I understand how to do factoring, but I do not understand what is the point of it - essentially what is the goal of factoring an expression? There are lots of ways to factor eg difference of 2 squares, difference of 2 cubes, sum of squares, sum of cubes etc..but what is the point in doing that? How do I know when to use each method? For example in the equation $3p^2-3p-36$, which I want to factor, I'm not sure were to start because I'm not sure what the question is even asking me to do.
Regards,
$\endgroup$ 32 Answers
$\begingroup$First off, factoring is useful in solving equations. For example, to solve $x^2-4x+3=0$, you would $x^2-4x+3$ as $(x-1)(x-3)$, so now your equation becomes $(x-1)(x-3)=0$. Now, if any one of those factors equals $0$, then the product equals $0$. That means that either $x-1$ or $x-3$ must be equal to $0$. From that you find out that the solutions are $x=1, \ 3$.
Second, completing the square uses factoring (who says that $a^2+2ab+b^2=(a+b)^2$ is not factoring). Completing the square really helps when graphing circles and hyperbolas.
Third, it is useful when graphing functions like this: $$f(x)=\dfrac{x^2-5x+6}{x+1}$$ Here you would factor $x^2-5x+6$ to give you $(x+1)(x-6)$. Now you can cancel out $x+1$ in the numerator and the denominator. This leaves you with $f(x)=x+6$, which is easy to graph. But just remember that there is a hollow point, or a hole, at $x=-1$. There are probably some other uses of factoring, but this is all I can think of right now.
Fourth, factoring helps to simplify fractions in general. That is done by finding factors in the numerator and denominator that cancel out. This really helps when evaluating limits like: $$\lim\limits_{x\to 1} \dfrac{x^2+12x-13}{x-1}$$ The numerator can be factored as $(x+13)(x-1)$. Cancelling the numerator and the denominator out, we are left with: $$\lim\limits_{x\to 1} x+13$$ This is easy to solve. Just plug the value $x=1$ into $x+13$. That gives you $\lim\limits_{x\to 1} x+13 = 14$. So the solution to the limit is: $$\displaystyle \boxed{\lim\limits_{x\to 1} \dfrac{x^2+12x-13}{x-1} = 14}$$
I hope this post was helpful
$\endgroup$ $\begingroup$The point of factoring polynomials is in some sense the same as factoring integers into primes. Some times it's more convenient to know what adds up to your function (in this case $3p^2$, $-3p$ and $-36$), and some times it's more convenient to know what multiplies to your function (in your case $3$, $(p + 3)$ and $(p - 4)$).
As for what rules to use... if you have a second-degree polynomial, the solution formula always gets you there, but there are some special cases you are expected to do quicker than that (like $x^2 + 6x + 9$). Other than that, it's like with primes: It doesn't matter what rules you use to get a factorization, or what order you do it in. As long as you do it correctly, you get the same answer anyways. Of course, some ways are easier than others, but until you get some experience it's more a matter of throwing every trick you know at it and hope something sticks.
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