# 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.

$------------------------------$

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

$--------------------------$

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.

Note by Trevor Arashiro
5 years, 9 months ago

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Nice find Trevor!

Example for cubics:

Factor $x^3-9x+6\sqrt{3}$

We divide the $x^2$ term by $\sqrt{3}$, the $x$ term by $\sqrt{3}^2=3$, and the constant by $\sqrt{3}^3=3\sqrt{3}$ to get this polynomial:

$x^3-3x+2$

This factors easily as $(x-1)^2(x+2)$

Now put the $\sqrt{3}$ back in to get the final answer of $\boxed{(x-\sqrt{3})^2(x+2\sqrt{3})}$

EDIT:

I have a feeling that your find is basically equivalent to this problem solving trick:

Factor $x^3-9x+6\sqrt{3}$

Plug in $x=\sqrt{3}u$ to get $3\sqrt{3}u^3-9\sqrt{3}u+6\sqrt{3}$.

Now factor out a $3\sqrt{3}$ to get $3\sqrt{3}(u^3-3u+2)$ which factors as $3\sqrt{3}(u-1)^2(u+2)$.

Now replace the $u$ with $\dfrac{x}{\sqrt{3}}$ to get $3\sqrt{3}\left(\dfrac{x}{\sqrt{3}}-1\right)^2\left(\dfrac{x}{\sqrt{3}}+2\right)=(x-\sqrt{3})^2(x+2\sqrt{3})$ and we're done.

- 5 years, 9 months ago

In general:

A cubic can always be represented by the form $a(x-r_1)(x-r_2)(x-r_3)$ (this is its factored form). Let's also say that the general form for the above factored cubic is $ax^3+bx^2+cx+d$.

Let's say that a certain cubic can be represented as $a(x-kr_1)(x-kr_2)(x-kr_3)$ for some constant $k$. Then, when you expand its factored form to get its general form, you get $ax^3+kbx^2+k^2cx+k^3d$ and we're done.

This easily generalizes to all polynomials.

- 5 years, 9 months ago

Nice, I was thinking something similar to that as well. But the only thing that confused me at first was if the equation can only be represented by $(ax+b)(cx^2+dx+e)$. Which isn't simplifyable (nicely) if the equation is something like $x^3-8)$. But then I realized that it can be written as $a(x-r_1)(x-r_2)(x-r_3)$, Even though it may be very ugly and not factorable unless you use wolframalpha, it follows the same principle that you talk about, so it would be $a(x+kr_1)(x^2+xkr_2+xkr_3+k^2r_2r_3)$. Thanks for the tip, I will add it to my note shortly.

- 5 years, 9 months ago

I think I remember seeing this on one of your other problems, did you find this as well?

- 5 years, 9 months ago

Yea, I also remember a problem of mine like this, although I can't find it anymore. It had to do with cubics. The intended solution for that problem was the alternate solution I wrote after the EDIT in my first post above.

- 5 years, 9 months ago

Nice Note, @Trevor Arashiro

- 5 years, 9 months ago

I admire that a method is found by Trevor Arashiro . This is a particular application of "Method of Substitution." First let us take its application to say Eq.(X,Y), equation in X and Y . Say $aX^4+bX^3Y+cX^2Y^2+dXY^3+fY^4=0.$ If m is greatest common factor of the coefficients a to d, and n of b to f substitute $p=\dfrac X m ~for~ X,~ q=\dfrac Y n ~for ~Y$ and solve the simplified equation and for p and q. Get back by using the result $X=m*P_i, ~Y=n * Q_i,~where~ P_i,~ Q_i~ are~ the~ solutions. \\ \color{#D61F06}{\text{Note, for an equation of single variable, as we are talking of, put Y=1.} }\\\text{ "Method of Substitution" is used in many area. Say we want to solve }16*(X+14)=(X+2)^4\\Substitute~ t=X+2.~\implies~16*(t+12)=t^4~\therefore~f(t)=t^4-16t-192=0.~~f(4)=0. \\\therefore~t=4, X=t-2=2.~~~~~~~~~~~~(X+2)^4~would~ be~ long.\\ \text{We use it much in calculus, specialy in integrations.}\\ \text{I invite others to give more examples so that it becomes a reference for all}$

- 5 years ago

I noticed this similarly a while ago and was thinking about including this in the note. When I learned about graphical transformations, it occurred to me that this method was a horizontal stretch of the graph. You could of course perform a shift as well if you like.

- 5 years ago

What is the factoring trick for the second degree equation , if the coefficient of second term in the equation is a prime number ?

- 5 years, 8 months ago

Yeah. How to do it @Trevor Arashiro ?

- 5 years, 7 months ago

As I have mentioned in my comment, in this case , there has to be a common factor of second and the third term or second and first term. It apply if the coefficient of second term in the equation is a prime number and first or third term is a multiple of that prime.

- 5 years ago

Thanks A TON..!! u r genius (y)

- 5 years, 8 months ago

Can we apply this to a cubic polynomial ?

- 5 years, 8 months ago

Yes, Daniel Liu posted a very good generalization for cubics. And at the bottom of my note I discuss a general form that works for polynomials of all degree and give a fourth degree example.

- 5 years, 8 months ago

I think this trick may not be applied to the product of Quartics .

- 5 years ago

It can be applied to functions of all degree. Assuming it's a polynomial with coefficients, real or imaginary or trancendental.

- 5 years ago

How to find factors of cubical equation??

- 4 years, 11 months ago

So, essentially, you used a common technique known to the world of didactic Calculus as "u-substitution" to compute values for imaginary numbers and excessively large exponents in the process of factoring. As much as I'd hate to burst your bubble, the concept is simple enough for others to derive it as well as you did. But the process of thought you used is, to the average human, brilliant, and to the intelligent human, a practice that should be kept private, a neat trick you can use for yourself when calculating quickly. People will marvel most when you solve a problem with a mysterious technique (the application) than if you taught it to others (revealing the technique).

Wonder how you're doing these days? I want to keep you as a personal statistic, so kudos...what college are you attending, and what are you majoring in?

- 2 years ago

Great way to use sum of roots and product of roots...you derived it using that or anything else....because I proved it using that...as you divide sum of roots by a number the product is automatically divided by that no. Squared...how did you came to this.....??

- 5 years, 9 months ago

I just randomly stumbled upon it. Like, I wasn't looking for any thing special, just kinda made it up.

- 5 years, 9 months ago