# Factorizing a polynomial using complex numbers

1. Nov 23, 2008

### alpha01

x3+4x2+9x+10

finding 1 root and using synthetic division we can factorize to:

(x+2)(x2+2x+5)

using complex numbers to factorize (x2+2x+5) we have (x+1)(x-2i)(x+2i), and so our final solution is:

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

is this correct?

Last edited: Nov 23, 2008
2. Nov 23, 2008

### Staff: Mentor

So far, so good.
It should be obvious that x^2 + 2x + 5 doesn't factor to (x + 1)(x - 2i)(x + 2i). In any case, this is not a calculus question.

3. Nov 23, 2008

### alpha01

i jus started in complex numbers today and thats what i came up with.. if someone could please move the thread to the correct place that would be appreciated..

Mark44, Can you please advise how i should go about finding the solution to that part of the question?

4. Nov 23, 2008

### rock.freak667

For x2+2x+5, how would you find the roots? If they aren't rational, is there a formula you can use?

5. Nov 23, 2008

### alpha01

there is an example in my notes where the quadratic equation has been used for polynomials with no real solutions. Using the same principal here we have:

x = (-2 +- sqrt(-16)) / 2

now, the square root of 16 is 4 so we have

x = (-2 +- 4i)/2
= -1 +- 2i

and thats how i got my roots.

6. Nov 23, 2008

### alpha01

is this not correct?

7. Nov 23, 2008

### LogicalTime

(post corrected)
all the roots you got are correct but the way you reconstructed the polynomial is not:
it should be

f(x)= (x+2)( x-(-1+2i) )( x-(-1-2i) ) (post was initially incorrect, but is now correct)

notice that x only appears 3 times, which is the same as the degree of the polynomial, this will always be true

lemme know if you have any other questions

Last edited: Nov 23, 2008
8. Nov 23, 2008

### alpha01

thank you logicalTime. It makes perfect sense now.

9. Nov 23, 2008

### alpha01

actually.. i just noticed something when i actually went back and wrote it... shouldnt it really be (x+(1-2i))(x+(1+2i))? because x = - 1 is x + 1 = 0 for both solutions so the only thing that should change in the outcomes is the sign in front of the 2i

why does one of the factors in your solution have x - 1?

10. Nov 23, 2008

### LogicalTime

(post corrected)
whoops, I messed up since the roots are -2, -1-2i, -1+2i then reconstructing the polynomial is:

f(x)= (x+2)(x-(-1-2i))(x-(-1+2i))

so when you plug in a root, one of the terms becomes zero making the whole thing equal to zero

should be correct now, sry bout that

Last edited: Nov 23, 2008
11. Nov 23, 2008

### alpha01

actually i have x=-1 so should be (x+(1-2i))(x+(1+2i))... not (x-(1-2i))(x-(1+2i))

Last edited: Nov 23, 2008
12. Nov 23, 2008

### LogicalTime

agh, it's late and I'm being sloppy, time for bed
I corrected my previous mistakes they are a little different format from yours, just because I like to make it obvious that plugging in the root leads to zero.