SOLUTION: f(x)=x^3+6x^2-25x+18 Find all the zeros of the given polynomial. can anyone help me solve this, I am new to this type of question. Jessica

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: f(x)=x^3+6x^2-25x+18 Find all the zeros of the given polynomial. can anyone help me solve this, I am new to this type of question. Jessica      Log On


   

Question 150029: f(x)=x^3+6x^2-25x+18
Find all the zeros of the given polynomial.
can anyone help me solve this, I am new to this type of question.
Jessica
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
A zero is where the function touches the x axis or in other words f(x)=0.
Usually I first graph it to see what's what using EXCEL or graph paper.
+graph%28+300%2C+300%2C+-10%2C+10%2C+-50%2C+50%2C+x%5E3%2B6x%5E2-25x%2B18+%29
As you can see it looks like 3 real zeros at x=1, x=2, and x=-9.
Plug the values into the equation to verify.
x=1
f%28x%29=x%5E3%2B6x%5E2-25x%2B18%29
f%281%29=1%5E3%2B6%281%29%5E2-25%281%29%2B18
f%281%29=1%2B6-25%2B18
f%281%29=0
That's a good one.
x=2
f%28x%29=x%5E3%2B6x%5E2-25x%2B18%29
f%282%29=2%5E3%2B6%282%29%5E2-25%282%29%2B18
f%282%29=8%2B24-50%2B18
f%282%29=0
That's a good one too.
x=-9
f%28x%29=x%5E3%2B6x%5E2-25x%2B18%29
f%28-9%29=%28-9%29%5E3%2B6%28-9%29%5E2-25%28-9%29%2B18
f%28-9%29=-729%2B486%2B225%2B18
f%28-9%29=0
That's a good one too.
You have found all three roots.
You could then represent the function as,
f(x)=(x-1)(x-2)(x+9)