Question

Who knows how to do this ughhh I need help!

Answer 1

Answer:

$\large\boxed{\sqrt{56x^{17}}=2x^8\sqrt{14x}}$

Step-by-step explanation:

$Domain:\ x\geq0\\\\\sqrt{56x^{17}}=\sqrt{4\cdot14\cdot x^{16+1}}\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=\sqrt{4\cdot14\cdot x^{16}\cdot x^1}\\\\\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt4\cdot\sqrt{14}\cdot\sqrt{x^{16}}\cdot\sqrt{x}=2\cdot\sqrt{14}\cdot\sqrt{x^{8\cdot2}}\cdot\sqrt{x}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=2\cdot\sqrt{14}\cdot\sqrt{(x^8)^2}\cdot\sqrt{x}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=2\cdot\sqrt{14}\cdot x^8\cdot\sqrt{x}=2x^8\sqrt{14x}$