Question

What are the solutions to this?
3x⁴ + 4x³ - 36x² + 64 = 0

Answer 1

when you're not able to factor by grouping then you should use p/q method
\[\rm \frac{ p }{ q }=\frac{ \pm~all~factors ~of~constant~term }{ \pm~all~factors~of~leading ~coefficient }\]

Answer 2

^to find all possible zeros
and then we have to test each possible zero by either using synthetic division or
replacing x with each number if you get 0 as final answer then the number u used would be the real zero

Answer 3

and yes that's correct

Answer 4

That seems like a tedious task...

Answer 5

why did you delete ur comment?
and the arrow poimting out my first comment

Answer 6

n*

Answer 7

because i thought it was wrong sorry

Answer 8

so do I have to plug in all the numbers till I get 0?

Answer 9

yes correct.
but first write all factors of 64 and 3

Answer 10

ohh noooo is there a easy way?

Answer 11

you can do synthetic division.

Answer 12

and the other easy way is graphing....

Answer 13

oh okay, I know how to do synthetic division, but which value do I divide by?

Answer 14

that would be the 2nd step
first divide factors of constant by factors of leading coefficient

Answer 15

hat are the factors of 64 ?

Answer 16

what**

Answer 17

64 32 16 8 4 2 1

Answer 18

correct and the factors of 3 are 1 and 3 that's it
\[\frac{ \pm 1 , \pm 2, \pm 4 ,\pm 8 ,\pm 16,\pm 18 , \pm 32 , \pm 64}{ \pm 1 , \pm 3 }\] we don't know if its negative or positive so we should keep both sign
and now divide all numbers at the numerator by each number in the denoinator

Answer 19

denominator **
and cross out duplicates numbers u get after division :)

Answer 20

What do I do once I find all the numbers?

Answer 21

then you have to test each number
which number would give you the value of 0 as a final answer ?

Answer 22

there is a typo 18 which isn't a factor
it should be
\[\frac{ \pm 1 , \pm 2, \pm 4 ,\pm 8 ,\pm 16,\pm 32 , \pm 64}{ \pm 1 , \pm 3 }\] divide
all the top numbers by p/m 1 and then by 3
let m know what you get ?
what would be the possible zeros ?

Answer 23

1, 2, 4, 8, 16, 32, 64, 1/3, 2/3, 4/3, 8/3, 16/3, 32/3, 64/3

Answer 24

What do i do from here, please help!

Answer 25

test each number
remember all of them are plus/minus
so start with p/m 1
\[\rm f(\color{ReD}{-1})=3x^2+4x^3-36x^2+64\]\[\rm f(\color{ReD}{1})=3x^2+4x^3-36x^2+64\]\[\rm f(\color{ReD}{-2})=3x^2+4x^3-36x^2+64\]\[\rm f(\color{ReD}{2})=3x^2+4x^3-36x^2+64\]

or do synthetic division

Answer 26

NEGATIVES TOO? That is going to take forever :( I think i am going to give up on this problem.

Answer 27

i don't which one is easy for you

if you get 0 remainder then that would be the real root