Respuesta :
answer : A , C , D
[tex]f(x) = 10x^3 + 29x^2 - 66x + 27[/tex]
WE plug in each root in f(x) and check whether we get value =0
When f(a) =0 then 'a' is the actual root
Lets start with -9/2, plug in -9/2 for x
[tex]f(x) = 10x^3 + 29x^2 - 66x + 27[/tex]
[tex]f(-9/2)= 10(\frac{-9}{2})^3 + 29(\frac{-9}{2})^2 - 66(\frac{-9}{2})+ 27=0[/tex]
Hence -9/2 is one of the actual root
now plug in -9/10 for x
[tex]f(x) = 10x^3 + 29x^2 - 66x + 27[/tex]
[tex]f(-9/10)= 10(\frac{-9}{10})^3 + 29(\frac{-9}{10})^2 - 66(\frac{-9}{10})+ 27=\frac{513}{5}[/tex]
Hence -9/10 is not an actual root
now plug in 3/5 for x
[tex]f(x) = 10x^3 + 29x^2 - 66x + 27[/tex]
[tex]f(3/5)= 10(\frac{3}{5})^3 + 29(\frac{3}{5})^2 - 66(\frac{3}{5})+ 27=0 [/tex]
So 3/5 is one of the actual root
Now plug in 1 for x
[tex]f(1) = 10(1)^3 + 29(1)^2 - 66(1)+ 27=0[/tex]
So 1 is one of the actual root
Now plug in 3 for x
[tex]f(3) = 10(3)^3 + 29(3)^2 - 66(3)+ 27=360[/tex]
So 3 is not the actual root
Answer is -9/2 , 3/5 and 1
