How would you factorise the following cubic, it only has one real root, but I can't find a factor, using the factor theorem:
$$ 3x^3-x^2+3x-1 $$
Could you please provide a step by step working out, or an explanation as how to tackle these sort of questions,
Thanks
On
$3x^3-x^2+3x-1=(3x-1)(x^2+1),$ by "grouping". If you're in complexes the factor $x^2+1=(x+i)(x-i).$
On
Try the https://en.wikipedia.org/wiki/Rational_root_theorem to find the root $\frac{1}{3}$. So you can write you cubic as $$3x^3 -x^2 +3x -1=(x-\frac{1}{3})q(x)= (3x-1)p(x)$$ then use polynomial division (or just guess) to get $p(x)=x^2+1$
You can do partial factorisations, with the hope of finding a common factor to these factorisations: \begin{align*}3x^3-x^2+3x-1&=x^2(3x-1)+3x-1&&\text{(factoring $x^2$ in the first two terms)}\\&=(3x-1)(x^2+1)&&\text{(irreducible factorisation)}.\end{align*}