Question 698436
4u^2 - 1


factors to


(2u + 1)(2u - 1)


using the difference of squares rule


The second term "(2u - 1)" cancels with the last term "(2u - 1)" on the very right.


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u^3 - 16u


factors to


u(u^2 - 16)


which further factors to


u(u-4)(u+4)


using the difference of squares rule


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u^2 + 4u


factors to


u(u + 4)


Notice the last two final factorizations have a 'u' and a 'u+4' in common. They cancel


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So after all of the factorizations and cancellations, we go from 


(4u^2 - 1)/(u^3 - 16u) * (u^2 + 4u)/(2u - 1)


to


((2u - 1)(2u + 1))/(u(u-4)(u+4)) * (u(u+4))/(2u - 1)


which simplifies to


(2u + 1) / (u-4)