1

Factorization of :

$$x^4-x^3+3x^2+3x+5$$

$$x^4-x^3+3x^2+3x+5=(x^4-x^3+2)+3(x^2+x+1)$$

what do i do ? please help me

Community
  • 1
Fricul38
  • 711
  • 1
    It wont be easy, i think those roots are all complex numbers... – Atmos Dec 26 '17 at 22:33
  • @Atmos Why do you say that ? Complex roots of polynomials with real coefficients come always by conjugate pairs : thus the issue is by no means more complicated in this case. – Jean Marie Dec 26 '17 at 22:53
  • I meant that if one real root was clearly obvious, the factorisation would be easy. Here, i would be curious to see how to do the factorization. – Atmos Dec 26 '17 at 23:00
  • @Atmos If one of the roots were rational, it might be easy. In this case, all of the roots are complex, hence it is going to be a royal PITA. Perhaps we should ask Cardano? – Xander Henderson Dec 26 '17 at 23:02

2 Answers2

3

A possible error in the question, it might explain things. If we change 5 to 54, we get $$ x^4 - x^3 + 3x^2 + 3x + 54 = ( x^2 -4 x + 9) ( x^2 + 3 x + 6) $$

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

First you look for rational roots. There are none. Then we need only check, because of GAUSS,

The second result states that if a non-constant polynomial with integer coefficients is irreducible over the integers, then it is also irreducible if it is considered as a polynomial over the rationals.

$$ (x^2 + a x + b) ( x^2 + c x + 5 b), $$ with $$ b = \pm 1 $$ For both the case $b=1$ and $b=-1,$ we easily get correct coefficients for $x^3$ and $x,$ but then the $x^2$ term comes out to something incorrect. So the thing is irreducible. That is really all that is needed, try two cases.

$$ ( x^2 + x + 1) ( x^2 - 2 x + 5) = x^4 - x^3 + 4x^2 + 3x + 5 $$ Next comes out non-integer, and still wrong anyway. Gauss says we did not need to continue once $a,c$ came up not being integers. $$( x^2 - \frac{x}{2} - 1) (x^2 - \frac{x}{2} - 5)= x^4 - x^3 -\frac{23x^2}{4} + 3x + 5$$

Perhaps I should add that, although there are no real roots, there are real numbers $p,q,r,s$ such that $$ (x^2 + px+q)(x^2 + rx + s) = x^4 - x^3 + 3 x^2 + 3 x + 5 $$ where both quadratic factors are positive for any chosen $x$

Will Jagy
  • 139,541
1

This is not an efficient way to show irreducibility, but it's kind of fun:

Theorem: If $P(x)\in\mathbb{Z}[x]$ is a reducible polynomial of degree $n$, then $|P(m)|$ is prime (or $1$) for at most $2n$ different integers $m$.

Proof: If $P(x)=A_k(x)B_{n-k}(x)$ where $A_k$ is of degree $1\le k\lt n$ (and $B_{n-k}$ is of degree $1\le n-k\lt n$) and $|P(m)|=|A_k(m)||B_{n-k}(m)|$ is prime (or $1$), then one (or both) of the factors is $1$ or $-1$. But $A_k$ takes each value at most $k$ times, while $B_{n-k}$ takes value $n-k$ times. So $|A_k(m)|=1$ for at most $2k$ values of $m$ and $|B_{n-k}(m)|=1$ for at most $2(n-k)$ values of $m$, for a total of at most $2n$ possible values of $m$.

So to show that $P(x)=x^4-x^3+3x^2+3x+5$ is irreducible, it suffices to find $9$ integers $m$ such that $|P(m)|$ is prime (or $1$). A little experimentations does the rest. The following numbers are all prime:

$$\begin{align} P(0)&=5\\ P(1)&=11\\ P(-1)&=7\\ P(2)&=31\\ P(-3)&=131\\ P(4)&=257\\ P(-6)&=1607\\ P(11)&=13711\\ P(-13)&=31231 \end{align}$$

Barry Cipra
  • 79,832