I am struggling to factor a larger expression. I seem to have forgotten how to do it, hence I am coming here for help, so that I can refresh the concept.
The equation is $$x^3+2x+3=0$$
How do solve it by splitting the middle term? I tried the total and error method, and subsequent long division to obtain $$(x+1)(x^2-x+3)$$
So I have indeed found the factors, but how to do it without lengthy calculations? The middle term can be split, I have just forgotten how to.
Edit: I see that splitting the middle term is the wrong term. Forgive me.
What I meant was using formulas like $a^3+b^3=(a+b)(a^2-ab+b^2)$
I am fairly confident that’s how it’s done, because I have used this before.
It’s extremely embarrassing to ask this, but it had to be done.
When I was at university, we did this kind of factorisation thru the rational root theorem.
In your example $$x^3+2x+3$$ we must have $p|3$ and $q|1$. We only have to check for $\pm1$ and $\pm3$.
You asked if there was a way to split the middle terms (there are two middle terms). We could develop something similar to quadratic factorisation. Let's work backward. $$(x+1)(x^2-x+3)$$ We distribute the first factor thru the second $$x^2(x+1)-x(x+1)+3(x+1)\tag{1}$$ distribute again $$x^3+x^2-x^2-x+3x+3\tag{2}$$ If we want to split the middle terms, we need to find equation $(2)$.
Starting with your polynomial $$x^3+2x+3$$ We are looking for integers $A$, $B$, $C$ and $D$ to split the middle term like this $$x^3+Ax^2+Bx^2+Cx+Dx+3$$ with conditions $$A+B=0\tag{Cond1}$$ $$C+D=2\tag{Cond2}$$ Futhermore, we want to be able to factorise it to look like equation $(1)$. With the first two terms, there is a $x^2$ in common $$x^2(x+A)+Bx^2+Cx+Dx+3$$ We want to have the exact same parenthesis with the other four terms, so we need $$x^2(x+A)+Bx\left(x+\tfrac CB\right)+D\left(x+\tfrac3D\right)$$ For the parenthesis to be the same, we need $$A=\frac CB\tag{Cond3}$$ $$A=\frac3D\tag{Cond4}$$ From $(\text{Cond}4)$, $D=\pm1$ or $D=\pm3$. From $(\text{Cond}1)$, $A=-B$, combined with $(\text{Cond}3)$, $C=-A^2$. We now try a value of $D$, which force the other value, finally check if it fits in $(\text{Cond}2)$
The one that work is $D=3$, then $A=1$, $C=-1$ and $B=-1$. It is easy to see $C+D=2$. $$x^3+Ax^2+Bx^2+Cx+Dx+3$$ $$x^3+x^2-x^2-x+3x+3$$ $$x^2(x+1)-x(x+1)+3(x+1)$$ $$(x+1)(x^2-x+3)$$
It work realatively well since the polynomial is simple. It could be a lot more complicated if the leading coefficient was different of $1$.