Question

Which of the following represent the zeros of f(x) = 6x^3 – 31x^2 + 4x + 5 ?

Answer 1

choices:
–5, one–third, one–half

5, one–third, one–half

5, one–third, – one–half

5, –one–third, one–half

Answer 2

the zeros are the x intercepts I believe

Answer 3

yes

Answer 4

So we need to find the x intercepts lol

Answer 5

yes heard of 'rational root theorem' before ?

Answer 6

Yeah I think so, and I got it narrowed down to two answers, the first x intercept is + 5 not -5

Answer 7

cool :)
according to rational root theorem, possible rational zeroes will be of form :-
p/q

p = factors of 5 = \(\pm1, \pm 5\)
q = factors of 6 \(\pm1, \pm2, \pm3, \pm6\)

Answer 8

so 5, and then 1 over something and 1 over someting basically?

Answer 9

thats right !

Answer 10

so now we need to figure out what the denominators will be

Answer 11

yes, there is a standard method to find the remaining zeroes,
u familiar wid synthetic division also ?

Answer 12

since u knw that 5 is a zero, just divide (x-5), then u wil get a quadratic.

Answer 13

knw synthetic division ?

Answer 14

Kind of know synthetic

Answer 15

okay, lets divide : \((6x^3 – 31x^2 + 4x + 5) \div (x-5)\)

Answer 16

5 | 6 -31 4 5
|
----------------------
6

Answer 17

see if u can complete the rest :)

Answer 18

5| 6 -31 4 5
| 30 -5 -5
--------------------
6 -1 -1 0

Answer 19

Excellent !
so the depressed quadratic is ? (look at bottom row)

Answer 20

6 - 1 - 1?

Answer 21

yes, depressed quadratic wud be :-
\(6x^2 - 1x - 1\)

Answer 22

which u can solve using any of ur favorite method.

Answer 23

factor it may be...

Answer 24

factoring it...

-1x(5x - 1)?

Answer 25

Still there?

Answer 26

\(6x^2 - 1x - 1\)

\(6x^2 - 3x + 2x - 1\)

\(3x(2x - 1) + 1(2x - 1)\)

\((3x + 1)(2x - 1)\)

set it to 0

x = -1/3, x = 1/2

Answer 27

sorry my phone rang :|

Answer 28

you still here ha ?

Answer 29

yeah lol

Answer 30

so 5, -1/3, 1/2?

Answer 31

yes :)