TI-30Xa Algorithm: Synthetic Division

Introduction

Synthetic division is a fairly simple division algorithm that divides a polynomial by the nominal (x-a):

p(x) / (x - c) where

p(x) = a_n * x^n + a_n-1 * x^(n-1) + a_n-2 * x^(n - 2) + … + a_1 * x + a_0

c = a numerical constant

the result, q(x), is a polynomial of order n-1:

q(x) = b_n-1 * x^(n-1) + b_n-2 * x^(n - 2) + … + b_1 * x + b_0 + b_r / (x - c)

b_r = remainder term

where

b_n-1 = a_n

b_i-1 = b_i * c + a_i for i = n-2 down to r (“-1”, see the example below)

For example, for the polynomial:

p(x) = a3 * x^3 + a2 * x^2 + a1 * x + a0

q(x) = p(x) / (x - c)

p(x) has the order n = 3, so q(x) will have the order n = 3 - 1 = 2:

q(x) = b2 * x^2 + b1 * x + b0 + br / (x - c)

b2 = a3

b1 = b2 * c + a2

b0 = b1 * c + a1

br = b0 * c + a0 (the algorithm stops, this is the remainder term)

If br = 0, the x - c divides p(x) evenly, and x = c is a root of p(x).

Note: for any “missing” terms, fill the term with 0.

Example: x^3 + 4 * x - 5 becomes x^3 + 0 * x^2 + 4 * x - 5.

Calculator Algorithm

Store c in one of the memory registers 1, 2, or 3. We’ll call this memory register m for the purpose of the blog.

Note the first coefficient of q(x): a_n

Compute the rest of the coefficients as follows: [ × ] [ RCL ] m [ + ] a_ni [ = ]

Examples

Remember: m is memory register 1, 2, or 3, your choice.

Example 1: (21 * x^2 + 42 * x + 144) / (x - 12)

c = 12

a2 = 21

a1 = 42

a0 = 144

b1 = a2 = 21

12 [ STO ] m

Enter 21

b0:

[ × ] [ RCL ] m [ + ] 42 [ = ]

Result: 294

br:

[ × ] [ RCL ] m [ + ] 144 [ = ]

Result: 3672

Stop.

q(x) = 21 * x + 294 + 3672 / (x - 12)

Example 2: (2 * x^3 + x - 3) / (x - 3) = (2 * x^3 + 0 * x^2 + x - 3) / (x - 3)

c = 3

a3 = 2

a2 = 0

a1 = 1

a0 = -3

b2 = a3 = 2

3 [ STO ] m

Enter 2

b1:

[ × ] [ RCL ] m [ + ] 0 [ = ]

Result: 6

b0:

[ × ] [ RCL ] m [ + ] 1 [ = ]

Result: 19

br:

[ × ] [ RCL ] m [ + ] 3 [ +/- ] [ = ]

Result: 54

q(x) = 2 * x^2 + 6 * x + 19 + 54 / (x - 3)

Example 3: (4 * x^3 + 8 * x^2 - 5 * x + 3) / (x + 2)

c = -2

a3 = 4

a2 = 8

a1 = -5

a0 = 3

b2 = a3 = 4

2 [ +/- ] [ STO ] m

Enter 4

b2:

[ × ] [ RCL ] m [ + ] 8 [ = ]

Result: 0

b3:

[ × ] [ RCL ] m [ + ] 5 [+/-] [ = ]

Result: -5

br:

[ × ] [ RCL ] m [ + ] 3 [ = ]

Result: 13

q(x) = 4 * x^2 - 5 + 13 / (x + 2)

I think this is a fundamental algorithm for students and math enthusiasts to learn, and it’s fairly simple to get a hang of it.

Until next time,

Eddie

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