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Factor completely. 32a 5011


please show your work and make sure the answers are correct.

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Answer #1

We have 32x-50x11 = 2x(16-25x4) = 2x[ 42 –(5x2)2] = 2x(4-5x2)(4+5x2) = 2x (4+5x2) [ 22 – (x√5)2] = 2x (4+5x2) (2 + x√5)(2- x√5).

Even 4+5x2 can be factored further, if we use complex numbers. Since i2 = -1, we have 4+5x2 = [22-(xi√5)2] = (2+ xi√5)(2- xi√5). Therefore, 32x-50x11 =2x (2 + x√5)(2- x√5) (2+ xi√5)(2- xi√5).   

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