Factoring by Extraction

Today, I will discuss how to factor polynomials with large coefficients such as $3x^2+10x-100$ with ease. I know that this is a long note but I feel that it is worth reading everything including the generalized form at the bottom except for the proof (unless you want to)

While sitting in my math class, I discovered a trick to factoring second degree polynomial's with large or irrational second and third coefficients. For example, try factoring $3x^2+10x-1000$. It's relatively simple to factor it to $(3x-50)(x+20)$ but that would take a little while or at least longer than the way that I'm about to discuss.

We begin with the expression $3x^2+10x-1000$. Then, we devide the second coefficient by 10 and the third by 100 and we are left with the expression $3x^2+x-10$ which we can easily factor to $(3x-5)(x+2)$. Finally, we multiply the second term in each coefficient by 10 and we have $(3x-50)(x+20)$. Looks familiar doesn't it? $\color{#20A900}{\text{Basically, what I have done is I divided the second coefficient by any}}$ $\color{#20A900}{\text{one of its factors (in this case 10) and divided the third coefficient by the}}$ $\color{#20A900}{\text{square of that factor while leaving the first untouched.}}$

--This method applies to irrational and imaginary coefficients as well. For example take $4x^2+8\sqrt2x+8$. We can factor out $2\sqrt2$ from the second coefficient and 8 from the third and are left with $4x^2+4x+1\Rightarrow(2x+1)^2\Rightarrow(2x+2\sqrt2)^2$

--This also works in reverse from the first coefficient. Say we have $25x^2-60x+36$. We can factor out 5 from the middle coefficient and 25 from the first and are left with $x^2-12x+36$. Next, we can factor out a 6 from the second coefficient and 36 from the final coefficient and are left with $x^2-2x+1\Rightarrow (x-1)^2$. Finally, we factor back in 5 to the first coefficient and a 6 into the last one to get $(5x-6)^2$

--This also works for unfactorable expressions. This method will not make unfactorable equations factorable, however, it will make the quadratic formula much easier to do. This is a little tougher to do because depending upon which way you factor a number out, the formula changes.

--Case 1: $ax^2+bx+c\Rightarrow ax^2+\frac{b}{d}+\frac{c}{d^2}$. This changes the quadratic equation to

$\dfrac{-\frac{b}{d}\pm \sqrt{\frac{b^2}{d^2}-\frac{4ac}{d^2}}}{2a}$

$\dfrac{-\frac{b}{d}\pm \frac{\sqrt{b^2-4ac}}{d}}{2a}\Rightarrow \dfrac{-b\pm \sqrt{b^2-4ac}}{2a\color{#3D99F6}{\textbf{d}}}$.

Thus once our answer is achieved, we must multiply our answer by the number d that we extracted at the start.

--Exg: $x^2+60x+2025\Rightarrow \boxed{\color{#3D99F6}{15}} x^2+4x+9$

$\dfrac{-4\pm \sqrt{16-36}}{2}\Rightarrow 2\pm \sqrt{5}i$

$\therefore x=2(15)\pm (15)\sqrt5i\Rightarrow \boxed{30\pm 15\sqrt5i}$

--Case 2: $ax^2+bx+c\Rightarrow \frac{ax^2}{d^2}+\frac{b}{d}+c$. This changes the quadratic to

$\dfrac{-\frac{b}{d}\pm \sqrt{\frac{b^2}{d^2}-\frac{4ac}{d^2}}}{\frac{2a}{d^2}}$

$\dfrac{d^2(-\frac{b}{d}\pm \frac{\sqrt{b^2-4ac}}{d})}{2a}\Rightarrow \frac{\color{#3D99F6}{\textbf{d}}(-b\pm \sqrt{b^2-4ac})}{2a}$.

Thus once our answer is achieved, we must divide our answer by the number d that we extracted at the start.

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PROOF:

--Now, we shall prove why this works.

--Take the general expression $ax^2+bx+c$ which can be factored into $(dx+e)(fx+g)$. This means that a=df, b=dg+ef, and c=eg. The last step of our method requires us to multiply both second coefficients in our binomials by n (n being the number that we factored out of b). So our expression becomes $(dx+\frac{e}{n})(fx+\frac{g}{n})$ which means that $a=df, b=\frac{dg+ef}{n}, c=\frac{eg}{n^2}$. This is why we factor out n and n^2 from the second and third coefficients respectively.

--IF THIS HAS BEEN POSTED ANYWHERE ONLINE (not necessarily on brilliant) please tell me because I would like to find out more about this. Because I haven't been able to find this anywhere, I'm considering myself an OG...lol

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EDIT: thank you to Daniel Liu for helping me to generalize this

--This rule can actually apply to polynomials of all degree. Unfortunately, the higher the degree of the polynomial, the lower the chances are that this will work. But say we have f(x) with degree n. It can be factored into $(a_1x+k_1)(a_2x+k_2)(a_3x+k_3)...(a_nx+k_n)$. Just not necessarily with integer or real coefficients. Next we assume that all coefficients after the second (inclusive), are divisible by $r^t$ where t represents the location of r relative to the second coefficient. This means that we have $\displaystyle\prod_{t=0}^n (a_tx+b_tr)$ which is equivalent to $\displaystyle\sum_{t=0}^n (r^{n-t}(p_tx^t))$ Upon expansion, we get $p_nx^n+r(p_{n-1}x^{n-1})+r^2(p_{n-2}x^{n-2})+r^3(p_{n-3}x^{n-3})...r^{n-1}(p_1x^1)+r^n(p_0)$. As we can see, this method works whenever every coefficient after the first is divisible by $r^t$.

--A useful method for seeing if this works is by prime factoring every coefficient after the second and looking for a common term. That has multiple powers throughout. Say we have $x^4-3x^3-63x^2+27x+486$. First we prime factor the numbers to get 3:3 | 63:3^2,7 | 27:3^3 | 486: 3^5,2. As we can see, they share 3 in increasing powers. Therefore, we can eliminate 3 in increasing powers from each coefficient and are left with

$(x^4-x^3-7x^2+x+6)\Rightarrow x^3(x-1)-(7x+6)(x-1)$

$[x(x^2-1)-6(x+1)](x-1)\Rightarrow (x^2-x-6)(x+1)(x-1)$

$(x-3)(x+2)(x+1)(x-1)$.

--Finally, we factor back in the three that we took out at the start and are left with $\boxed{(x-9)(x+6)(x+3)(x-3)}$

Daniel Liu posted a great generalization for cubics below.