I recently, while problem -solving came accross various circumstances where I must factorise expressions and proceed. These things are generally taught to high school/middle school students.
However, it was up until recently, I found huge expressions say,
- $$p^4-(x+2y+1)p^3+(x+2y+2xy)p^2-2xyp=0$$
Now, this really looks simple and it is! It is nothing but a polynomial in $p.$
Things turn out more simple, if we observe $1$ and $0$ as roots of the polynomial. Then, dividing the given expression by $p(p-1)$ we get, $(p-x)(p-2y),$ and we are done. So, $$p^4-(x+2y+1)p^3+(x+2y+2xy)p^2-2xyp=p(p-1)(p-x)(p-2y).$$
This was easy.
Now, I encountered another expression,
- $$xyp^2+(x^2+xy+y^2)p+x^2+2y=0$$
This is where things start to get troublesome, if not complicated.
To be honest, I had a hard time dealing with this.
I broke up all the products and found a huge expression:$$xyp^2+x^2p+xyp+y^2p+x^2+2y=0$$
I tried all possible ways, of grouping the terms say, by $(xyp^2,x^2p)$, anticipating that we would get $(p+1)$ as a factor but then, it was not the case. This turned out to be misadventure. Next, I tried grouping $(xyp^2,xy)$ and this time, I hoped $(p^2+1)$ comes up in each grouping, but again, things went down the hill. Next, I tried grouping $(xyp^2,xyp,xy)$. My intuition for this, was simply that $xy$ was common in these $3$ terms and I felt it might be a good idea to start by grouping these terms. But it was not fruitful. The thing was indeed deceptive.
Suddenly, $15$ minutes later, it struck me that $yp+x$ is a factor of this "thing." Then things turned simpler and I was able to factorise the expression as $$xyp^2+(x^2+xy+y^2)p+x^2+2y=(xp+x+y)(yp+x).$$
Another expression was
- $$(x^2+x)p^2+(x^2+x-2xy-y)p+y^2-xy=0$$
This was probably the hardest. I was completely tired of grouping stuffs and factoring out things. I gave up, and found the answer to be, $[(x+1)p-y][xp+x-y]=0,$ using a calculator.
This was least expected by me!
Now, of course the question ,"Factorise this expressions…" were not given to me. They are pretty much intended towards middle/high-school students. But this terms came out while solving a problem, or better say, while solving a problem, I found that factorising some expressions would be really helpful and would simplify the approach.
However, the thing is, I want to know, if there are any standard "tricks" to factorise expressions. Since this was a part of a numerical calculation, I shouldn't ideally get bogged down with these stages of it. It would be really a waste of time. Any standard/commonly known strategies to factorise these sort of expressions(which I mentioned above) if shared will be much helpful in my case.
Best Answer
There is no guarantee that an expression of the form $$Ax^2+Bx+C$$ can be factorized.
In high school we usually do it by trial and error. Most problems in high school are so set that they can be done by inspection.
Thus to factorize $6x^2+19x+10$,
we note that $6=2 \times 3$ and $10 = 2 \times 5$. By inspection, we find that $3\times 5 + 2 \times 2 = 19$. We come up with the factorization: $$6x^2+19x+10=(2x+5)(3x+2).$$
In your problem, to factorize $$(x^2+x)p^2+(x^2+x-2xy-y)p+y^2-xy,$$ we cannot guarantee that factorization is possible.
We can only hope that it can be done.
Taking the expression as a quadratic expression of $p$ and applying the same strategy as above, we can work as follows:
$(1)$ $x^2+x=x(x+1),$
$(2)$ $y^2-xy=y(y-x)$.
$(3)$ We try to manipulate the factors $x, x+1$ and $y, y-x$ and see if we can come out with the coefficient of $p$, namely $x^2+x-2xy-y.$
$(4)$ By inspection, it turns out that $$x \times (-y) + (x+1) \times [-(y-x)]=x^2+x-2xy-y$$
$(5)$ Thus the final answer is:
$$(x^2+x)p^2+(x^2+x-2xy-y)x+y^2-xy=\left[ xp-(y-x)\right]\left[(x+1)p-y \right]$$