SOLUTION: State the degree of each polynomial equation. Find all of the real and imaginary roots of each equation, stating multiplicity when it is greater than one. x^3 - 4x^2 - 6x = 0

Algebra ->  Radicals -> SOLUTION: State the degree of each polynomial equation. Find all of the real and imaginary roots of each equation, stating multiplicity when it is greater than one. x^3 - 4x^2 - 6x = 0       Log On


   

Question 728317: State the degree of each polynomial equation. Find all of the real and imaginary roots of each equation, stating multiplicity when it is greater than one.
x^3 - 4x^2 - 6x = 0
I know that the degree is 3 because of the highest power shown. but i was thinking that i should use the quadratic equation to find the multiplicity. i was wrong according to the answer section...2 +/- sqrt 10 please, can you show the work so that i can understand this problem
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
x³ - 4x² - 6x = 0

Factor x out of the left side:

x(x² - 4x - 6) = 0

We use the zero factor property:

x = 0;   x² - 4x - 6 = 0

         x = %28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
         x = %28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%2A%281%29%2A%28-6%29+%29%29%2F%282%2A%281%29%29+
         x = %284+%2B-+sqrt%2816%2B24+%29%29%2F2+         
         x = %284+%2B-+sqrt%2840+%29%29%2F2+
         x = %284+%2B-+sqrt%284%2A10%29%29%2F2+
         x = %284+%2B-+2sqrt%2810%29%29%2F2+
         x = %282%282+%2B-+sqrt%2810%29%29%29%2F2+
         x = %28cross%282%29%282+%2B-+sqrt%2810%29%29%29%2Fcross%282%29+
         x = 2±√10

So there are three solutions:

0, 2+√10, 2-√10

All have multiplicity 1, because there are the same number of
different solutions as the degree, which is 3.
 
Edwin