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

Respuesta :

Answer:

a) -5/2.

Step-by-step 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:

a) [tex]x= -5/2[/tex]

Step-by-step explanation:

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

a) [tex]x= -5/2[/tex]

the function will be:

[tex]f(-\frac{5}{2} )=6(-\frac{5}{2} )^4 + 5(-\frac{5}{2} )^3-33(-\frac{5}{2} )^2-12(-\frac{5}{2})+20[/tex]

simplifying:

[tex]f(-\frac{5}{2} )=6(\frac{625}{16} )-5(\frac{125}{8} )-33(\frac{25}{4} )+30+20[/tex]

[tex]f(-\frac{5}{2} )=234.375-78.125-206.25+50\\f(-\frac{5}{2} )=0\\\\[/tex]

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