Final Answer
$\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{\left(2x-1\right)\left(4x^2-4x+1\right)}$
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Step-by-step Solution
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1
Multiplying the fraction by $5\left(2x^2-2x-13\right)$
$\frac{\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{2x-1}}{4x^2-4x+1}$
2
Divide fractions $\frac{\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{2x-1}}{4x^2-4x+1}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$
$\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{\left(2x-1\right)\left(4x^2-4x+1\right)}$
Final Answer
$\frac{5\left(x^2+3x+5\right)\left(2x^2-2x-13\right)}{\left(2x-1\right)\left(4x^2-4x+1\right)}$