SOLUTION: find all real zeros(if any) and state the multiplicity of each f(x)=6x^(4)-24x^(3)+24x^(2)

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Question 657435: find all real zeros(if any) and state the multiplicity of each f(x)=6x^(4)-24x^(3)+24x^(2)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+6x%5E4-24x%5E3%2B4x%5E2
To find the zeros we factor the expression. And when we factor we start with the greatest common factor (GCF). The GCF of the right side is 6x%5E2:
f%28x%29+=+6x%5E2%28x%5E2-4x%2B4%29
Next we factor the trinomial. The trinomial fits the patterns of a%5E2-2ab%2Bb%5E2+=+%28a-b%29%5E2 with the "a" being "x" and the "b" being 2:
f%28x%29+=+6x%5E2%28x-2%29%5E2

Now that f(x) is factored we can find the zeros. The zeros of f(x) are the values of x that make f(x) equal to zero. And now that we have f(x) factored (i.e. expressed as a product) we can use the Zero Product Property to find the zeros. Only the values of x that make the factors zero will make f(x) be zero. So our solutions will come from:
6+=+0
x+=+0
and
x-2+=+0
The first equation in untrue and has no solution. The solution to the second equation is x = 0 and the solution to the third equation is x = 2.

Finally, the multiplicity of each zero is the number of times each zero's factor is a factor of f(x). x = 0 is a zero of f(x) because of the factors of "x". Since f(x) has x%5E2 as a factor, it has two factors of x. So the multiplicity of the zero of 0 is two. Similarly, since there are two factors of (x-2) in f(x) the multiplicity of the zero of 2 is also two.

In summary, the zeros of f(x) are 0 and 2, each with a multiplicity of 2.