SOLUTION: Find the factorized form of cubic polynomial f(x) satisfying the following conditions, f(1)=0, f(3)=0,the coefficient of x^3 is 1 and f(2)=6

Algebra ->  Radicals -> SOLUTION: Find the factorized form of cubic polynomial f(x) satisfying the following conditions, f(1)=0, f(3)=0,the coefficient of x^3 is 1 and f(2)=6      Log On


   

Question 678149: Find the factorized form of cubic polynomial f(x) satisfying the following conditions, f(1)=0, f(3)=0,the coefficient of x^3 is 1 and f(2)=6
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Some things we need to know about this problem:
  • A zero of a polynomial is a value for the variable that makes the whole polynomial zero.
  • We are given that f(1) and f(3) are both equal to zero. This means that 1 and 3 are zeros of our polynomial.
  • A polynomial will have as many zeros as the degree of a polynomial
  • So a cubic polynomial will have three zeros.
  • If a polynomial has complex or imaginary zeros then they will come in conjugate pairs.
  • So our cubic polynomial will either have three real zeros or one real zero and two complex/imaginary zeros. Since we already have two real zeros, our polynomial must have three real zeros.
  • A polynomial can be expressed as a product involving factors of the form: (x-z) where "z" is a zero of the polynomial. For our cubic polynomial this would be:
    f%28x%29+=+a%28x-z%5B1%5D%29%28x-z%5B2%5D%29%28x-z%5B3%5D%29
So we can use the following equation (and the given information) to find our polynomial:
f%28x%29+=+a%28x-z%5B1%5D%29%28x-z%5B2%5D%29%28x-z%5B3%5D%29
As pointed out above we already know two of the zeros, 1 and 3. So we can write:
f%28x%29+=+a%28x-1%29%28x-3%29%28x-z%5B3%5D%29
All we have to do now is find values for "a" and the third zero. To find these we will be using the fact that the coefficient of x%5E3 is 1 and the fact that f(2) = 6. First we multiply out our equation. Using FOIL on (x-1)(x-3):
f%28x%29+=+a%28x%5E2-4x%2B3%29%28x-z%5B3%5D%29
Next we'll multiply each term of x%5E2-4x%2B3 by each term of x-z%5B3%5D:
f%28x%29+=+a%28x%5E3-x%5E2z%5B3%5D-4x%5E2%2B4xz%5B3%5D%2B3x-3z%5B3%5D%29
And last we will distribute the a:
f%28x%29+=+ax%5E3-ax%5E2z%5B3%5D-4ax%5E2%2B4axz%5B3%5D%2B3ax-3az%5B3%5D
Since the coefficient of x%5E3 is 1 and the equation has "a" as the coefficient, "a" must be 1. Replacing all the a's with 1's we get:
f%28x%29+=+x%5E3-x%5E2z%5B3%5D-4x%5E2%2B4xz%5B3%5D%2B3x-3z%5B3%5D
Now we will use f(2) = 6. Replacing f(x) with 6 and x with 2 we get:

which simplifies as follows:
6+=+8-4z%5B3%5D-4%2A4%2B8z%5B3%5D%2B6-3z%5B3%5D
6+=+-2+%2Bz%5B3%5D
Adding 2 to each side:
8+=+z%5B3%5D

Now that we know our third zero, 8, we can complete the equation by replacing the z%5B3%5D's with 8's in:
f%28x%29+=+x%5E3-x%5E2z%5B3%5D-4x%5E2%2B4xz%5B3%5D%2B3x-3z%5B3%5D
f%28x%29+=+x%5E3-x%5E2%288%29-4x%5E2%2B4x%288%29%2B3x-3%288%29
which simplifies as follows:
f%28x%29+=+x%5E3-8x%5E2-4x%5E2%2B32x%2B3x-24
This is the desired cubic polynomial. You can see that the coefficient of x%5E3 is correct. And you can try f(1), f(3) and f(2) to see if you get the given values: 0. 0 and 6.