<s>[INST] Factorizing $(x+a)(x+b)(x+c)$ I was solving questions related to polynomial factorization. I have learnt the remainder and factor theorems, and some basic identities.
There was a question like this one:
$$p(x)=x^3+8x^2+19x+12$$
The method we have been taught is to first find the factors of the last number, here it is $12$, so the factors are $\pm1$, $\pm2$, $\pm3$, $\pm4$, $\pm6$, $\pm12$. Then we have to substitute each of these values for $x$ till $p(x)=0$. Then, using factor theorem, we get a factor. For example, if $p(2)=0$, $x-2$ is a factor. Then, we use long division to get the product of the other two factors. Finally, we split the middle term and factor that expression.
I noticed that all of the answers were of the form $(x+a)(x+b)(x+c)$, which is equal to $x^3+x^2(a+b+c)+x(ab+bc+ca)+abc$. So, for the given question, if you find three integers such that $a+b+c=8$ (coefficient of $x^2$) and $abc=12$, can the factorization be simply written as $(x+a)(x+b)(x+c)$? In this case, it is $1+3+4=8$ and $1\times3\times4=12$, so factorization is $(x+1)(x+3)(x+4)$
I tried explaining this to my teacher, but she said that we must use the prescribed method of the NCERT book only.
So, is my method right? (I think so, but I'm not 100% sure) Also, is it necessary to confirm that the coefficient of $x$ is $ab+bc+ca$, or will that always hold true?
 [/INST] Since $$P(-1)=(-1)^3+8(-1)^2+19(-1)+12=0$$
Therefore $x+1$ is one of the factors. After dividing $p(x)$ by $x+1$ we have
$$x^3+8x^2+19x+12=(x+1)(x^2+7x+12)$$
Since $$x^2+7x+12=(x+3)(x+4)$$ We are going to have
$$p(x)=x^3+8x^2+19x+12=(x+1)(x+3)(x+4)$$
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