Question

According to the rational root theorem, the numbers below are some of the potential roots of f(x) = 10x3 + 29x2 – 66x + 27. Select all that are actual roots. A) -9/2 B)-9/10 C) 3/5 D) 1 E) 3

Answer 1

answer : A , C , D

$f(x) = 10x^3 + 29x^2 - 66x + 27$

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

$f(x) = 10x^3 + 29x^2 - 66x + 27$

$f(-9/2)= 10(\frac{-9}{2})^3 + 29(\frac{-9}{2})^2 - 66(\frac{-9}{2})+ 27=0$

Hence -9/2 is one of the actual root

now plug in -9/10 for x

$f(x) = 10x^3 + 29x^2 - 66x + 27$

$f(-9/10)= 10(\frac{-9}{10})^3 + 29(\frac{-9}{10})^2 - 66(\frac{-9}{10})+ 27=\frac{513}{5}$

Hence -9/10 is not an actual root

now plug in 3/5  for x

$f(x) = 10x^3 + 29x^2 - 66x + 27$

$f(3/5)= 10(\frac{3}{5})^3 + 29(\frac{3}{5})^2 - 66(\frac{3}{5})+ 27=0$

So 3/5 is one of the actual root

Now plug in 1 for x

$f(1) = 10(1)^3 + 29(1)^2 - 66(1)+ 27=0$

So 1 is one of the actual root

Now plug in 3 for x

$f(3) = 10(3)^3 + 29(3)^2 - 66(3)+ 27=360$

So 3 is not the actual root

Answer is -9/2  , 3/5  and 1

Answer 2

Answer:

-9/1, 3/5, 1

Step-by-step explanation: