# Sextic Polynomials — A Practical Approach

At a young age a builder taught me how to mark out a right angle using the 3,4,5 triangle rule with a tape measure. That math practicality has stuck with me and recently I have been working to decipher it into its ‘LEGO block’ constituents to increase intuitive understanding.

This post adopts a ‘Tool Kit’ of methods of finding polynomial roots using basic math including Similar Triangles, ‘internal Quadratic architecture’ and formula adaptations and creation of twin function ‘SOSO’ nodes as application points for root approximation methods. Its purpose is to foster intuitive thinking over convenience solutions.

The intention isn’t to beat a calculator. It is to give students confidence to play the function as they see it, to be creative and in the process strengthen their math abilities.

This post assumes math at a high school level.

# T**he Architecture**

Consider the following Sextic function:

** y= x⁶+x⁵-13x⁴-0.9x³+34x²-14.5x-5**.

# T**he 5 Steps to Finding 6 Roots:**

- Find
using**Root B***Similar Triangles**,*supported by, a constructed twin function with*‘SOSO’***S**ame**O**pposite**S**ame**O**pposite +/- coefficients to provide application nodes. Refer to*Quintic Polynomials-Finding Roots From Primary and Secondary Nodes; a Double Shot!*for an appreciation of the scope of SOSO nodes. - Find
using the quadratic content to approximate the principal function. Refer to*Root C**Cubic Polynomials — Managing the Architecture to Calculate Roots**,*to get the ideas. This also provides an initial application node for.*Root D* - Find
using**Root D***Similar Triangles,*amethod presented in*Cubic Polynomials — Using Similar Triangles to Approximate Roots*. - Find
using a variant of**Root E***Similar Triangles*discussed in*The Simplest Cubic Root* - Finally
and*Roots A*are simply downloaded from an Extended Quadratic Equation I presented in*F**Cubic Polynomials — A Simpler Approach*

Where the Factors ** K**,

**,**

*L***and**

*M***represent**

*N***the known roots.**

** A, B,** are usual

**coefficients and**

*x^n***the constant**

*G***.**

*=-5*# S**tep 1** — *Root B*

*Root B*

Consider the following function ** y= x⁶+x⁵-13x⁴-0.9x³+34x²-14.5x-5**, shown in blue and twin

**derived from**

*y= x⁶-x⁵-13x⁴+0.9x³+34x²+14.5x-5***SOSO**shown in red.

Three Nodes ** H, V** and

**are created by the intersection of the SOSO twin and the primary function as follows:**

*J*** y= x⁶+x⁵-13x⁴-0.9x³+34x²-14.5x-5; **subtract the twin.

** y= x⁶-x⁵-13x⁴+0.9x³+34x²+14.5x-5 **hence:

** y= 2x⁵-1.83x³-29x=x(2x⁴-1.8x²-29); **solve by standard quadratic equation letting

**from which**

*z=x², z=+4.28,***Hence; Node**

*x=+-2.07. x=0, +-2.07.***, Node**

*H=(-2.07, -19.32)***and Node**

*J=(+2.07, -19.32)***.**

*V=(0, -5)*To apply *Similar Triangles* to approximate ** Root B**. We first need to raise

**above the x axis by see-saw action.**

*Node H***See-Saw Rotation to raise the Node**

A) Change the coefficient of ** x** to avoid introducing a constant term into the simultaneous SOSO Quartic function. This see-saws the twin function proportionally with

**so it is necessary to divide the required height shift at**

*x***and reduce the current twin**

*Node H(x)***coefficient**

*x***accordingly as follows:**

*F=14.5*B) Reduce coefficient of ** x **by

**hence**

*19.31/2.07=9.33;***thereby transposing**

*Coefficient F=14.5–9.33=5.17***to**

*Node H***and**

*(-1.9, 4.07)***to**

*Node J***as shown in Graph 6.**

*(1.9, -13.68)*** Node H** coordinates are again easily calculated by simultaneous equations between the new function:

**and the principal function:**

*y= x⁶-x⁵-13x⁴+0.9x³+34x²+5.17x-5,*

*y= x⁶+x⁵-13x⁴-0.9x³+34x²-14.5x-5*Hence: ** y=x(2x⁴-1.8x²-19.67)** which can readily be solved by letting

**.**

*z=x²*Hence:** x=0, +-1.907 **and

*y=4.07.**Note: there is no known rule to determine a shift that would transpose a node exactly onto the root! The overshoot to y=4.07 occurred because the twin gradient is positive at x=-2.07*.

*Similar Triangles*

*Similar Triangles*

Referring to my earlier Post and Diag 1 below, ** Nodes H** and

**strike a chord across the segment containing the target**

*H’***shown in red.**

*Root B*** Node H** has transposed from

**to**

*(–2.07, -19.31)***hence:**

*(-1.9, 4.07)*** Δx=k*J/(k+m)=4.07*(-2.07+1.9)/(4.07+19.31)=-0.03**.

* Approx Root B=-1.9+Δx=-1.93*,

*which equals actual*

**.**

*x=-1.93*# S**tep 2** — *Root C*

*Root C*

Referring to Graph 2 below; separating the function into (a) Sextic-cubic ** Ya=x⁶+x⁵-13x⁴-0.9x³** and (b) Quadratic

**shown in dotted black which closely approximates the principal function from the turning point**

*Yb=34x²-14.5x-5***well**

*Xtp***through**

**.**

*Root C*The Quadratic equation content directly delivers quadratic ** Root C=0.23 **and also provides Quadratic

**which is the starting point for creation of a Node for application**

*Root k=0.65***approximation to**

*Similar Triangles***.**

*Root D**That was the easy one!*

# S**tep 3 **— *Root D*

*Root D*

Referring to Graph 3 below; construct a 1st node intercept on the function using a Sextic-Cubic and Quadratic Contents.

**1st Node Intercept**

Again consider the Quadratic content ** Yb=34x²-14.5x-5 **with

**. Since**

*Root K(J)=0.65***the Sextic-cubic content**

*y=Ya+Yb***(shown in dotted green)**

*Ya=x⁶+x⁵-13x⁴-0.9x³***must intersect the principal function at**

**where**

*Node K,***. Therefore**

*x=0.65***.**

*Ya=-2.4***2nd Node Intercept**

Refer to graph 6

Establish the 2nd node as close as possible to the root with y positive. A simple method is to construct a segment from ** TP=(0.21, -6.55) **through

**and using similar triangles, intercept the x axis at**

*Node K=(0.65, -2.4)***. Use that intercept as a root of the Quadratic content from which the**

*I=(0.9, 0)***value of the Sextic-cubic content can be calculated.**

*y*Refer to Graph 7

Next calculate the constant shift ** H** of the Quadratic required to create

*Root I.*Let *Yb=34x²-14.5x+H.*

Since roots are ** +0.9 **and

**as shown at**

*-0.9–2*0.21=-0.48***, Constant**

*Root J***.**

*C=-0.432*34=-14.69***Adding this to**

**to maintain**

*Ya=x⁶+x⁵-13x⁴-0.9x³***we get**

*y=Ya+Yb***from which intercept**

*Ya=x⁶+x⁵-13x⁴-0.9x³+14.69***.**

*Node K2=(0.9, 1.43)***Using Similar Triangles**

** Node K** has transposed from

**to**

*(0.65, -2.4)***hence:**

*(0.9, 1.43)**Δx=1.43*(0.9–0.65)/(1.43+2.4)=0.09*

* Approx Root B=0.9-Δx=0.81*,

*compared with actual*

**.**

*0.8*# S**tep 4 — ***Root E*

*Root E*

Referring to Graph 8, even after the shift to find ** Root B**, the remoteness of

*coupled with its negative twin gradients makes overshoot in 1 iteration for*

**Node J=(1.9, -13.68)***Similar Triangles*as used in Steps 1 and 3 too complex so we will adopt a variation for

**.**

*Root E*In my post *The Simplest Cubic Root, *Similar Triangles are adopted using a Cubic polynomial Inflection point ** Ip** and the Constant term to construct a triangle hypotenuse which intercepts the x axis close to the target root. Unlike the preceding examples, the nodes don’t strike a cord across the x axis.

In this Sextic example, ** Node J=(2.07, -19.3) **coordinates prior to shift, are to represent the constant term and after shift

**, assumed to be proximal to**

*Node J=(1.9, -13.68)***(not defined).**

*Ip***Using Similar Triangles**

** Δx=0.17*13.68/5.62=0.414**.

** Approx Root E=1.9–0.414=1.486 **versus

**actual.**

*1.49*Note 1: The assumption to use a proximal ** Ip **in this particular case is borne out when it is considered that the worst case scenario for

**results in an approximation**

*Ip=B<-13.69***as shown in Graph 9 below. This is comparable with a result using Newton’s approximation from**

*A=1.53***of;**

*Node J=(1.9, -13.68)**A=1.9-(-13.68/-37.99)=1.54.*

Note 2: Using an ** Ip** value very close to the initial

**would be akin to using Newton’s approximation from that same point which would not result in a good 1st approximation from either method.**

*Node J*# S**tep 5**— **Roots A and F by Formula**

Given ** Roots B, C, D** and

**and their related Factors**

*E***,**

*k, l, m, n***and**

*Roots A***are simply derived from the Extended Qudaratic Equation I presented in an earlier post**

*F**Cubic Polynomials — A Simpler Approach*

*as follows:*

Giving ** Root A=-3.68 **compared with

**and**

*-3.7*

*Root***compared with**

*F=2.54***actual.**

*2.57*I hope this post has shown that even with higher order polynomials, a partnership of node management and basic architectural content can lead to very good results. I have tried to adopt its simplicity, to play with the architecture and use lego block math to solve a complex function with no paternal rules specific to it. Making math work for you, not you for it!