How do you find all zeros with multiplicities of f(x)=9x^3-5x^2-x?
1 Answer
Aug 11, 2018
all with multiplicity
Explanation:
0 = 36f(x)
color(white)(0) = 36(9x^3-5x^2-x)
color(white)(0) = x(324x^2-180x-36)
color(white)(0) = x((18x)^2-2(18x)(5)+25-61)
color(white)(0) = x((18x-5)^2-(sqrt(61))^2)
color(white)(0) = x(18x-5-sqrt(61))(18x-5+sqrt(61))
Hence:
x=0
or:
18x=5+-sqrt(61)" " so" "x = 5/18+-sqrt(61)/18