Question

which of the following represents the zeros of f(x) = 6x^3 - 31x^2 + 4x + 5?

Answer 1

@phi ?

Answer 2

are there options that you're given?

Answer 3

yes

a. -5, 1/3, 1/2
b. 5, 1/3, 1/2
c. 5, 1/3, -1/2
d. 5, -1/3, 1/2

Answer 4

In that case, you can just plug each value into the equation and see if it gives you zero

ie)
\[ a)
\\ f(-5) = 6(-5)^3 - 31(-5)^2 + 4(-5) + 5 = ?
\\ f(1/3) = 6(1/3)^3 - 31(1/3)^2 + 4(1/3) + 5=?
\\ f(1/2) = 6(1/2)^3 - 31(1/2)^2 + 4(1/2) + 5=?
\]

Answer 5

a makes it equal 0?

Answer 6

You have to do the arithmetic to find out ^_^

Answer 7

hmmm i got b as the answer.

Answer 8

but i think i mess up on 1/3 let me try it again

Answer 9

yeah i got b

Answer 10

I think b is incorrect - plugging in x=1/3 does not make that equal 0.

Answer 11

i knew i messed up on 1/3. ok lemme try c

Answer 12

well, off the bat, if b doesn't work, neither will c - they both have x=1/3 as a root, and 1/3 we found is not a root.

Answer 13

so it has to be d then because its the only one without 1/3

Answer 14

correct! ^_^

Answer 15

yayyyy thanks man. could you help me with a couple more

Answer 16

sure

Answer 17

ok one is
which of the following is a polynomial with the roots

Answer 18

the choices are

a. x^3 - 2x^2 - 3x + 6

b. x^3 - 3x^2 - 5x + 15

c. x^3 + 2x^2 - 3x - 6

d. x^3 + 3x^2 - 5x - 15

Answer 19

So for this, it's pretty much the same process, you just have to plug in the roots into each of the equations.

Answer 20

yeah, but im not sure how to do that with radicals :/

Answer 21

what's
\[(\sqrt 3 )^2=?\]

Answer 22

81