Question

According to the Rational Root Theorem, the following are potential roots of f(x) = 6x4 + 5x3 - 33x2 – 12x + 20.

Which is an actual root of f(x)?

a) -5/2
b) -2
c) 1
d) 10/3
​

Answer 1

a) -5/2.

Explanation:

Try substituting (-5/2) into f(x):

6(-5/2)^4 + 5(-5/2)^3 - 33(-5/2)^2 - 12(-5/2) + 20

= 234.375 + -78.125 - 206.25 + 30 + 20

= 0.

So it is -5/2.

Answer 2

a)
`x= -5/2`

Explanation:

We substitute in the function the values given in the options to confirm whether they are roots or not

a)
`x= -5/2`

the function will be:

`f(-(5)/(2) )=6(-(5)/(2) )^4 + 5(-(5)/(2) )^3-33(-(5)/(2) )^2-12(-(5)/(2))+20`

simplifying:

`f(-(5)/(2) )=6((625)/(16) )-5((125)/(8) )-33((25)/(4) )+30+20`

`f(-(5)/(2) )=234.375-78.125-206.25+50\\f(-(5)/(2) )=0\\\\`

since the function when
`x= -5/2` is equal to zero, this is an actual root of f(x).