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 of irreducible 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$f\left(x\right)=x^7+2x^6-6x^4-15x^2-33x-9$ 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 9.

Clearly, by trial and error method, we have, 3 and -1 are the roots of polynomial.

To find product of polynomials using Synthetic Division

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

$x^7+2x^6-6x^4-15x^2-33x-9=\left(x-3\right)\left(x^6+x^5+3x^4+3x^3+9x^2+12x+3\right)$

Again, by Synthetic Division, we factorize,$\left(x^6+x^5+3x^4+3x^3+9x^2+12x+3\right)$ which gives,

$\left(x^6+x^5+3x^4+3x^3+9x^2+12x+3\right)=\left(x+1\right)\left(x^5+3x^3+9x+3\right)$

Therefore, we get,

$x^7+2x^6-6x^4-15x^2-33x-9=\left(x-3\right)$$\left(x+1\right)\left(x^5+3x^3+9x+3\right)$

The remaining polynomial is irreducible by Eisenstein’s Criterion for .

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