SOLUTION: Hi. May I have some help finding solutions to the equation? Find all real solutions of the following equation. (If there are extra answer boxes, enter NONE.) x^3 - x^2 + 2x -

Algebra ->  Equations -> SOLUTION: Hi. May I have some help finding solutions to the equation? Find all real solutions of the following equation. (If there are extra answer boxes, enter NONE.) x^3 - x^2 + 2x -      Log On


   

Question 121207: Hi. May I have some help finding solutions to the equation?
Find all real solutions of the following equation. (If there are extra answer boxes, enter NONE.)
x^3 - x^2 + 2x - 3 = x^2 + 1
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

Hi. May I have some help finding solutions to the equation? 
Find all real solutions of the following equation. 
(If there are extra answer boxes, enter NONE.) 

 x³ - x² + 2x - 3 = x² + 1

Get 0 on the right by subtracting x² + 1 from both sides:

x³ - 2x² + 2x - 4 = 0

Factor the first two terms on the left x³ - 2x² as x²(x - 2)

Factor the last two terms on the left  + 2x - 4 as + 2(x - 2)

x²(x - 2) + 2(x - 2) = 0

This is NOT factored because there is a + sign
which is NOT inside parentheses.

However there is a common factor of (x - 2), so we'll factor that
out:

x²(x - 2) + 2(x - 2) = 0
     (x - 2)(x² + 2) = 0

Now this is factored because now there are no + signs or - signs
which are not inside parantheses.  It is also as completely
factored as it can be factored using integers.

Set each factor = 0

x - 2 = 0 gives the solution x = 2 immediately

x² + 2 =  0    requires some more work. Subtract 2 from both sides:

    x² = -2    take square roots of both sides, remembering ±
                __
     x = ±Ö-2  simplify as imaginary numbers
                 _
     x = ±iÖ2
                                                       _               _
So the three solutions are 2, iÖ2, and -iÖ2

Edwin