Question

Use the Rational Root Test to write each polynomial as a product of irreducible polynomials in :$\mathbf{\mathbb{Q}}\left(x\right)$

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

Short answer

$\mathbf{-}{\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{x}}^{\mathbf{3}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{-}\left(x+1\right)\left(x-2\right)\left(x^2+1\right)$

Step-by-step solution

Rational root test

Consider $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+.......+a_{1}x+a_0$is a polynomial with integer coefficients. If $r\ne 0$and if the rational number is $\frac{r}{s}$(in lowest terms), then it is a root of $f\left(x\right)$, also $\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$a_0$}\right.$and$\raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$a_0$}\right.$$\raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$a_0$}\right.$.

Use the rational root test to write each polynomial as the product of irreducible polynomials in ℚ(x)

a)

If $\frac{r}{s}$is a root for $\raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$and $\raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$-1$}\right.$, then the only possible rational roots are$\pm 1,\pm 2$ . The fact that $-1$is a root may be easily verified; therefore, $x+1|-x^4+x^3+x^2+x+2$.

$\frac{-x^4+x^3+x^2+x+2}{x+1}=-x^3+2x^2-x+2$

Again, it can be observed that 2 is a root of the quotient $-x^3+2x^2-x+2$, and as a result, it is .$x-2|-x^3+2x^2-x+2$

$\frac{-x^3+2x^2-x+2}{x-2}=-x^2-1$

Now, $x^2+1$does not have a rational root;the factorization is as follows:

$\mathbf{-}{\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{x}}^{\mathbf{3}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}$

Thus, the factorization is

$\mathbf{-}{\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{x}}^{\mathbf{3}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}$