4

Is it possible to express $x^4-x^3+3x^2-4x+6$ as a product of polynomials of smaller degree with integer coefficients?

My attempt: By equating the polynomial to $0$, one obtains $x=1\pm i, \pm\frac{1}{2}i(\sqrt{11}\mp i)$. From this, one can write the polynomial given in the question as the

$$(x^2-2x+2)(x^2+x+3)$$ by using Viete formula.

However, I'm not sure whether I can solve the polynomial before hand and use the answers to express it as product of polynomial of smaller degree.

Idonknow
  • 15,643
  • 1
    You have solved the question already, because the quadratic polynomials here are irreducible, i.e., have no integer roots. So the answer is: yes, it is possible. The complete factorization over $\mathbb{Z}$ is given by $x^4-x^3+3x^2-4x+6=(x^2+x+3)(x^2-2x+2)$. – Dietrich Burde May 21 '15 at 11:06
  • Yes, it is possible because given polynomial is not prime. – Sufyan Naeem May 21 '15 at 11:14
  • Thats right ,Post your own answer @idonknow #udoknow – Ilaya Raja S May 21 '15 at 11:29

1 Answers1

2

Your polynomial has no integer root since such a root must be a divisor of $6$. More over it can have no negative root since, if $x>0$, $p(x)> 0$. Thus one has to test the values $\,1,2 ,3,6$ and none is a root.

Thus a factorisation must have the form: $$x^4-x^3+3x^2-4x+6=(x^2+ax+b)(x^2+a'x+b')$$ which leads to the system: $$\begin{cases} a+a'=-1\\aa'+b+b'=3\\ab'+ba'=-4\\bb'=6 \end{cases}$$ Furthermore, $b$ and $b'$ have to be positive: if they were negative, the factors, hence $p(x)$, would have a negative root.

Theere remains two possibilities: $b=2,\enspace b'=3\,$ or $\, b=1,\enspace b'=6$. From the second equation, we deduce that $aa'=-2$ or $aa'=-4$.

Now we have to solve for the classical problem of finding two numbers, knowing their sum, $-1$ and their product, $-2$ or $-4$. They're roots of one of these equations: $$t^2+t-2=0,\quad t^2+t-4=0.$$ It happens that the first one has integer roots, $1$ and $-2$, while the second has not.

Thsi third equation will let us know which root is $a$ and which is $a'$. It can be written as $\,3a+2a'=-4$, which enforces $a=-2,\ a'=1$. Finally we obtain the factorisation: $$x^4-x^3+3x^2-4x+6=(x^2-2x+2)(x^2+ x+3). $$

Bernard
  • 175,478