How do you simplify $\frac{15 \sqrt{8 x^{16}}}{5 \sqrt{2 x^{4}}}$ $(15 sqrt (8x ^16))/( 5 sqrt (2x ^4))$ ?

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How do you simplify $\frac{15 \sqrt{8 x^{16}}}{5 \sqrt{2 x^{4}}}$ $(15 sqrt (8x ^16))/( 5 sqrt (2x ^4))$ ?

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Answer 1 · 2018-06-21T07:17:10

divide by 5

$\frac{3 \sqrt{8 x^{16}}}{\sqrt{2 x^{4}}}$ $[3sqrt(8x^16)]/[sqrt(2x^4)]$

square root the two $x$ $x$ terms

$\frac{3 x^{8} \sqrt{8}}{x^{2} \sqrt{2}}$ $[3x^8sqrt(8)]/[x^2sqrt(2)]$

cancel the $x^{2}$ $x^2$

$\frac{3 x^{6} \sqrt{8}}{\sqrt{2}}$ $[3x^6sqrt(8)]/[sqrt(2)]$

$\sqrt{8} = \sqrt{4} \times \sqrt{2} = 2 \sqrt{2}$ $sqrt8=sqrt4xxsqrt2=2sqrt2$

so $\frac{3 x^{6} \sqrt{8}}{\sqrt{2}}$ $[3x^6sqrt(8)]/[sqrt(2)]$ = $\frac{3 x^{6} 2 \sqrt{2}}{\sqrt{2}} = \frac{6 x^{6} \sqrt{2}}{\sqrt{2}}$ $[3x^(6)2sqrt(2)]/[sqrt(2)] =[6x^6sqrt(2)]/[sqrt(2)]$

cancel the $\sqrt{2}$ $sqrt2$

$6 x^{6}$ $6x^6$

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How do you simplify $\frac{15 \sqrt{8 x^{16}}}{5 \sqrt{2 x^{4}}}$ $(15 sqrt (8x ^16))/( 5 sqrt (2x ^4))$ ?

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How do you simplify $\frac{15 \sqrt{8 x^{16}}}{5 \sqrt{2 x^{4}}}$ $(15 sqrt (8x ^16))/( 5 sqrt (2x ^4))$ ?