How do you reduce (8z^3)-1/(2z^2)+(5z)-3?

Question

math hw help!!

Answer 1

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Seriously, the numerator is a difference of cubes and has a standard factor (one is 2z-1, the other is a quadradic), and the denominator is factorable to (2z-1)(z+3).  So the 2z-1 term cancels out, leaving the quadratic compliment in the numerator/(z+3)

Answer 2

The numerator is the difference of two cubes:

8z^3 - 1 = (2z - 1)(4z^2 + 2z + 1)

(You can multiply to check this.)

The denominator factors also:

2z^2 + 5z - 3 = (2z - 1)(z + 3)

Common factor of (2z - 1) in numerator and denominator = 1

So, the expression simplifies to:  (4z^2 + 2z + 1)/(z + 3)