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How do you find the zeros of the function f of x equals -x to the power of 4 plus 6x to the power of 3 -9x to the power of 2 and their multiplicities?
FaeEvrnytgp3758
Rewrite as f(x)=-x^4 + 6x^3 - 9x^2 (where ^ means to the power of)
= -x^2(x^2 - 6x + 9)
= -x^2 (x -3)(x-3)
So the zeroes are 0, 0, 3, 3.
I am not sure what you mean by multiplicities.
Wiki User
∙ 8y ago
Wiki User
∙ 8y agoYou start by factoring the polynomial. Factoring a polynomial of degree 3 or higher can be very tricky; in this case it's almost trivial, because of the common factor (x squared). After separating this factor, you can factor whatever remains, perhaps by using the common rules for factoring a trinomial. If nothing else works, you can use the quadratic formula.Once you have the factors, you set the polynomial = 0, which means that the product of all the factors is equal to zero. Remember that this is only possible if one of the factors is zero.
Wiki User
∙ 8y agoThis is what I think you are asking: f(x) = - (x^4) + 6(x³) - 9(x²) = 0. So you can factor out x² to get this: f(x) = (-x² + 6x - 9)(x²) = 0. So you have x=0 as a root (twice). Then you can factor the quadratic to get (-x² + 6x - 9) = (-1)((x² - 6x + 9) = (-1)(x - 3)(x - 3), which has x = 3 as a root (twice)
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What are the zeros of a function?
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