Question

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

e)$\mathbf{2}{\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{7}{\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{5}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathit{x}\mathbf{+}\mathbf{3}$

Short answer

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

Step-by-step solution

Rational root test

Consider $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+........+a_{1}x+a_o$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 $2x^4+7x^3+5x^2+7x+3$, then $r|3$and $s|2$. Therefore, the only possible rational roots are $\pm 1,\pm 3,\pm \frac{1}{2},\pm \frac{3}{2}$. The fact that -3 is a root can be easily verified. Therefore, $x+3|2x^4+7x^3+5x^2+7x+3$and we get the following:

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

Moreover, $-\frac{1}{2}$will be a root. Therefore, is as follows:

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

Neither of the candidates’ roots is the root of this final quadratic polynomial. Therefore, it will be irreducible, and the factorizationis as follows:

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

Thus, the factorization is .

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