Question

solve x^4-12x-5=0 given that -1+2i is a root!

Answer 1

-1 - 2i is also a root. Polynomial division now

Answer 2

if \(-1+2i\) is a root then one of the factors is \(x^2+2x+5\)

Answer 3

yeah! i know that -1-2i is also a root!

Answer 4

how did you get that?! :o

Answer 5

if that is not clear, we can work backwards
\[x=-1+2i\]
\[x+1=2i\]
\[(x+1)^2=-4\]
\[x^2+2x+5=0\]

Answer 6

\[(-1 - 2i)(-1+2i)\times p(x) = x^4 - 12x - 5\]Your job is to find \(p(x)\), which'd turn out to be what sat said.

Answer 7

But yeah, the guy is a genius. You should listen to him before doing polynomial division.

Answer 8

or if you don't want to work backwards you can memorize that if \(a+bi\) is a root your quaddratic is \[x^2-2ax+a^2+b^2\]

Answer 9

Omg, I messed up.

Answer 10

okay i understand how you got x2+2x+5=0!

Answer 11

That will not work @ParthKohli
\(p(x) = \cfrac{x^4 - 12x - 5 }{5}\)

Answer 12

or if you really want to be annoyed, multiply
\[(x-(-1+2i))(x-(-1-2i))\] which is not as hard as it seems, although it is easier just to memorize it

Answer 13

\[(x + 1 - 2i)(x + 1 + 2i)\]

Answer 14

Yep, its now right.

Answer 15

nah, leave in the parentheses

Answer 16

yeah my teacher taught me (x−(−1+2i))(x−(−1−2i))

Answer 17

yeah that works if you want to do it that way

Answer 18

so um is x2+2x+5=0 the answer? or are theyre more steps?

Answer 19

first outer inner last
first is \(x^2\) last is \(1^2+2^2=5\) and outer, inner add up to \(-2x\)

Answer 20

You should use that quadratic to get your solutions.

Answer 21

how do i use this quadratic to get my solutions?

Answer 22

Just get that quadratic's solutions!

Answer 23

oops i meant "inner" adds up to \(2\) then divide or just think

Answer 24

\((x-(-1 + 2i) )(x- (-1-2i) )\)
\((x+1 - 2i )(x+1 + 2i) \)
\(((x+1)-2i)((x+1) + 2i)\)

Use (a+b)(a-b) identity now.

Solution of quadratic equation :

\(ax^2 + bx + c = 0 \) --> general form

\(x = \cfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Answer 25

\[x^4-12x-5=(x^2+2x+5)(something)\]

Answer 26

pretty clear that the first term of the "something" is \(x^2\) and the last term is \(-1\) just need the middle term

Answer 27

sorry um, where did you get the x4−12x−5?

Answer 28

i looked at the original question

Answer 29

x2+2x+5=0 do i just plug this into the quadratic formual?

Answer 30

no no

Answer 31

Sorry, I messed up again :-P

Answer 32

You should get that something and its solutions

Answer 33

that is the one that has zeros of \(-1+2i\) and \(-1-2i\) we already know those zeros

Answer 34

right!
so i divide x4−12x−5
by (x2+2x+5)

Answer 35

you want the other two zeros, so you have to factor \[x^4-12x-5\]

Answer 36

you can do that if you want a pain

Answer 37

i dont want a pain :(

Answer 38

i would do what i said above, write
\[x^4-12x-5=(x^2+2x+5)(something)\] and think about what the something is

Answer 39

it is easier than dividing by long division
you know it has to look like \[x^4-12x-5=(x^2+2x+5)(x^2+bx-1)\] the only think you don't know is \(b\)

Answer 40

i hope that is clear
the constant is \(-5\) and you have a \(5\) so the constant must be \(-1\) for the second factor

Answer 41

okay in the text book it says that SOMETHING is x^2-2x-1

Answer 42

fine, we can find the \(-2\) now

Answer 43

the \(x^2\) part is obvious i hope, as is the \(-1\) part

Answer 44

yes i understand the -1 and the x!

Answer 45

for example, when you multiply out in what i wrote you will get \(-2x+5bx\) and you know the answer has to be \(-12x\) so if \(-2x+5bx=-12x\) then \(-2+5b=-12\) and so \(b=-2\) that is one way

Answer 46

or another way:
if you multiply you will get \(bx^2+2x^3\) but you know you have no \(x^3\) and so \(b=-2\)

Answer 47

typo there, i meant \(bx^3+2x^3=0\) and so \(b=-2\)

Answer 48

in any case you must have \(b=-2\) making it
\[x^4-12x-5=(x^2+2x+5)(x^2-2x-1)\]

Answer 49

or if thinking is too annoying, you can always use long division
which i think it totally annoying

Answer 50

i think i understand your way a bit, but just not very well. I probably will try your way and use long division to check

Answer 51

Thanks a lot Satellite!