Question

What are the zeros of f(x) = x^3 + 6x^2 + 3x – 10 ?

Answer 1

use rational roots theorem

Answer 2

When I do that I get \[\pm1.\pm2,\pm5, and \pm10\] But there are only 3 zeros. What am I doing wrong?

Answer 3

those are POSSIBLE roots...they're not automatically the zeros

Answer 4

so you have to try them one by one...three of them are the roots to the equation

Answer 5

do you want to know a shortcut?

Answer 6

I would love to know a shortcut!

Answer 7

pick a number from those numbers you listed

Answer 8

\[\pm2 \]

Answer 9

that means +2 and -2 <--those are two numbers...pick one of those

Answer 10

Okay, +2.

Answer 11

okay...now substitute +2 into EVERY x of x^3 + 6x^2 + 3x - 10

then tell me what the result is

Answer 12

I get to 28.

Answer 13

IF you get a result that is 0 then that is the ZERO..in this case you got 28 so it's not a zero

Answer 14

Other way is to jus factorize. A factor of x+a implies -a is a zero.

Answer 15

for example.. +1

(1)^3 + 6(1)^2 + 3(1) - 10
= 1 + 6(1) + 3 - 10
= 4 + 6 - 10
= 10 - 10
= 0

therefore x = 1 is a zero

i gave you a bonus already...can you solve the other two zeroes now? ;)

Answer 16

I have it from here, thank you. I've got 1, and -2 so far.

Answer 17

welcome ^_^ and nice going

Answer 18

actually...since it's a cubic function, once you get a zero you can just divide to get a quadratic equation then just factor