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What are the zeros of f(x)=x^4-7x^3+15x^2-9x
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well one zero is x =0 :D
zero, one, and a double root at 3
so f(x)= x(x-1)(x-3)^2
do you know how to factor it out yourself though?
no
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\[f(x)=x^4-7x^3+15x^2-9x=x(x^3-7x^2+15x-9)\] but after that i am stuck because i don't know how to solve a cubic equation
possible rational zeros are \(\pm1,\pm3,\pm9\) but no one says the zeros are rational, so no i do not know how to factor it unless it was cooked up by some math teacher to make the zeros easy to find, which is what you have to do, check the easy ones
however, one thing i am very good at is cheating, so here is the answer as written by saljudie http://www.wolframalpha.com/input/?i=0%3Dx^4-7x^3%2B15x^2-9x
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