How do i know when the second factor when factoring a cubic expression is also factorizable?

Question

Let's say I have to factor $x^3 + 1$.

Now I know that the first factor will be $(x+1)$ because when you input -1 into the function you get $0$. And the rest is all about the factor-remainder theorem right.

Now i will obtain another factor when i divide $x^3 +1$ by $(x+1)$, which is $x^2 - x + 1$.

How do I know if the third factor is then factorizable? I'm asking because if I had to perform partial fraction decomposition, I would have to know.

I cannot test the factorizability of it by finding the discriminant, because the discriminant also tells me that the expression is factorizable when the solutions are not whole numbers.

And when spreading an expression into partial fractions , i believe that you are only asked to do that when the solutions are whole numbers.

So is there another way i can know?

Answer 1

If the discriminant is less than $0$, then the quadratic cannot be factorised. If it is greater than or equal to $0$, then the quadratic can be factorised, although this factorisation might be messy. The quadratic $ax^2+bx+c$ can be factorised over the integers if you can find two numbers $p$ and $q$ such that $p\cdot q = a \cdot c$ and $p+q=b$. Here is an example: \begin{align} 3x^2+10x+8 &= 3x^2+6x+4x+8 \\[5pt] &= 3x(x+2)+4(x+2) \\[5pt] &= (3x+4)(x+2) \, . \end{align} This is sometimes called the "AC method".

Answer 2

It fully depends upon the districiminant. If the discriminant $b^2-4ac>0$ then the polynomial can be factorised and it will give rational or irrational roots. However if the discriminant $< 0$ then the equation cannot be factorised.

Learn more about discriminant - https://en.wikipedia.org/wiki/Discriminant

Also if this might be of any help Vieta formula can also help. It states that if in polynomial $ax^2+bx+c=0$ and roots are $\alpha, \beta$ then $ \alpha+\beta=\frac{-b}{a} , \alpha\beta=\frac{c}{a}$