Question

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

(d)$\mathbf{2}{\mathit{x}}^{\mathbf{4}}\mathbf{-}\mathbf{5}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathit{x}\mathbf{-}\mathbf{6}$

Short answer

The factorization is$2x^4-5x^3+3x^2+4x-6=\left(x+1\right)\left(2x-3\right)\left(x^2-2x+2\right)$

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 in  ℚ(x)

If $\frac{r}{s}$is a root of $2x^4-5x^3+3x^2+4x-6$, then $r|-6$and $s|-2$. Therefore, the only possible rational roots are ,$\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2}$. The fact that -1 is a root can be easily verified. Therefore, $x+1|2x^4-5x^3+3x^2+4x-6$and we get the following:

$\frac{2x^4-5x^3+3x^2+4x-6}{x+1}=2x^3-7x^2+10x-6$

Neither of the candidates’roots $\frac{r}{s}$is the roots of the final quadratic polynomial. Therefore, itwillbe irreducible,and the factorizationisas follows:

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

Thus, the factorization is$2x^4-5x^3+3x^2+4x-6=\left(x+1\right)\left(2x-3\right)\left(x^2-2x+2\right)$