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Cubic Polynomial “Zero-Sum” Factoring Strategy, Part 2

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by

Mars Tonry
in

I talked about the motivation and derivation of this factoring strategy in my July 27th post (hyperlink here), and here I will prove some other facts about it. I assume the fact that the product and sum of integers is also an integer. Also, I apologize in advance for the lack of academic rigor in these proofs.

To begin with, if a cubic has a factored form of linear terms with integer coefficients, its roots will be rational numbers, and the coefficients of the cubic will be integers.

ax3+bx2+cx+d=(m1x−b1)(m2x−b2)(m3x−b3), xn=bnmn; mn,bn∈ℤ⟹xn∈ℚax^3+bx^2+cx+d=(m_1x-b_1)(m_2x-b_2)(m_3x-b_3),\ x_n=\frac{b_n}{m_n};\ m_n,b_n \in\mathbb{Z} \implies x_n\in\mathbb{Q}
=(m1m2m3)x3−(b1m2m3+m1b2m3+m1m2b3)x2+(b1b2m3+b1m2b3+b1b2m3)x−b1b2b3=(m_1m_2m_3)x^3-(b_1m_2m_3+m_1b_2m_3+m_1m_2b_3)x^2+(b_1b_2m_3+b_1m_2b_3+b_1b_2m_3)x-b_1b_2b_3
a=m1m2m3, b=−b1m2m3−m1b2m3−m1m2b3, c=m1b2b3+b1m2b3+b1b2m3, d=−b1b2b3⟹a,b,c,d∈ℤa=m_1m_2m_3,\ b=-b_1m_2m_3-m_1b_2m_3-m_1m_2b_3,\ c=m_1b_2b_3+b_1m_2b_3+b_1b_2m_3,\ d=-b_1b_2b_3\implies a,b,c,d\in\mathbb{Z}

Next, we can prove that the roots of the simplified cubic will be integers if it can be factored.

z3+(9ac−3b2)z+(2b3−9abc+27a2d)=(z−r1)(z−r2)(z−r3), zn=rnz^3+(9ac-3b^2)z+(2b^3-9abc+27a^2d)=(z-r_1)(z-r_2)(z-r_3),\ z_n=r_n
xn=rn−b3ax_n=\frac{r_n-b}{3a}
x1=r1−b3ax_1=\frac{r_1-b}{3a}
b1m1=r1−(−b1m2m3−m1b2m3−m1m2b3)3(m1m2m3)\frac{b_1}{m_1}=\frac{r_1-(-b_1m_2m_3-m_1b_2m_3-m_1m_2b_3)}{3(m_1m_2m_3)}
3b1m2m3=r1+b1m2m3+m1b2m3+m1m2b33b_1m_2m_3=r_1+b_1m_2m_3+m_1b_2m_3+m_1m_2b_3
2b1m2m3−m1b2m3−m1m2b3=r1⟹r1∈ℤ2b_1m_2m_3-m_1b_2m_3-m_1m_2b_3=r_1 \implies r_1\in\mathbb{Z}
x2=z2−b3ax_2=\frac{z_2-b}{3a}
b2m2=r2−(−b1m2m3−m1b2m3−m1m2b3)3(m1m2m3)\frac{b_2}{m_2}=\frac{r_2-(-b_1m_2m_3-m_1b_2m_3-m_1m_2b_3)}{3(m_1m_2m_3)}
3m1b2m3=r2+b1m2m3+m1b2m3+m1m2b33m_1b_2m_3=r_2+b_1m_2m_3+m_1b_2m_3+m_1m_2b_3
−b1m2m3+2m1b2m3−m1m2b3=r2⟹r2∈ℤ-b_1m_2m_3+2m_1b_2m_3-m_1m_2b_3=r_2 \implies r_2\in\mathbb{Z}
x3=r3−b3ax_3=\frac{r_3-b}{3a}
b3m3=r3−(−b1m2m3−m1b2m3−m1m2b3)3(m1m2m3)\frac{b_3}{m_3}=\frac{r_3-(-b_1m_2m_3-m_1b_2m_3-m_1m_2b_3)}{3(m_1m_2m_3)}
3m1m2b3=r3+b1m2m3+m1b2m3+m1m2b33m_1m_2b_3=r_3+b_1m_2m_3+m_1b_2m_3+m_1m_2b_3
−b1m2m3−m1b2m3+2m1m2b3=r3⟹r3∈ℤ-b_1m_2m_3-m_1b_2m_3+2m_1m_2b_3=r_3 \implies r_3\in\mathbb{Z}

From this, we have proven:

rn∈ℤr_n\in\mathbb{Z}

Next, we can prove that if the product of the three roots of the simplified cubic is positive, two roots must be negative and one must be positive. By contrast, if the product is negative, two roots must be positive and one must be negative.

First, we have that:

r1+r2+r3=0r_1+r_2+r_3=0
r1r2r3≠0⟹r1,r2,r3≠0r_1r_2r_3\neq0\implies r_1,r_2,r_3\neq0

Because we have proven that each root is an integer, we can say that:

r1,r2,r3>0⟹r1,r2,r3≥1⟹r1+r2+r3≥3r_1,r_2,r_3>0\implies r_1,r_2,r_3\geq1\implies r_1+r_2+r_3 \geq3
r1,r2,r3<0⟹r1,r2,r3≤−1⟹r1+r2+r3≤−3r_1,r_2,r_3<0\implies r_1,r_2,r_3\leq-1\implies r_1+r_2+r_3 \leq-3

Because this contradicts our original statement, that the sum of roots is zero, we may not have all three roots be negative or positive. Now, let’s focus on the case when the product is positive.

r1r2r3>0r_1r_2r_3>0

Let’s incorrectly assume that one root is negative and the other two are positive. Without loss of generality, we can assign a placeholder variable to the first root.

r1<0,r2>0,r3>0, r1=−q, q>0r_1<0, r_2>0, r_3>0,\ r_1=-q,\ q>0
qr2r3>0, q=−r1qr_2r_3>0,\ q=-r_1
−r1r2r3>0⟺r1r2r3<0-r_1r_2r_3>0\iff r_1r_2r_3<0

We can reach a similar contradiction we assume the product is negative but two roots are negative and one root is positive.

r1r2r3<0r_1r_2r_3<0
r1>0,r2<0,r3<0, r1=−q, q>0r_1>0, r_2<0, r_3<0,\ r_1=-q,\ q>0
qr2r3<0, q=−r1qr_2r_3<0,\ q=-r_1
−r1r2r3<0⟺r1r2r3>0-r_1r_2r_3<0\iff r_1r_2r_3>0
Divisor bounds

Closer inspection of our original system of equations yields this simplification:

r1+r2+r3=0r_1+r_2+r_3=0
r3=−r1−r2r_3=-r_1-r_2
r1r2r3=9abc−2b3−27a2dr_1r_2r_3=9abc-2b^3-27a^2d
r1r2(−r1−r2)=9abc−2b3−27a2dr_1r_2(-r_1-r_2)=9abc-2b^3-27a^2d

Without loss of generality, let’s assume the following:

r1>0, r2>0, r3=−r1−r2<0, r1≥r2r_1>0,\ r_2>0,\ r_3=-r_1-r_2<0,\ r_1\geq r_2

Note that:

|r2|≤|r1|≤|r3||r_2|\leq |r_1| \leq |r_3|

Therefore,

r1r2r3<0r_1r_2r_3<0
0<r2≤r1⟺1≤r2≤r10<r_2\leq r_1 \iff 1 \leq r_2 \leq r_1

Let’s define the product as P.

P=9abc−2b3−27a2d=r1r2r3=r1r2(−r1−r2)P=9abc-2b^3-27a^2d=r_1r_2r_3=r_1r_2(-r_1-r_2)

We can then create bounds for r1 based on r2 using P.

r1(r1)(−r1−r1)≤P≤r1(1)(−r1−1)r_1(r_1)(-r_1-r_1)\leq P\leq r_1(1)(-r_1-1)
−2r13≤P≤−r1(r1+1)-2r_1^3\leq P\leq -r_1(r_1+1)
−2r13≤P-2r_1^3\leq P
r1≤−P23r_1\leq \sqrt[3]{-\frac{P}{2}}
−r12−r1≥P-r_1^2-r_1 \geq P
−r12−r1−14≥P−14-r_1^2-r_1-\frac{1}{4}\geq P-\frac{1}{4}
−(r1−12)2≥P−14-\left(r_1-\frac{1}{2}\right)^2\geq P-\frac{1}{4}
(r1−12)2≥14−P\left(r_1-\frac{1}{2}\right)^2\geq \frac{1}{4}-P
r1−12≥14−Pr_1-\frac{1}{2}\geq \sqrt{\frac{1}{4}-P}
r1≥12+14−Pr_1\geq \frac{1}{2}+\sqrt{\frac{1}{4}-P}

Therefore, if P is negative,

−P23≤r1≤12+14−P\sqrt[3]{-\frac{P}{2}}\leq r_1\leq \frac{1}{2}+\sqrt{\frac{1}{4}-P}

If P is positive and r1 and r2 are negative (with the inequality between them flipped as well), a similar calculation yields:

P23≤−r1≤12+14+P\sqrt[3]{\frac{P}{2}}\leq -r_1\leq \frac{1}{2}+\sqrt{\frac{1}{4}+P}

Generally, we can say

±P23≤∓ r1≤12+14±P, r1r2>0, |r1|≥|r2|, r3=−r1−r2\sqrt[3]{\pm\frac{P}{2}}\leq \mp\ r_1\leq \frac{1}{2}+\sqrt{\frac{1}{4}\pm P},\ \frac{r_1}{r_2}>0,\ |r_1|\geq|r_2|,\ r_3=-r_1-r_2

These are useful bounds for calculating the roots if the simplified cubic.

Let’s finish this post with an example.

2x3+9x2−2x−24=02x^3+9x^2-2x-24=0
9abc−2b3−27a2d=−324−1458+2592=8109abc-2b^3-27a^2d=-324-1458+2592=810

Next, we calculate the bounds for |r1| with 810:

7.399≤r1≤28.9657.399\leq r_1\leq 28.965

We only need to search for divisors of 810 in between these bounds. Let’s list the divisors of 810:

1,2,3,5,6,9,10,15,18,27,30,45,54…1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54…

We can stop there, as |r3| cannot be larger than 2|r1|. We only need to check 9, 10, 15, 27, and 30 for candidates as |r1|.

−15=r1,−3=r2,18=r3-15=r_1, -3=r_2,18=r_3

Are the solutions to this equation, with the solutions of the original cubic being -4, -2, and 1.5.

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