Answer:
Therefore the roots of equation 1 are 1,3 1/2,-i,+i
Step-by-step explanation:
in this question we have given
[tex]f(x) = 2x^5 - 9x^4 + 12x^3 - 12x^2 + 10x - 3 = 0[/tex].......1
we have to find the roots of this equation
put x=1 in equation 1
[tex]f(1)=2-9+12-12+10-3\\f(1)=0[/tex]
It means x=1 is root of equation 1
now put x=3 in equation 1
[tex]f(3)=2\times3^5-9\times3^4+12\times3^3-12\times3^2+10\times3-3\\f(3)=0[/tex]
therefore,x=3 is root of equation 1
Now put x=[tex]\frac{1}{2}[/tex] in equation 1
[tex]f(\frac{1}{2})=2\times (\frac{1}{2} )^5-9\times(\frac{1}{2})^4+12\times(\frac{1}{2})^3-12*(\frac{1}{2})^2+10\times \frac{1}{2}-3\\f(\frac{1}{2})=0[/tex]
It means x=[tex]\frac{1}{2}[/tex] is also root of equation 1
therefore equation one can be written as
[tex]f(x)=(x-1)(x-3)(2x-1)(x^2+1)[/tex]
therefore remaining 2 roots can be calculated by solving
[tex]x^2+1=0\\x=\sqrt(-1)\\x=i,-i[/tex]
Therefore the roots of equation 1 are 1,3 1/2,-i,+i
