Factorizing a polynomial using complex numbers

[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?

 

[Mark44]

x³+4x²+9x+10

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

(x+2)(x²+2x+5)

So far, so good.

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?

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.

 

[alpha01]

i jus started in complex numbers today and that's 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?

 

[rock.freak667]

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

 

[alpha01]

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

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 that's how i got my roots.

 

[alpha01]

is this not correct?

 

[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

 

[alpha01]

thank you logicalTime. It makes perfect sense now.

 

[alpha01]

actually.. i just noticed something when i actually went back and wrote it... shouldn't 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?

 

[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

 

[alpha01]

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

 

[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.