How do you factor #x^3-7x^(3/2)-8=0#?

1 Answer
Jim H
Apr 10, 2015

The key to factoring this is to notice (guess and verify) that the square of #x^(3/2)# is #x^3#

So I can think of this as:

#["something"]^2 - 7 ["something"] - 8 = 0#

Until you are more comfortable with the process, do the substitution: the "something" here is #x^(3/2)# so we'll use a different variable to rename #x^(3/2)#. The traditional variable in this situation is #u#.

Let #u = x^(3/2)#. That makes #u^2 = (x^(3/2))^2 =x^3#

So we have:
#u^2-7u-8=0#

#(u-8)(u+1)=0#

#u-8=3# or #u+1=0#, so

#u=8# or #u=-1#

Now go back to the original variable:

#x^(3/2)=8# or #x^(3/2) = -1#

#x=8^(2/3)# or #x = (-1)^(2/3)#

#x=(root(3)8)^2=4# or #x=(root(3)(-1))^2=-1#

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