Question

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

b)role="math" localid="1649891924413" $x^5+4x^4+x^3-x^2$

Short answer

The factorization is$x^5+4x^4+x^3-x^2=x^2\left(x^3+4x^2+x\right)$

Step-by-step solution

Rational root test

Consider$f\left(x\right)=a_0{x}^n+a_{n-1}{x}^{n-1}+.....a_{1}x+a_0$ 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$}\right.$.

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

Certainly, $x^2$canbe a factor, and we getthe following:

$\frac{x^5+4x^4+x^3-x^2}{x^2}$

If is $\frac{r}{s}$a root of $x^3+4x^2+x-1$, then $r|1$, and $s|1$. Therefore, the only possible rational roots are $\pm 1$. Here, neither 1 nor -1 roots cannot be easily verified; therefore, $x^3+4x^2+x-1$is irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$. Hence, the factorization is as follows:$x^5+4x^4+x^3-x^2=x^2\left(x^3+4x^2+x\right)$

Thus, the factorization is .$x^5+4x^4+x^3-x^2=x^2\left(x^3+4x^2+x\right)$