How do you find all zeros with multiplicities of f(x)=x^6-3x^3-10f(x)=x6−3x3−10?
1 Answer
The six zeros are:
x = root(3)(5)x=3√5
x = -1/2root(3)(5)+-1/2sqrt(3)root(3)(5)ix=−123√5±12√33√5i
x = -root(3)(2)x=−3√2
x = 1/2root(3)(2)+-1/2sqrt(3)root(3)(2)ix=123√2±12√33√2i
all with multiplicity
Explanation:
x^6-3x^3-10 = (x^3-5)(x^3+2)x6−3x3−10=(x3−5)(x3+2)
Then:
x^3-5 = (x^3-(root(3)(5))^3)x3−5=(x3−(3√5)3)
color(white)(x^3-5) = (x-root(3)(5))(x^2+root(3)(5)x+root(3)(25))x3−5=(x−3√5)(x2+3√5x+3√25)
color(white)(x^3-5) = (x-root(3)(5))((x+1/2root(3)(5))^2+(1/2sqrt(3)root(3)(5))^2)x3−5=(x−3√5)((x+123√5)2+(12√33√5)2)
color(white)(x^3-5) = (x-root(3)(5))((x+1/2root(3)(5))^2-(1/2sqrt(3)root(3)(5)i)^2)x3−5=(x−3√5)((x+123√5)2−(12√33√5i)2)
color(white)(x^3-5) = (x-root(3)(5))(x+1/2root(3)(5)-1/2sqrt(3)root(3)(5)i)(x+1/2root(3)(5)+1/2sqrt(3)root(3)(5)i)x3−5=(x−3√5)(x+123√5−12√33√5i)(x+123√5+12√33√5i)
and:
x^3+2 = (x^3+(root(3)(2))^3)x3+2=(x3+(3√2)3)
color(white)(x^3+2) = (x+root(3)(2))(x^2-root(3)(2)x+root(3)(4))x3+2=(x+3√2)(x2−3√2x+3√4)
color(white)(x^3+2) =(x+root(3)(2))((x-1/2root(3)(2))^2+(1/2sqrt(3)root(3)(2))^2)x3+2=(x+3√2)((x−123√2)2+(12√33√2)2)
color(white)(x^3+2) =(x+root(3)(2))((x-1/2root(3)(2))^2-(1/2sqrt(3)root(3)(2)i)^2)x3+2=(x+3√2)((x−123√2)2−(12√33√2i)2)
color(white)(x^3+2) =(x+root(3)(2))(x-1/2root(3)(2)-1/2sqrt(3)root(3)(2)i)(x-1/2root(3)(2)+1/2sqrt(3)root(3)(2)i)x3+2=(x+3√2)(x−123√2−12√33√2i)(x−123√2+12√33√2i)
So the six zeros are:
x = root(3)(5)x=3√5
x = -1/2root(3)(5)+-1/2sqrt(3)root(3)(5)ix=−123√5±12√33√5i
x = -root(3)(2)x=−3√2
x = 1/2root(3)(2)+-1/2sqrt(3)root(3)(2)ix=123√2±12√33√2i
