$x^3-3x^2+4x-2$ cannot be factored over $\mathbb R$ I'm new to the site, and I need a bit of help from you.
How can I prove that the polynomial:  $f(x)=x^3-3x^2+4x-2$
cannot be factored as a product of polynomials of degree 1 with real coefficients?
Thanks.
 A: If $f$ had $3$ real roots, then by Rolle's theorem its derivative would have $2$ real roots, but $f'(x) = 3x^2 - 6x + 4$ has no real roots because its discriminant is negative. 
A: By the rational root theorem, the possible rational roots are $\pm1$, and $\pm2$. Since $f(1)=0$, $(x-1)$ must be a factor of $f$. After dividing we have that
$$f(x)=x^3−3x^2+4x−2=(x-1)(x^2-2x+2).$$ 
Now we can take the discriminant of the quadratic factor $(x^2-2x+2)$, that being 
$$b^2-4ac=(-2)^2-4\cdot 1 \cdot 2 = -4,$$ 
to note that $f$ has one real root and two complex roots. Hence $f$ can only be factored as the product of one linear factor and the two complex conjugate linear factors, that being
$$f(x)=(x-1)(x-(1+i))(x-(1-i)).$$
A: First note that $(x-1)$ is a factor of $f(x)$ since $f(1)=1-3+4-2=0$. Therefore, we have
$f(x)=(x-1)(Ax^2+Bx+C)$. Compare the coefficients, we can find that 
$$f(x)=(x-1)(x^2-2x+2).$$
However, the factor $x^2-2x+2$ cannot be factored as a product of polynomials of degree 1 with real coefficients, because it has discriminant of $(-2)^2-4(2)=-4<0$.
A: Observe that$$\begin{align*}
f(x)=x^3-3x^2+4x-2&=(x^3-3x^2+3x-1)+(x-1)\\\\
&=(x-1)^3+(x-1)\\\\
&=(x-1)((x-1)^2+1)\\\\
&=(x-1)(x^2-2x+2)
\end{align*}$$
By the quadratic formula, the roots of $x^2-2x+2$ are complex, so $f$ cannot be factored any further into polynomials with real coefficients.
A: The derivative $3x^2-6x+4$ has no real roots, so $f$ can't have two distinct real roots nor a multiple real root.
