Question 678149
Some things we need to know about this problem:<ul><li>A zero of a polynomial is a value for the variable that makes the whole polynomial zero.</li><li>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.</li><li>A polynomial will have as many zeros as the degree of a polynomial</li><li>So a cubic polynomial will have three zeros.</li><li>If a polynomial has complex or imaginary zeros then they will come in conjugate pairs.</li><li>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.</li><li>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(x) = a(x-z[1])(x-z[2])(x-z[3])}}}</li></ul>So we can use the following equation (and the given information) to find our polynomial:
{{{f(x) = a(x-z[1])(x-z[2])(x-z[3])}}}
As pointed out above we already know two of the zeros, 1 and 3. So we can write:
{{{f(x) = a(x-1)(x-3)(x-z[3])}}}
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^3}}} is 1 and the fact that f(2) = 6. First we multiply out our equation. Using FOIL on (x-1)(x-3):
{{{f(x) = a(x^2-4x+3)(x-z[3])}}}
Next we'll multiply each term of {{{x^2-4x+3}}} by each term of {{{x-z[3]}}}:
{{{f(x) = a(x^3-x^2z[3]-4x^2+4xz[3]+3x-3z[3])}}}
And last we will distribute the a:
{{{f(x) = ax^3-ax^2z[3]-4ax^2+4axz[3]+3ax-3az[3]}}}
Since the coefficient of {{{x^3}}} 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(x) = x^3-x^2z[3]-4x^2+4xz[3]+3x-3z[3]}}}
Now we will use f(2) = 6. Replacing f(x) with 6 and x with 2 we get:
{{{(6) = (2)^3-(2)^2z[3]-4(2)^2+4(2)z[3]+3(2)-3z[3]}}}
which simplifies as follows:
{{{6 = 8-4z[3]-4*4+8z[3]+6-3z[3]}}}
{{{6 = -2 +z[3]}}}
Adding 2 to each side:
{{{8 = z[3]}}}<br>
Now that we know our third zero, 8, we can complete the equation by replacing the {{{z[3]}}}'s with 8's in:
{{{f(x) = x^3-x^2z[3]-4x^2+4xz[3]+3x-3z[3]}}}
{{{f(x) = x^3-x^2(8)-4x^2+4x(8)+3x-3(8)}}}
which simplifies as follows:
{{{f(x) = x^3-8x^2-4x^2+32x+3x-24}}}
This is the desired cubic polynomial. You can see that the coefficient of {{{x^3}}} 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.