Question

Find all the roots for:

Answer 1

\[4x^4-32x-x^3+8 =0 \]

Answer 2

You can rewrite this as:\[(4x^4 - x^2)-(32x-8)\]

Answer 3

\[4x(x^3-8)-(x^3-8)=0\]

Answer 4

or that too

Answer 5

same result

Answer 6

but, anyway, use the distributive property in reverse

Answer 7

\[(4x-1)(x^3-8)=0\]

Answer 8

Yes

Answer 9

\[(4x-1)(x-2)(x^2+2x+4)\]

Answer 10

seems about right

Answer 11

okay i dont know what to do from here on, i cant seem to factor the third bracket

Answer 12

oh

Answer 13

ok

Answer 14

well, you can't factor it into real numbers

Answer 15

would i use the quadratic formula ?

Answer 16

Yes

Answer 17

i get x1=\[-1+\sqrt{5}\]

Answer 18

x2=

Answer 19

\[-1-\sqrt{5}\]

Answer 20

no

Answer 21

sorry

Answer 22

you get:
\[-1 \pm i \sqrt{5}\]

Answer 23

but how would i write that in brackets ?

Answer 24

like if x=3 we would write x-3

Answer 25

\[(x+1-i \sqrt{5})(x+1+i \sqrt{5})\]

Answer 26

You essentially do the same thing, except the number you are subtracting is complex

Answer 27

ohhh thanks :D

Answer 28

You are welcome

Answer 29

But i got 5's not 3 :S

Answer 30

http://www.wolframalpha.com/input/?i=%28x%2B1-i3%5E1%2F2%29%28x%2B1%2Bi3%5E1%2F2

Answer 31

4-4(4) = -12