Answer:
Please check the explanation.
Step-by-step explanation:
Given the function
[tex]f(x)= 5x^4 -10x^3 +4x^2 -8x[/tex]
To determine the zeros, set f(x) = 0
[tex]0=\:5x^4\:-10x^3\:+4x^2\:-8x[/tex]
switch sides
[tex]5x^4-10x^3+4x^2-8x=0[/tex]
as
[tex]5x^4\:-10x^3\:+4x^2\:-8x=x\left(x-2\right)\left(5x^2+4\right)[/tex]
so the equation becomes
[tex]x\left(x-2\right)\left(5x^2+4\right)=0[/tex]
Using the zero factor principle:
[tex]\mathrm{\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex]
[tex]x=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:5x^2+4=0[/tex]
so solving
[tex]x-2=0[/tex]
Therefore, the real zeros are:
[tex]x=0[/tex] and [tex]x=2[/tex]
[tex]5x^2+4=0[/tex] has all the imaginary zeros.
solving
[tex]5x^2+4=0[/tex]
[tex]5x^2=-4[/tex]
[tex]x^2=-\frac{4}{5}[/tex]
[tex]\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]x=\sqrt{-\frac{4}{5}},\:x=-\sqrt{-\frac{4}{5}}[/tex]
[tex]x=i\frac{2\sqrt{5}}{5},\:x=-i\frac{2\sqrt{5}}{5}[/tex] ∵ [tex]\sqrt{-1}=i[/tex]
Therefore, the total zeros are:
[tex]x=0,\:x=2,\:x=i\frac{2\sqrt{5}}{5},\:x=-i\frac{2\sqrt{5}}{5}[/tex]
