Question

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

(c)$x^5+5x^2+4x+7$

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 the definitions 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 a reducible polynomial.

Let $f\left(x\right)=x^5+5x^2+4x+7$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 $\pm 7$, 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:

$\begin{array}{rcl}f\left(x\right)& =& x^5+5x^2+4x+7=\left(x^2+ax+b\right)\left(x^3+c{x}^2+dx+e\right)\\ & \Rightarrow & x^5+0x^4+0x^3+5x^2+4x+7=x^5+c{x}^4+x^{3}d+e{x}^2+a{x}^4+ac{x}^3\\ & & +ad{x}^2+aex+b{x}^3+bc{x}^2+bdx+be\\ & \Rightarrow & x^5+0x^4+0x^3+5x^2+4x+7=x^5+\left(a+c\right)x^4+\left(d+ac+b\right)x^3+\left(ad+bc+e\right)x^2\\ & & +\left(ae+bd\right)x+be \cdots \cdots \left(A\right)\\ & & \end{array}$

To show fxis irreducible in ℚx

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

$$\begin{gathered} a+c=0 \cdots \cdots \left(1\right), \\ ac+b+d=0 \cdots \cdots \left(2\right), \\ ad+bc+e=5 \cdots \cdots \left(3\right), \\ ae+bd=4 \cdots \cdots \left(4\right) \\ be=7 \cdots \cdots \left(5\right) \end{gathered}$$

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

From (5) , either$b=1$and $e=7$or $b=7$and $e=1$(Without loss of generality we may assume that both are positive)

If b=1 and e= 7, then from (4), we get,

$7a+d=4\Rightarrow d=4-7a$

Now, from (2),

$\begin{array}{rcl}a\left(-a\right)+1+4-7a& =& 0\\ & \Rightarrow & -a^2-7a+5=0\\ & \Rightarrow & a^2+7a-5=0\\ & & \end{array}$

Which by quadratic equation has no rational root.

If b=7 and e=1 , then then from (4), we get ,

$$\begin{gathered} a+7d=4 \\ \Rightarrow 7d=4-a \\ \Rightarrow d=\frac{4-a}{7} \end{gathered}$$

Now, from (2),

$$\begin{gathered} a\left(-a\right)+7+\frac{4-a}{7}=0 \\ \Rightarrow -7a^2+49+4-a=0 \\ \Rightarrow -7a^2-a+53=0 \\ \Rightarrow 7a^2+a-53=0 \\ \Rightarrow a^2+\frac{a}{7}-\frac{53}{7}=0 \end{gathered}$$

Which by quadratic equation has no rational root.

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