Question

Prove that $\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{x}\mathbf{+}\mathit{c}\mathbf{\in }\mathbf{\square }\left[x\right]$with${\mathit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathit{a}\mathit{c}\mathbf{<}\mathbf{0}$ is irreducible in$\mathbf{\square }\left[x\right]$ .[Hint see Exercise 6].

Short answer

The polynomial $a{x}^2+bx+c\in \square \left[x\right]$with $b^2-4ac<0$ is irreducible in $\square \left[x\right]$.

Step-by-step solution

Reducible Polynomial

A quadratic polynomial in $\square \left[x\right]$is reducible if and only if it has real roots.

fx is irreducible

Consider the function $f\left(x\right)=a{x}^2+bx+c\in \square \left[x\right]$with $b^2-4ac<0$.

If $b^2-4ac<0$ then the roots of a quadratic polynomial $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$are non-real complex numbers.

Here, the roots of $f\left(x\right)$are complex, which is not real. It implies that $f\left(x\right)$ is irreducible in $\square \left[x\right]$.

Therefore, $a{x}^2+bx+c\in \square \left[x\right]$with $b^2-4ac<0$is irreducible in $\square \left[x\right]$.