SOLUTION: Will someone please help me with the process on finding the real zeros of: 3x^4 + 3x^3 - 17x^2 + x - 6 = 0. Thanx

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Will someone please help me with the process on finding the real zeros of: 3x^4 + 3x^3 - 17x^2 + x - 6 = 0. Thanx      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 149981: Will someone please help me with the process on finding the real zeros of:
3x^4 + 3x^3 - 17x^2 + x - 6 = 0.
Thanx
Answer by Fombitz(13823) About Me  (Show Source):
You can put this solution on YOUR website!
Usually I first graph it to see what's what using EXCEL.
+graph%28+300%2C+200%2C+-5%2C+5%2C+-20%2C+20%2C+3x%5E4%2B3x%5E3-17x%5E2%2Bx-6%29+
As you can see it looks like 2 real zeros at x=-3 and x=2.
Plug the values into the equation to verify.
x=-3
3x%5E4%2B3x%5E3-17x%5E2%2Bx-6=0
3%28-3%29%5E4%2B3%28-3%29%5E3-17%283%29%5E2%2B%28-3%29-6=0
3%2881%29-81-17%289%29-3-6=0
243-81-153-3-6=0
0=0
True.
x=2
3x%5E4%2B3x%5E3-17x%5E2%2Bx-6=0
3%282%29%5E4%2B3%282%29%5E3-17%282%29%5E2%2B%282%29-6=0
3%2816%29%2B3%288%29-17%284%29%2B2-6=0
48%2B24-68%2B2-6=0
0=0
True.
.
.
.
Both solutions led to true statements.
They are good solutions.
They are the real zeros of the equation.
x=2, x=-3