document.write( "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 \n" ); document.write( "

Algebra.Com's Answer #421351 by jsmallt9(3758) $\"\"$ $\"About$

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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:
\n" ); document.write( " $\"f%28x%29+=+a%28x-z%5B1%5D%29%28x-z%5B2%5D%29%28x-z%5B3%5D%29\"$

So we can use the following equation (and the given information) to find our polynomial:
\n" ); document.write( " $\"f%28x%29+=+a%28x-z%5B1%5D%29%28x-z%5B2%5D%29%28x-z%5B3%5D%29\"$
\n" ); document.write( "As pointed out above we already know two of the zeros, 1 and 3. So we can write:
\n" ); document.write( " $\"f%28x%29+=+a%28x-1%29%28x-3%29%28x-z%5B3%5D%29\"$
\n" ); document.write( "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):
\n" ); document.write( " $\"f%28x%29+=+a%28x%5E2-4x%2B3%29%28x-z%5B3%5D%29\"$
\n" ); document.write( "Next we'll multiply each term of $\"x%5E2-4x%2B3\"$ by each term of $\"x-z%5B3%5D\"$ :
\n" ); document.write( " $\"f%28x%29+=+a%28x%5E3-x%5E2z%5B3%5D-4x%5E2%2B4xz%5B3%5D%2B3x-3z%5B3%5D%29\"$
\n" ); document.write( "And last we will distribute the a:
\n" ); document.write( " $\"f%28x%29+=+ax%5E3-ax%5E2z%5B3%5D-4ax%5E2%2B4axz%5B3%5D%2B3ax-3az%5B3%5D\"$
\n" ); document.write( "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:
\n" ); document.write( " $\"f%28x%29+=+x%5E3-x%5E2z%5B3%5D-4x%5E2%2B4xz%5B3%5D%2B3x-3z%5B3%5D\"$
\n" ); document.write( "Now we will use f(2) = 6. Replacing f(x) with 6 and x with 2 we get:
\n" ); document.write( "

\n" ); document.write( "which simplifies as follows:
\n" ); document.write( " $\"6+=+8-4z%5B3%5D-4%2A4%2B8z%5B3%5D%2B6-3z%5B3%5D\"$
\n" ); document.write( " $\"6+=+-2+%2Bz%5B3%5D\"$
\n" ); document.write( "Adding 2 to each side:
\n" ); document.write( " $\"8+=+z%5B3%5D\"$

\n" ); document.write( "Now that we know our third zero, 8, we can complete the equation by replacing the $\"z%5B3%5D\"$ 's with 8's in:
\n" ); document.write( " $\"f%28x%29+=+x%5E3-x%5E2z%5B3%5D-4x%5E2%2B4xz%5B3%5D%2B3x-3z%5B3%5D\"$
\n" ); document.write( " $\"f%28x%29+=+x%5E3-x%5E2%288%29-4x%5E2%2B4x%288%29%2B3x-3%288%29\"$
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( " $\"f%28x%29+=+x%5E3-8x%5E2-4x%5E2%2B32x%2B3x-24\"$
\n" ); document.write( "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. \n" ); document.write( "

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