdot 7 \cdot p \cdot q^4 \cdot r \cdot \sqrt{66qr}$

Step4: Multiply the constant terms

$6 \cdot 7 = 42$, so:
$42pq^4r\sqrt{66qr}$

Answer:

$42pq^4r\sqrt{66qr}$