SOLUTION: Find a third degree polynomial function f(x) with real coefficients that has -3,-2,and 2 as zeros such that f(1)=-12

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find a third degree polynomial function f(x) with real coefficients that has -3,-2,and 2 as zeros such that f(1)=-12      Log On


   

Question 812405: Find a third degree polynomial function f(x) with real coefficients that has -3,-2,and 2 as zeros such that f(1)=-12
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
A general formula for a 3rd degree polynomial in factored form would be:
f%28x%29=a%28x-z%5B1%5D%29%28x-z%5B2%5D%29%28x-z%5B3%5D%29
where the z's are zeros of the polynomial and the "a" is a non-zero constant. Using the given zeros we get:
f%28x%29=a%28x-%28-3%29%29%28x-%28-2%29%29%28x-2%29
which simplifies to:
f%28x%29=a%28x%2B3%29%28x%2B2%29%28x-2%29
Now we must multiply this out and find the right value for "a", either one first. First I will find the "a". For this we will use the fact that f(1) = -12:
-12=a%28%281%29%2B3%29%28%281%29%2B2%29%28%281%29-2%29+=+0
Simplifying...
-12=a%284%29%283%29%28-1%29
-12=-12a
Dividing by -12:
1+=+a
So:
f%28x%29=1%28x%2B3%29%28x%2B2%29%28x-2%29
Now we multiply this out. I'll start by using the %28a%2Bb%29%28a-b%29=a%5E2-b%5E2 pattern to multiply the last two factors:
f%28x%29=%28x%2B3%29%28%28x%29%5E2-%282%29%5E2%29
which simplifies to:
f%28x%29=%28x%2B3%29%28x%5E2-4%29
Using FOIL on what's left:
f%28x%29=x%5E3-4x%2B3x%5E2-12
Reordering into standard form:
f%28x%29=x%5E3%2B3x%5E2-4x-12