Irreducible polynomial in Q[x]?

[pivoxa15]

Homework Statement

Determine whether y(x)=x^3-7x^2+14x-4 is irreducible in Q[x] with only pen and paper.

The Attempt at a Solution

The factor theorem would be good to use as if it isn't irreducible there must be a linear factor (x-a) with a in Q. I have worked out that at leat one zero does exist because for x<0, y(x)<0 & x>0, y(x)>0. But is this x value in Q or R?

 

[steven10137]

Homework Statement

Determine whether y(x)=x^3-7x^2+14x-4 is irreducible in Q[x] with only pen and paper.

The Attempt at a Solution

The factor theorem would be good to use as if it isn't irreducible there must be a linear factor (x-a) with a in Q. I have worked out that at leat one zero does exist because for x<0, y(x)<0 & x>0, y(x)>0. But is this x value in Q or R?

From what I can understand, there are no basic linear factors (x-a) where a is an integer. The only way I can possibly think you can solve -not even reduce it, for real values, is by finding the numerical factors i.e.
set it to zero;
$$x^3 - 7x^2 + 14x - 4 = 0$$
$$(x-0.341032918083) (x^2-6.65896708192 x + 11.7290730246) = 0$$
hence,
$$x-0.341032918083 = 0$$
$$x^2-6.65896708192 x + 11.7290730246 = 0$$
cannot have a negative discriminant for real values
therefore
$$x = 0.341032918083$$

In answer to your question, "it" is irreducible in Q[x] with only pen and paper.

Steven
- this is quite tricky, really got me thinking ...

 

[matt grime]

All those things you wrote down are rational, so you just factored it over the rationals.

There is an elementary result that states any rational solution to a monic integer polynomial is an integer.

 

[StatusX]

There's a simple test to see if a polynomial with integer coefficients has rational roots. Assume the polynomial

$$f(x) = a_n x^n + ... +a_1 x +a_0$$

has r=p/q as a root, where p/q is in lowest terms (ie, (p,q)=1). Then plugging in r and multiplying both sides by qⁿ, we have:

$$a_n p^n + a_{n-1} p^{n-1} q ... + a_1 p q^{n-1} + a_0 q^n = 0$$

Rearranging we get:

$$a_n p^n = -q(a_{n-1} p^{n-1} ... + a_1 p q^{n-2} + a_0 q^{n-1})$$

In other words, q divides a_n pⁿ. Since (p,q)=1, this is only possible if q divides a_n.
A similar argument shows that p divides a₀.

Thus the only possible rational roots of f(x) are of the form:

$$r = \pm \frac{p}{q}$$

where p is a positive divisor of a₀ and q is a positive divisor of a_n.

In your case, we see the only possible rational roots are $\pm 1, \pm 2, \pm 4$. So just plug these in and see if they're roots. If none of them are, then you know it has no rational roots.

 

[drpizza]

I agree wholeheartedly with StatusX's explanation of how to do this problem. But, I'll point out that synthetic division using the roots is usually quicker than plugging in those values. If your remainder is zero, then it's a root.