# Why do we factor polynomials the way we do?

Today, I was working on some limit practice problems and came across two that I had to factor.

The first limit had this polynomial in the denominator: $$x^2+2x-15$$

which I factored down to: $$(x-3)(x+5)$$

The second limit had this polynomial in the numerator: $$2z^2-17z+8$$

which I factored down to: $$(2z-1)(z-8)$$

As I was looking over these problems, I realized I don't know why polynomials factor down like this. I was just taught what to do when I come across each type. When I factor them down the answers "make sense", but I just can't see a reason why they do as I look at it from my current perspective. Could someone shed some light on this? Are there proofs for things like this?

• Just use the quadratic equation to find the roots, $r_1$ and $r_2$, then the factorization is $(x- r_1)(x-r_2)$. – David G. Stork Sep 4 '18 at 2:14
• @DavidG.Stork What if the polynomial isn't monic? – John Coleman Sep 4 '18 at 12:05
• For the limit exercises in which the denominator causes a difficulty (by tending to zero), any polynomial doing so must have a root there (so you get at least one factor quite easily, but check for possible multiplicity of that root). – hardmath Sep 4 '18 at 12:34

Perhaps you're noticing the streetlight effect:

A policeman sees a drunk man searching for something under a streetlight and asks what the drunk has lost. He says he lost his keys and they both look under the streetlight together. After a few minutes the policeman asks if he is sure he lost them here, and the drunk replies, no, and that he lost them in the park. The policeman asks why he is searching here, and the drunk replies, "this is where the light is."

What's the point of factoring a polynomial? It's to undo polynomial multiplication. Our streetlight is our knowledge of polynomial multiplication. How do we get a quadratic that we can factor "nicely" (ie over the integers)? Like this: $$(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd$$ So if we have a quadratic that we can factor, we had better be able to find four numbers $a,b,c,d$ such that $ac$ is the coefficient of $x^2$, $bd$ is the constant term, and $ad + bc$ is the coefficient of $x$.

That's the ultimate root of the techniques you learned for factoring different types of quadratics over the integers. There are many quadratics that aren't under the streetlamp, so to speak, such as $x^2 - 2$ and $x^2 + 1$; in broadening the circle of light, we meet the real numbers, the quadratic formula, and the complex numbers.

• Wow, I really appreciate your perspective on the question. The fact that you posted this after @Siong Thye Goh, actually helped further explain his purely mathematical approach. I've never heard of the streetlight effect before, but I think it kind of explains the difficulties I've recently had writing proofs as well. – CaptainAmerica16 Sep 4 '18 at 2:42

Consider the polynomial $Ax^2+Bx+C$

$$(ax+b)(cx+d)=acx^2+(bc+ad)x+bd$$

Comparing the coefficients, $A=ac, C=bd$ and we want them to satisfies $bc+ad=B$.

Hence we factor $A$ and $C$ and find factors such that $bc+ad=B$.

• Thank you for this, but I got a little lost at the part where you compare the coefficients. Why exactly does $A=ac$ and $C=bd$? – CaptainAmerica16 Sep 4 '18 at 2:27
• the coefficient of $x^2$ for the first form is $A$ and the coefficient of $x^2$ for the second form is $ac$. Hence if the two polynomials are the same, they should be equal. – Siong Thye Goh Sep 4 '18 at 2:30
• Ah, ok. Thank you! This was very helpful! – CaptainAmerica16 Sep 4 '18 at 2:35

See fundamental theorem of algebra https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

Note over 2 variables or if you wish to factor over integers only this no longer holds