SOLUTION: The polynomial equation x(x^2+4)(x^2-x-6)=0. Has how many real roots? We are very rusty on our Algebra skills, and we are reviewing for a mat

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The polynomial equation x(x^2+4)(x^2-x-6)=0. Has how many real roots? We are very rusty on our Algebra skills, and we are reviewing for a mat      Log On


   

Question 24416: The polynomial equation
x(x^2+4)(x^2-x-6)=0. Has how many real roots?
We are very rusty on our Algebra skills, and we are reviewing for a math placement exam, b/c we are returning back to college.
Thanks,
K
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You can apply the zero product principle to this problem: Ifa%2Ab+=+0 then either a+=+0 or b+=+0 or both.
x%28x%5E2%2B4%29%28x%5E2-x-6%29+=+0
x+=+0 and/or x%5E2%2B4+=+0 and/or x%5E2-x-6+=+0
One of the real roots is x = 0
If x%5E2%2B4+=+0 then x%5E2+=+-4 and x = +or-sqrt%28-4%29 so these two roots are:
x = 2i and x = -2i These are not real roots
If x%5E2-x-6+=+0 then %28x%2B2%29%28x-3%29+=+0 so x+=+-2 and x+=+3 These two roots are real.
The real roots are:
x = 0
x = -2
x = 3