Question

Which of these polynomials are irreducible in$\mathrm{\mathbb{Q}}\left[x\right]$ :

(c)${\mathit{x}}^{\mathbf{5}}\mathbf{+}\mathbf{4}{\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{2}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathit{x}\mathbf{+}\mathbf{5}$

Short answer

The given polynomial is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$.

Step-by-step solution

Step 1:To factorization of the polynomial as a product of quadratic and cubic polynomial

Let us see thedefinitions of Reducible polynomial and Irreducible polynomial,

A nonconstant polynomial $f\left(x\right)\in F\left[x\right]$, where F is a field, is said to be irreducible if its only divisors are its associates and the nonzero constant polynomials(units).

A nonconstant polynomial that is not irreducible is called reducible polynomial.

Let $f\left(x\right)=x^5+4x^4+2x^3+3x^2-x+5$be the given polynomial.

We claim that $f\left(x\right)$is irreducible in .$\mathrm{\mathbb{Q}}\left[x\right]$

By contradiction, let if possible,$f\left(x\right)$ is reducible, then it can be factored as the product of two nonconstant polynomials in $\mathrm{\mathbb{Q}}\left[x\right]$.

If either of these factors have degree 1, then polynomial$f\left(x\right)$ has a root in $\mathrm{\mathbb{Q}}\left[x\right]$.

But by the Rational Root Test, $f\left(x\right)$has no roots in$\mathrm{\mathbb{Q}}\left[x\right]$ .

Then rational roots are $\pm 1$or $\pm 5$, and neither is a root.

Thus, by Theorem 4.2, if $f\left(x\right)$ is reducible, then it can be factored as the product of quadratic polynomial and cubic polynomial.

Therefore, the given polynomial can be factored as follows:

$\begin{array}{rcl}f\left(x\right)& =& x^5+4x^4+2x^3+3x^2-x+5=\left(x^2+ax+b\right)\left(x^3+c{x}^2+dx+e\right)\\ & \Rightarrow & x^5+4x^4+2x^3+3x^2-x+5=x^5+c{x}^4+x^{3}d+e{x}^2+a{x}^4+ac{x}^3\\ & & +ad{x}^2+aex+b{x}^3+bc{x}^2+bdx+be\\ & \Rightarrow & x^5+4x^4+2x^3+3x^2-x+5=x^5+\left(a+c\right)x^4+\left(d+ac+b\right)x^3+\left(ad+bc+e\right)x^2\\ & & +\left(ae+bd\right)x+be \cdots \cdots \left(A\right)\\ & & \end{array}$

To show fx is irreducible in ℚx

From $\left(A\right)$in step 1, equal polynomials have equal coefficients, therefore, on comparing coefficients, we get,

$$\begin{gathered} a+c=4 \cdots \cdots \left(1\right), \\ ac+b+d=2 \cdots \cdots \left(2\right), \\ ad+bc+e=3 \cdots \cdots \left(3\right), \\ ae+bd=-1 \cdots \cdots \left(4\right) \\ be=5 \cdots \cdots \left(5\right) \end{gathered}$$

Then from (1) , $c=4-a$.

From $\left(5\right)$, either $b=1$and $e=5$or $b=5$ and $e=1$(Without loss of generality we may assume that both are positive)

If $b=1$and $e=5$, then from (4), we get ,

$5a+d=-1\Rightarrow d=-1-5a$

Now, from (2),

$\begin{array}{rcl}a\left(4-a\right)+1-1-5a& =& 2\\ & \Rightarrow & 4a-a^2-5a=2\\ & \Rightarrow & a^2+a+2=0\\ & & \end{array}$

Which by quadratic equation has no rational root.

If $b=5$and $e=1$, then then from (4), we get ,

$$\begin{gathered} a+5d=-1 \\ \Rightarrow 5d=-1-a \\ \Rightarrow d=\frac{-1-a}{5} \end{gathered}$$

Now, from (2),

$\begin{array}{rcl}a\left(4-a\right)+5+\frac{-1-a}{5}& =& 2\\ & \Rightarrow & 4a-a^2+5-\frac{1+a}{5}=2\\ & \Rightarrow & 20a-5a^2+25-\left(1+a\right)=10\\ & \Rightarrow & -5a^2+19a+15=0\\ & \Rightarrow & 5a^2-19a-14=0\\ & \Rightarrow & a^2-\frac{19}{5}a-\frac{14}{4}=0\\ & & \end{array}$

Which by quadratic equation has no rational root.

Hence, $f\left(x\right)$is irreducible in .$\mathrm{\mathbb{Q}}\left[x\right]$