**Polynomials — Division by Vision with Remainders (Part B)**

*A further look at polynomial ‘division by vision’, covering non-unity lead coefficients and quadratic remainders.*

*A further look at polynomial ‘division by vision’, covering non-unity lead coefficients and quadratic remainders.*

This post expands on methods I’ve explored in *Polynomials—Division by Vision with Remainders (Part A)**, *and *Polynomials — Division by Vision*, which avoid long division by ‘downloading’ coefficients from the original polynomials into the solution.

My earlier posts were limited to simpler divisions resulting in linear and quadratic quotients.

With a little extra effort, I’ll now show how the ‘division by vision’ methodology applies to non-unity leading coefficients, and higher order quotients.

Admittedly, big quotients do stretch the ‘division by vision’ claim a little, but it will be handy for programmers and those who seek to understand and marvel at the architecture of polynomials.

You can follow my methodology in these previous posts:

Otherwise, read on, and hopefully by the end you’ll agree that this method:

where *m=B-b) and n=(C-c),*

is essentially this but doesn’t need or show the ‘underbelly’ workings!

The ‘division by vision’ formula for quadratic quotients is:

*y=x²+(B-b)x+(C-c)-b(B-b)+R,*

where ** R** can represent both

**and constant remainders i.e.**

*x***.**

*Rx and Rc*The divisor’s and quotient’s ** x** coefficients and constants generate remainders as per the two equations below.

Now, let’s apply this in three examples.

**Example 1) A non-unity coefficient of x⁴**

The product of the denominator and the quotient equals the original function.

Therefore, we can simply divide the function by the leading coefficient, and multiply the resultant quotient by this coefficient.

This first example produces no remainder, but we will extend this to division with remainder in Example 2.

Divide: *y=2x⁴+6x³-4x²-12x+4*

By: *y=x²+1.39x-0.54*

Divide numerator by: *2:y=x⁴+3x³-2x²-6x+2*

By vision solution: *y=x²+(B-b)x+(C-c)-b(B-b)+R*

Where: *(B-b)=1.61, (C-c)=-1.46, b(B-b)=2.24*

Therefore:* y=x²+1.61x-3.7+R,*

Where: ** Rx=-6x-((-0.54*1.61)-(-3.7*1.39))*x=0 **and

*Rc=2-(-0.54*-3.7)=0*

Hence:* y=x²+1.61x-3.7*

Multiple by the original leading coefficient of *2*

Hence quotient:* y=2x²+3.22x-7.4*

**Example 2) Quintic Polynomial Division by Cubic with Remainder**

This example deals with a non-unity coefficient of ** x²** in the denominator, which we then divide out of both the denominator and quotient.

This example also introduces a quadratic remainder, and also needs to use ** 0x²**.

Divide: *y=x⁵-5x⁴+5x³+5x²-5x-2*

By:* y=2x³+0x²+0.32x+1*

Divide denominator by 2:* y=x³+0x²+0.16x+0.5*

By vision solution after dividing by 2: *y=0.5x²-2.5x+2.42+R*

Calculate ** x²** remainders where we will use the previous

**formula as a guide, and aid ‘division by vision’ by simply cross multiplying, starting with coefficients of**

*R***as follows:**

*x²**(Numerator)=(Denominator)*(Quotient)*

*(x⁵-5x⁴+5x³+5x²-5x-2)=(x³+0x²+0.32x+1)*(0.5x²-2.5x+2.42)*

Cross multiplying by vision:

*Rx²=5x²-(0*2.42+0.32*-2.5+1*0.5)x²=5.3x²*

*Rx=-5x-(0.32*-2.5+1*-2.5)x=-3.2x*

*Rc=-2-(1*2.42)=4.42*

Hence quotient: ** y=0.5x²-2.5x+2.42+(5.3x²-3.2x+4.42)/(2x³+0.32x+**1)

**Example 3) Quintic Polynomial Division by Quadratic with Remainder**

The expanded ‘division by vision’ formula for a quintic with cubic quotient demonstrates the unlimited application of ‘division by vision’ as follows:

Notice the compounding repetition, e.g. where the coefficient of ** x** is

**the**

**in quadratic quotients, and the constant is the coefficient of**

*constant*

*x*b.*This can be simplified further by letting ** l=(C-c)-(B-b)b **as in the header formula of this post, which may be useful in

**higher order problems.**

Divide: ** y=x⁵-5x⁴+5x³+5x²-6x-1**,

**(where**

**)**

*D=+5*By:* y=x²+0.32x+1.22*

** (B-b)=-5.32** and

**and**

*(C-c)=3.78***and**

*(B-b)b=-1.7*

*(B-b)*c=-6.5*By vision solution:

*y=x³-5.32x²+5.48x+5+6.5–5.48*0.32+R*

*y=x³-5.32x²+5.5x+9.75+R*

Using the cross-multiplication method, calculate the remainders.

*(Numerator)=(Denominator)*(Quotient)*

*(x⁵-5x⁴+5x³+5x²-6x-1)=(x²+0.32x+1.22)*(x³-5.32x²+5.5x+9.75)*

*Rx²=5x-(1*9.75+0.32*5.5+1.22*-5.32)x=0*

*Rx=-6x-(1.22*5.5+9.75*0.32)x=-15.83x*

*Rc=-1-(9.75*1.22)=-12.9*

*y=x³-5.32x²+5.5x-(15.83x+12.9)/(x²+0.32x+1.22)*

**Example 4) Cubic Polynomial Division by Linear with Remainder**

I have included this example to demonstrate the flexibility of the ‘division by vision’ method in dealing with all manner of denominators.

Divide: *y=x³+6x²-x-30*

By:* y=x-1+0 (where 0=c)*

** (B-b)=6- -1=7** and

**and**

*(C-c)=-1- 0=-1*

*(B-b)b=7*-1=-7*By vision solution:

*x²+7x+6+R*

*Rx=-1-(-7+6)=0*

*Rc=-30-(-1*6)=-24*

*y=x²+7x+6–24/(x-1)*

I hope this post, in combination with my previous posts which derived from my polynomial factorisation studies, has helped you recognise how an understanding of the ‘architecture’ of polynomials can help simplify the workings and lead to visual solutions.