How do I get to the answer (z-2) from ((z^3)-8)/((z^2)+(2*z)+4)?

Question

I'm stuck at factoring out cubed z.

So far, I've factored the expression ((z^3)-8)/((z^2)+(2*z)+4) into ((z-2)^3)/z(z+2)+4. I'm not sure I've done this correctly in the first place, mostly because of my confusion with the cube.

 

In my frustration, I consulted a number of online calculators so I could know where I'm going with this problem, and all of them gave me the answer (z-2); however despite messing with the numbers, I cannot for the life of me get to this answer.

 

Help please

Answer 1

(z³-8)/(z²+2z+4)

 

z³-8 = z³-2³ is a "difference of cubes" binomial.  There is a standard factoring formula for factoring the difference of two cubes:

 

a³-b³ = (a-b)(a²+ab+b²)

 

So for your problem (a=z & b=2):

 

z³-2³ = (z-2)(z²+2z+4)

 

So:

 

(z³-8)/(z²+2z+4) = (z-2)(z²+2z+4)/(z²+2z+4) = (z-2)

Answer 2

Finn: The numerator is a difference of two cubes. It's factors are (z-2) and( zsquared plus 2z plus 4). The second factors cancels with the denominator, leaving just (z-2). Dattaprabhakar (Dr. G.)

Answer 3

hi Finn,

 

    This system does not have the graphics to be able to demonstrate polynomial long division, check out this site they take you through it step by step http://www.purplemath.com/modules/polydiv2.htm

 

 

z²+2z+4 divided into z³-8=z-2 because (z-2)*(z²+2z+4)=z³-8

 

Hope this helps

Jim