Question

Solve the polynomial equation.
f(x) = x^3 - 4x^2 + x + 6
How do I do this? ..

Answer 1

solve means? do you need to find roots/zeros ?

Answer 2

I think so.

Answer 3

The instructions are as above..

Answer 4

A bit vague though.

Answer 5

which topic does this belong to ? can we use rational root theorem ? heard about it ?

Answer 6

Pre-cal, polynomial and rational functions. Yeah, the rational root theorem is allowed.

Answer 7

good, so you know about it ?
the roots can be factors of \(\pm\)6/1
start by test 1, then -1
is any of these a root of f(x) ?
do you know how to test ?

Answer 8

I have no idea what it is . Testing would just be substituting the value into the function right?

Answer 9

yes
test +1 first

Answer 10

if you get a 0, then it is a root

Answer 11

f(x) = x^3 - 4x^2 + x + 6
f(1) = 1 - 4(1) + 1 + 6
f(1) = 1-4 + 1 + 6
f(1) = -3 + 7
= 4
so it isn't?

Answer 12

correct
now test -1

Answer 13

Jackpot in 3.... 2... 1....

Answer 14

lol

Answer 15

lol

Answer 16

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

Answer 17

so that's one.

Answer 18

finding all of these must be a pain.

Answer 19

yes
now you can continue doing this for
+2,-2, +3,-3 ....
but if you know synthetic division, it'll simplify things

Answer 20

I was suppose to learn that.. never got around to it.
Care to enlighten? lol.

Answer 21

http://www.youtube.com/watch?v=bZoMz1Cy1T4

Answer 22

ranga the video helps, but what am I suppose to divide it by? it's not a rational function

Answer 23

its not ? how ?
since we have -1 as on root,
take -1 on left (similar to polynomial division by x+1)

Answer 24

so you would have to do that for each possible value tho..

Answer 25

seems a bit tedious lol.

Answer 26

each ?
do it for -1 and you directly get a simple quadratic equation

Answer 27

and you can solve quadratic equations right ?

Answer 28

yes lol.

Answer 29

problem solved :P

Answer 30

so i tried synthetic division