When factoring any polynomial, it's generally agreed upon (at least in my math class) that all the factors you get are equal to $0$ (When you solve). I haven't really been able to myself build a serious argument against until recently, when we were introduced to synthetic division, and therefor, given factors.
Take for example, this problem which we did in class today.
Factor:
$f(x)=2x^3+3x^2-39x-20$ Given $f(4)=0$
Because $f(4)=0$, we know that one factor is $(x-4)$, and using synthetic division to solve the rest, we get factors of $(x-4)(2x^2+11x+5)$, and because we're solving, $(x-4)(2x^2+11x+5)=0$.
This is the point where I get confused. Because we know $x-4 = 0$, a multiplication by 0 always equals 0, how can we know that $2x^2+11x+5=0$? Couldn't $2x^2+11x+5 = \frac{35}{\pi(x-56)}$, because $\frac{35}{\pi(x-56)} \times 0 = 0$?
Why can we assume $(2x^2+11x+5)=0$, instead of anything else?
Best Answer
Making an answer out of Compulsive's comment...
Using the hint $f(4) = 0$, you found $$ f(x) = (x-4)(2x^2 + 11x +5). $$ Now, you are trying to find all numbers $x$ for which $f(x) = 0$. Assume for a second that you have found a number $x$ such that $f(x) = 0$. Because $f(x)$ is the product of $x-4$ and $2x^2 + 11x +5$, you have either $$x-4 = 0$$ or $$2x^2 + 11x +5 = 0$$ From there, considering the two possibilities, you can find what $x$ must be to indeed satisfy the equation $f(x) = 0$.