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Cubic Polynomial “Zero-Sum” Factoring Strategy

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by

Mars Tonry
in

In 9th grade, I discovered a method for factoring cubic polynomials with integer coefficients, which I will call the “Zero-Sum” method. This very likely has been discovered before, but I have never heard about it. It is equivalent to the “Diamond Method” for factoring quadratics, which involves finding two numbers that sum to a certain value and multiply to another. However, the Zero-Sum Method involves finding three numbers that sum to zero and multiply to a certain value. To begin with, we need to simplify the general cubic polynomial with a few substitutions.

ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0
x3+bax2+cax+da=0x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}=0
x=y−b3ax=y-\frac{b}{3a}
(y−b3a)3+ba(y−b3a)2+ca(y−b3a)+da=0\left(y-\frac{b}{3a}\right)^3+\frac{b}{a}\left(y-\frac{b}{3a}\right)^2+\frac{c}{a}\left(y-\frac{b}{3a}\right)+\frac{d}{a}=0
(y3+3(−b3a)y2+3(−b3a)2y+(−b3a)3)+ba(y2+2(−b3a)y+(−b3a)2)+ca(y−b3a)+da=0\left(y^3+3\left(-\frac{b}{3a}\right)y^2+3\left(-\frac{b}{3a}\right)^2y+\left(-\frac{b}{3a}\right)^3\right)+ \frac{b}{a}\left(y^2+2\left(-\frac{b}{3a}\right)y+\left(-\frac{b}{3a}\right)^2\right)+\frac{c}{a}\left(y-\frac{b}{3a}\right)+\frac{d}{a}=0
(y3−bay2+b23a2y−b327a3)+ba(y2−2b3ay+b29a2)+ca(y−b3a)+da=0\left(y^3-\frac{b}{a}y^2+\frac{b^2}{3a^2}y-\frac{b^3}{27a^3}\right)+\frac{b}{a}\left(y^2-\frac{2b}{3a}y+\frac{b^2}{9a^2}\right)+\frac{c}{a}\left(y-\frac{b}{3a}\right)+\frac{d}{a}=0
y3−bay2+b23a2y−b327a3+bay2−2b23a2y+b39a3+cay−bc3a2+da=0y^3-\frac{b}{a}y^2+\frac{b^2}{3a^2}y-\frac{b^3}{27a^3}+\frac{b}{a}y^2-\frac{2b^2}{3a^2}y+\frac{b^3}{9a^3}+\frac{c}{a}y-\frac{bc}{3a^2}+\frac{d}{a}=0
y3+(−ba+ba)y2+(b23a2−2b23a2+ca)y+(−b327a3+b39a3−bc3a2+da)=0y^3+\left(-\frac{b}{a}+\frac{b}{a}\right)y^2+\left(\frac{b^2}{3a^2}-\frac{2b^2}{3a^2}+\frac{c}{a}\right)y+\left(-\frac{b^3}{27a^3}+\frac{b^3}{9a^3}-\frac{bc}{3a^2}+\frac{d}{a}\right)=0
y3+(−b23a2+ca)y+(2b327a3−bc3a2+da)=0y^3+\left(-\frac{b^2}{3a^2}+\frac{c}{a}\right)y+\left(\frac{2b^3}{27a^3}-\frac{bc}{3a^2}+\frac{d}{a}\right)=0
y3+3ac−b23a2y+2b3−9abc+27a2d27a3=0y^3+\frac{3ac-b^2}{3a^2}y+\frac{2b^3-9abc+27a^2d}{27a^3}=0

We have eliminated the quadratic term in the cubic polynomial. This is also the first step in deriving the cubic formula (which we are not going to do). However, these terms are very messy and unpleasant to work with. To simplify further, we can make another substitution:

y=z3ay=\frac{z}{3a}
(z3a)3+3ac−b23a2(z3a)+2b3−9abc+27a2d27a3=0\left(\frac{z}{3a}\right)^3+\frac{3ac-b^2}{3a^2}\left(\frac{z}{3a}\right)+\frac{2b^3-9abc+27a^2d}{27a^3}=0
127a3z3+3ac−b29a3z+2b3−9abc+27a2d27a3=0\frac{1}{27a^3}z^3+\frac{3ac-b^2}{9a^3}z+\frac{2b^3-9abc+27a^2d}{27a^3}=0
z3+(9ac−3b2)z+(2b3−9abc+27a2d)=0z^3+\left(9ac-3b^2\right)z+\left(2b^3-9abc+27a^2d\right)=0

Our goal is to factor this into three terms. We will write out a factored form, expand, and collect like terms:

(z−r1)(z−r2)(z−r3)=0 [z=r1, r2, r3](z-r_1)(z-r_2)(z-r_3)=0\ [z=r_1,\ r_2,\ r_3]
z3−r1z2−r2z2−r3z2+r1r2z+r1r3z+r2r3z−r1r2r3=0z^3-r_1z^2-r_2z^2-r_3z^2+r_1r_2z+r_1r_3z+r_2r_3z-r_1r_2r_3=0
z3−(r1+r2+r3)z2+(r1r2+r1r3+r2r3)z−r1r2r3=0z^3-(r_1+r_2+r_3)z^2+(r_1r_2+r_1r_3+r_2r_3)z-r_1r_2r_3=0

Comparing to our simplified cubic polynomial, we can equate terms with the same degree to create a system of equations:

0=r1+r2+r30=r_1+r_2+r_3
9ac−3b2=r1r2+r1r3+r2r39ac-3b^2=r_1r_2+r_1r_3+r_2r_3
2b3−9abc+27a2d=−r1r2r32b^3-9abc+27a^2d=-r_1r_2r_3

We can ignore the second equation, as it adds unnecessary complication to the method. Additionally, we can multiply the third equation by -1 and rearrange terms:

0=r1+r2+r30=r_1+r_2+r_3
9abc−2b3−27a2d=r1r2r39abc-2b^3-27a^2d=r_1r_2r_3

To solve this system of equations, we must find three factors of

9abc−2b3−27a2d9abc-2b^3-27a^2d

Which sum to zero. These factors are the solutions to the simplified cubic equation. We must undo the substitutions that we made to solve for the solution to the general cubic equation.

x=y−b3a, y=z3ax=y-\frac{b}{3a},\ y=\frac{z}{3a}
x=z3a−b3ax=\frac{z}{3a}-\frac{b}{3a}
x=z−b3a, z=r1, r2, r3x=\frac{z-b}{3a},\ z=r_1,\ r_2,\ r_3
x=r1−b3a, r2−b3a, r3−b3ax=\frac{r_1-b}{3a},\ \frac{r_2-b}{3a},\ \frac{r_3-b}{3a}

From this, we have found the three solutions to the general cubic equation.

Example:
6x3−7x2−28x+20=06x^3-7x^2-28x+20=0
9abc−2b3−27a2d=9(6)(−7)(−28)−2(−7)3−27(6)2(20)9abc-2b^3-27a^2d=9(6)(-7)(-28)-2(-7)^3-27(6)^2(20)
=10584+686−19440=10584+686-19440
=−8170=-8170

So, we need to find three factors of -8170 which add to zero. Because -8170 is a negative number, we will need 1 negative factor and 2 positive factors. To approach this, let’s factor -8170:

−8170=−1∗2∗5∗19∗43-8170=-1*2*5*19*43

Here, it becomes obvious that -43 will be the negative factor and 5 and 38 will be the two positive factors, because they are the only combination which sum to zero (-43+5+38=0). Now, we can easily solve for the three solutions to our original cubic equation.

x=−43−(−7)3(6), 5−(−7)3(6), 38−(−7)3(6)x=\frac{-43-(-7)}{3(6)},\ \frac{5-(-7)}{3(6)},\ \frac{38-(-7)}{3(6)}
x=−3618, 1218, 4518x=-\frac{36}{18},\ \frac{12}{18},\ \frac{45}{18}
x=−2, 23, 52x=-2,\ \frac{2}{3},\ \frac{5}{2}

These solutions can be verified by plugging them in to the cubic equation. For legitimacy, I will verify the first solution.

6(−2)3−7(−2)2−28(−2)+20=06(-2)^3-7(-2)^2-28(-2)+20=0
−48−28+56+20=0-48-28+56+20=0
0=00=0

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