Question

According to the Rational Root Theorem, the following are potential roots of f(x) = 6x4 + 5x3 – 33x2 – 12x + 20. -5/2,-2, 1, 10/3 Which is an actual root of f(x)?

Answer 1

-5/2 is the answer. I just took the test and got it right.

Answer 2

Answer:

The actual root of f(x) is -5/2 (first option)

Step-by-step explanation:

A root of f(x) is a value of the variable "x" that makes f(x)=0

1) With x=-5/2

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

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

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

f(-5/2)=3(625/8)-625/8-825/4+50

f(-5/2)=1,875/8-625/8-825/4+50

f(-5/2)=[1,875-625-2(825)+8(50)]/8

f(-5/2)=(1,875-625-1,650+400)/8

f(-5/2)=0/8

f(-5/2)=0 then x=-5/2 is a root of f(x)

2) With x=-2

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

f(-2)=6(2^4)+5(-2^3)-33(2^2)+24+20

f(-2)=6(16)-5(8)-33(4)+44

f(-2)=96-40-132+44

f(-2)=-32 different of 0, then x=-2 is not a root of f(x)

3) With x=1

f(1)=6(1)^4+5(1)^3-33(1)^2-12(1)+20

f(1)=6(1)+5(1)-33(1)-12+20

f(1)=6+5-33+8

f(1)=-14 different of 0, then x=1 is not a root of f(x)

4) With x=10/3

f(10/3)=6(10/3)^4+5(10/3)^3-33(10/3)^2-12(10/3)+20

f(10/3)=6(10^4/3^4)+5(10^3/3^3)-33(10^2/3^2)-4(10)+20

f(10/3)=6(10,000/81)+5(1,000/27)-33(100/9)-40+20

f(10/3)=2(10,000/27)+5,000/27-11(100/3)-20

f(10/3)=20,000/27+5,000/27-1,100/3-20

f(10/3)=[20,000+5,000-9(1,100)-27(20)]/27

f(10/3)=(20,000+5,000-9,900-540)/27

f(10/3)=14,560/27 different of 0, then x=10/3 is not a root of f(x)