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Review Seth's steps for rewriting and simplifying an expression.

Given:
[tex]\[ 8 x^6 \sqrt{200 x^{13}} \div 2 x^5 \sqrt{32 x^7} \][/tex]

Step 1:
[tex]\[ 8 x^6 \sqrt{4 \cdot 25 \cdot 2 \cdot (x^6)^2 \cdot x} \div 2 x^5 \sqrt{16 \cdot 2 \cdot (x^3)^2 \cdot x} \][/tex]

Step 2:
[tex]\[ 8 \cdot 2 \cdot 5 \cdot x^6 \cdot x^6 \sqrt{2 x} \div 2 \cdot 16 \cdot x^5 \cdot x^3 \sqrt{2 x} \][/tex]

Step 3:
[tex]\[ 80 x^{12} \sqrt{2 x} \div 32 x^8 \sqrt{2 x} \][/tex]

Step 4:
[tex]\[ \frac{80 x^{12} \sqrt{2 x}}{32 x^8 \sqrt{2 x}} \][/tex]

Step 5:
[tex]\[ \frac{5}{2} x^4 \][/tex]

Answer :

We start with the expression

[tex]$$
\frac{8x^6 \sqrt{200x^{13}}}{2x^5 \sqrt{32x^7}}.
$$[/tex]

Below is a step-by-step simplification.

1. First, rewrite the radicals by factoring out perfect squares.

 • For the numerator, notice that
  [tex]$$
  200x^{13} = 4\cdot25\cdot2\,(x^6)^2\,x.
  $$[/tex]
  Thus,
  [tex]$$
  \sqrt{200x^{13}} = \sqrt{4\cdot25\cdot2\,(x^6)^2\,x} = 2\cdot5\,x^6\sqrt{2x} = 10x^6\sqrt{2x}.
  $$[/tex]

 • For the denominator, observe that
  [tex]$$
  32x^7 = 16\cdot2\,(x^3)^2\,x.
  $$[/tex]
  So,
  [tex]$$
  \sqrt{32x^7} = \sqrt{16\cdot2\,(x^3)^2\,x} = 4\,x^3\sqrt{2x}.
  $$[/tex]

2. Substitute these into the original expression:

 • The numerator becomes
  [tex]$$
  8x^6\cdot\left(10x^6\sqrt{2x}\right)=80x^{12}\sqrt{2x},
  $$[/tex]
 • The denominator becomes
  [tex]$$
  2x^5\cdot\left(4x^3\sqrt{2x}\right)=8x^{8}\sqrt{2x}.
  $$[/tex]

3. Divide the numerator by the denominator:

[tex]$$
\frac{80x^{12}\sqrt{2x}}{8x^8\sqrt{2x}} = \frac{80}{8}\cdot x^{12-8}\cdot \frac{\sqrt{2x}}{\sqrt{2x}}.
$$[/tex]

Here, [tex]$\sqrt{2x}$[/tex] cancels out, leaving

[tex]$$
10x^4.
$$[/tex]

Thus, the final simplified expression is

[tex]$$
10x^4.
$$[/tex]

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Rewritten by : Barada