Question

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

(c)$3x^5+2x^4-7x^3+2x^2$

Short answer

The factorization is$3x^5+2x^4-7x^3+2x^2=x^2\left(x-1\right)\left(x+2\right)\left(3x-1\right)$

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$}\right.$.

Use the rational root test to write each polynomial as the product of irreducible polynomials inℚx

Here, $x^2$will be a factor, and we get the following:

If is a root of , then , and . Therefore, the only possible rational roots are , and . The fact that 1 is a root can be easily verified; therefore, and we get:

$\frac{3x^5+2x^4-7x^3+2x^2}{x^2}=3x^3+2x^2-7x+2$

If $\frac{r}{s}$is a root of $3x^5+2x^4-7x^3+2x^2$, then , $r|2$and $s|3$. Therefore, the only possible rational roots are $\pm 1,\pm 2,\pm \frac{1}{3},\pm \frac{2}{3}$. The fact that 1 is a root can be easily verified; therefore, role="math" localid="1649897694618" $x-1|3x^3+2x^2-7x+2$and we get:

$\frac{3x^3+2x^2-7x+2}{x-1}=3x^2+5x-2$

Moreover, -2 will be a root. As a result, $x+2|3x^2+5x-2$.

$\frac{3x^2+5x-2}{x+2}=3x-1$

Thus, the factorization is$3x^5+2x^4-7x^3+2x^2=x^2\left(x-1\right)\left(x+2\right)\left(3x-1\right)$