Factoring

[Marie Eggplant]

Factoring bi/trinomials is kicking my butt ... could use some help on this one;

x cubed minus 5x squared minus 25x plus 125

 

[Unco]

Let P(x) = x³ - 5x² - 25x + 125

On a hunch, I substituted x=5 to find P(5) = 0.

By the factor theorum, (x-5) is thus a factor.

Long divide x³ - 5x² - 25x + 125 by (x-5):

        x^2  + 0    - 25
     _________________________
 x-5 \  x^3 - 5x^2 - 25x + 125
     /
        x^3 - 5x^2
        -------------
              0   -  25x
              0   +   0
              -----------
                    -25x + 125
                    -25x + 125
                    ----------
                            0     <- zero remainder confirms x-5 was a factor

Thus x³ - 5x² - 25x + 125 factorises to
(x-5)(x²-25)
ie. (x-5)(x+5)(x-5) = (x+5)(x-5)²

Of course, using the factor theorum with x+5 as well would have been a lot easier!

 

[Gene]

It should have been a little more than a hunch. The factors of 125 are +1,5,25,125 if there is an integer answer. The method is fine though just substituting those 8 possibilities might be faster before dividing.