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Factor the following sum of two cubes.

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1 Answer

Hi Jamal,

To factor this completely, you start by determining the first, most apparent factor, which you get by setting the expression equal to zero and solving for z.  

You get z^3 = -125   ------>>   z = -5

Therefore, one of your factors is (z+5).  

To get the other two, you need to divide this factor into the expression, so you set up the synthetic division like this:

-5    1  0  0     125                

          -5  25  -125

       1   -5  -125   0 

(The -5 is what you're dividing by. The other numbers are the coefficients of the 3rd order polynomial, with zeros in place for the missing x^2 and x terms.  Simply bring down the first coefficient unchanged, then multiply by the dividing factor and add to the next coefficient.)

These numbers become the coefficients of a polynomial one order less than the original, in this a quadratic:

x^2 -5x +125

This cannot be factored, so we use quadratic formula:

(5 +- sqrt(25 - 4*125))/2

= (5+- sqrt(-475))/2   =  (5 +-sqrt(-1)*sqrt(25)*sqrt(19))/2

= 2.5 +- 2.5i*sqrt(19)

So the fully factored polynomial is

(2.5 + 2.5i*sqrt(19)) (2.5 - 2.5i*sqrt(19)) (x + 5)


Hey Jamal,

There was actually a mistake here (though I posted a comment about it already, but don't see it)

The reduced quadratic equation from they synthetic division is:

x^2 - 5x + 25

Since this is prime, the solution is (x+5)(x^2 -5x + 25)