Math Is Fun Forum

freddogtgj

Member

Registered: 2006-12-02

Posts: 54

Factorising Polynomials

Hey could someone help me in this question?

Factorise the polynomial x^4 + 4 into two quadratics with real coefficients.

I have the answer from a Maths software package but i'm not too sure how the answer has come about.

(x^2-2*x+2)*(x^2+2*x+2)

Last edited by freddogtgj (2007-01-20 05:33:09)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Factorising Polynomials

If you asked me to just to factor this, I would look at you and laugh.  The first thing to do is to look for the real roots... there are none.  Then the next thing to do is to group like terms... there are none.

However, the way the question is posed, you aren't asked just to factor it.  You're given that:

a. It does factor, and
b. It factors into two quadratics.

So it must be that:

So we know that:

ad = 1
ae + bd = 0
af + cd + be = 0
bf + ce = 0
cf = 4

Now unfortunately, we're told that the coefficients must be real.  But lets, for the time being, assume that they are integers.

ad = 1, so it must be that a = 1 and d = 1.  Thus:

e + b = 0 or rather e = -b.

Now here is where you have to guess.  cf = 4.  So either:

c = 4 and f = 1
c = 1 and f = 4
c = 2 and f = 2

And you continue through each of the equations till you reach a solution.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Factorising Polynomials

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

you can use 0=-x+x ,this kind of idea to do it

Last edited by Stanley_Marsh (2007-01-20 10:49:21)

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Numbers are the essence of the Universe

freddogtgj

Member

Registered: 2006-12-02

Posts: 54

Re: Factorising Polynomials

i assumed you would use complex numbers...am i right?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Factorising Polynomials

Duh!  Why didn't I think of that?

The complex roots are:

1 + i
1 - i
-1 + i
-1 - i

So:

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

Multiplying these out will give two real quadratics.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

freddogtgj

Member

Registered: 2006-12-02

Posts: 54

Re: Factorising Polynomials

Ricky could you explain to me how I get those complex roots please?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Factorising Polynomials

Simply take the square root of both sides:

And then let x = a+bi, so:

So:

And solve those equations.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

freddogtgj

Member

Registered: 2006-12-02

Posts: 54

Re: Factorising Polynomials

Thanks appreciate it!