Question

Write each polynomial as a product of irreducible polynomials in $\mathrm{\mathbb{Q}}\left[x\right]$.

(a) ${\mathit{x}}^{\mathbf{5}}\mathbf{+}\mathbf{2}{\mathit{x}}^{\mathbf{4}}\mathbf{-}\mathbf{6}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{16}\mathit{x}\mathbf{-}\mathbf{8}$ (b)${\mathit{x}}^{\mathbf{7}}\mathbf{-}\mathbf{2}{\mathit{x}}^{\mathbf{6}}\mathbf{-}\mathbf{6}{\mathit{x}}^{\mathbf{4}}\mathbf{-}\mathbf{15}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{33}\mathit{x}\mathbf{-}\mathbf{9}$

Short answer

We proved that, given polynomial is written as a product ofirreducible polynomials in$\mathrm{\mathbb{Q}}\left[x\right]$ .

Step-by-step solution

To find roots by Trial and Error

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 reducible polynomial.

Let$f\left(x\right)=x^5+2x^4-6x^2-16x-8$ be the given polynomial.

We have to write each polynomial as a product of irreducible polynomials.

By Rational Root Test, the possible roots are the divisors of 8.

Clearly, by trial and error method, we have, 2 and -2 are the roots of polynomial.

To find product of polynomials using Synthetic Division

By Synthetic Division, we factorize the given polynomial as follows:

$x^5+2x^4-6x^2-16x-8=\left(x-2\right)\left(x^4-6x-4\right)$

Again, by Synthetic Division, we factorize $\left(x^4-6x-4\right)$, which gives,

$\left(x^4-6x-4\right)=\left(x+2\right)\left(x^3+2x^2+4x+2\right)$

Therefore, we get,

$x^5+2x^4-6x^2-16x-8=\left(x-2\right)$

The remaining polynomial is irreducible by Eisenstein’s Criterion for .$p=2$

Hence, given polynomial is written as a product of irreducible polynomials in $\mathrm{\mathbb{Q}}\left[x\right]$.