Question

x^3-2x^2+5x-4=0
solve for x

Answer 1

+- 1,2,4

x=2?

2 | 1 -2 5 -4
0 2 0 10
------------
1 0 5 6 nope, but since its all+ it aint bigger than 2

Answer 2

wait how'd you get 2 and 4?

Answer 3

x=1?

1 | 1 -2 5 -4
0 1 -1 4
-----------
1 -1 4 0 goody

(x+1)(x^2 -x +4)

Answer 4

the rule is:

factors of last#
-------------
factors of first#

Answer 5

typo

(x-1)(x^2 -x +4)

Answer 6

1,2,4
----- leads to our options to test
1

Answer 7

i dont understand that

Answer 8

4 factors to what?

1.4 ; 2.2

all the factors of 4 are: 1,2,4

Answer 9

1 factors to 1 ....

Answer 10

?????

Answer 11

what is our last number in the equation?

Answer 12

the last number is always a product of 2 other numbers ....

Answer 13

1(4) = 4
2(2) = 4

thats all we get right?

Answer 14

sorry i dont understand that..... im only in algebra 2..... all i know is you graph it and you get x=1, then i don't know what to do from there

Answer 15

well, if x=1; then that means that (x-1) is a factor and you divide it out of the original equation to get whats left

Answer 16

yea....

Answer 17

but i get a remainder and i dont know what to do with it

Answer 18

you dont get a remainder ....

Answer 19

i did

Answer 20

when i synthetically divide i got -2 as a remainder

Answer 21

ooooo. i see i forgot the negative sign okay so i got a remainder of 0

Answer 22

then i get (x-1)(x^2-x+4)
now what

Answer 23

x^2 -x +4
------------------
x-1 | x^3 -2x^2 +5x -4
+x^2
-------------
-x^2 +5x
- x
------
4x -4
+4
-----
0

Answer 24

then do i use the quadratic formula?

Answer 25

you can try; but your gonna end up with a sqrt(-) whih means it wont factor over the reals

Answer 26

i dont know what that is..... but do i get an imaganary number?

Answer 27

you get imaginary numbers yes

Answer 28

okay... can you help me with this?
((2)/(x+2))-((x)/(2-x))=((x^2+4)/(x^2-4))

Answer 29

2 x x^2+4
---- - ---- = ------
x+2 2-x x^2-4
^^^
-(x-2)


-2 x x^2+4
---- - ---- = ------
2-x 2-x x^2-4

-2(2-x) x(2-x) x^2+4
------ - ------- = ------
4 -x^2 x^2-4

Answer 30

yea and then i got 4/(x+2)
but i dont know how to solve for x from there

Answer 31

-4 +2x -2x +x^2 x^2+4
---------------- = ------
-(x^2-4) x^2-4


4-x^2 = x^2+4

-x^2 = x^2 when x=0

Answer 32

that almost works :) but 1 doesnt= -1

Answer 33

hold on let me check my work

Answer 34

wait what would the answer be then?

Answer 35

let me do it on paper to keep better track of it

Answer 36

wait how'd you get -(x^2-4) in the denom. i only got (x^2-4)

Answer 37

when i simplify i get 4/(x+2) but i dont know what to do from there

Answer 38

2(2-x) -x(2+x) = x^2 +4

4 -2x -2x -x^2 = x^2+4

0 = x^2+x^2 +4x +4-4

0 = 2x^2 +4x

0 = 2x(x+2)

x = 0, -2

Answer 39

how did you get rid of the denom.?

Answer 40

i multiplied evething by x^2-4; but i got an error still

Answer 41

.............

Answer 42

ok well can you just tell me how to solve 4/(x+2)

Answer 43

there is no solution; you end up with 0=8

Answer 44

how?

Answer 45

\[\frac{2}{x+2}-\frac{x}{2-x}=\frac{x^2 +4}{(x-2)(x+2)}\]

Answer 46

ok

Answer 47

\[\frac{2(x+2)(x-2)}{x+2} + \frac{x(x+2)(x-2)}{x+2}=x^2+4\]

Answer 48

shouldnt the right hand side have a denom.?

Answer 49

no denom, it canceled:
\[frac{(x^2+4)(x-2)(x+2)}{(x-2)(x+2)}\]
cancels
\[2(x-2)+x(x-2)=x^2+4\]
\[2x-4+x^2-2x=x^2+4\]
\[x^2 -4 = x^2+4\]
\[x^2-x^2 = 4+4\]
\[0=8 \implies \emptyset \]

Answer 50

\[\frac{(x^2+4)(x−2)(x+2)}{(x−2)(x+2)}\]

Answer 51

ooooooo...... ok

Answer 52

but i dont see how it cancels on the left hand side

Answer 53

omygosh thank you soooo much :)