## Sunday, August 21, 2022

### Rationalizing a Quadratic Polynomial

Rationalizing a Quadratic Polynomial

Problem

Rewrite the Quadratic Polynomial 1 + A*x + B*x^2 as a rational function of polynomials.  A and B are real numbers (however, this should work if A and B were complex numbers).

1 + A*x + B*x^2 → p(x) / q(x)

Attempt 1:  1 + A*x + B*x^2 → (1 + C*x) / (1 + D*x)

1 + A*x + B*x^2 = (1 + C*x) / (1 + D*x)

(1 + A*x + B*x^2) * (1 + D*x) = (1 + C*x) / (1 + D*x) * (1 + D*x)

A*x + B*x^2 + D*x + A*D*x^2 + D*B*x^3 = C*x

Comparing the powers of x:

constant:  0 = 0

x:  A + D = C

x^2:  B + A*D = 0

x^3:  B*D = 0

This leads to either B=0 or D=0

Assume B=0.

Then B + A*D = 0

A*D = 0

If A=0, then D = C, which leads to:

1 + A*x + B*x^2 = (1 + C*x) / (1 + D*x)

1 = (1 + C*x) / (1 + C*x)

1 = 1

If D=0, then A = C

1 + A*x + B*x^2 = (1 + C*x) / (1 + D*x)

1 + C*x = (1 + C*x)

If D=0, then B=0, and we get the same results as above.

Ultimately this transformation to (1 + C*x) / (1 + D*x) leads to nothing useful.

Attempt 2:  1 + A*x + B*x^2 → (1 + C*x^2) / (1 + D*x)

1 + A*x + B*x^2 = (1 + C*x^2) / (1 + D*x)

(1 + A*x + B*x^2) * (1 + D*x) = 1 + C*x^2

(A + D)*x + (B + A*D)*x^2 + B*D*x^3 = C*x^2

Comparing the powers of x:

constant:  0 = 0

x:  A + D = 0

x^2:  B + A*D = C

x^3:  B*D = 0

A + D = 0 implies that A = -D or D = -A

Also either B = 0 or D = 0.

Assume B = 0. Then with A = -D:

A*D = C

-D*D = C

C = -D^2

1 + A*x + B*x^2 = (1 + C*x^2) / (1 + D*x)

1 -  D*x = (1 - D*x^2) / (1 + D*x)

1 -  D*x = ((1 - D*x) * (1 + D*x))/ (1 + D*x)

1 - D*x = 1 - D*x

If we assume that D = -A, then:

C = -A^2 and

1 + A*x = (1 - A*x^2) / (1 - A*x)

1 + A*x = ((1 - A*x)  * (1 + A*x)) / (1 - A*x)

1 + A*x = 1 + A*x

Assume D = 0.

Then A = 0 and B = C:

1 + B*x^2 = 1 + B*x^2

1 + C*x^2 = 1 + C*x^2

Again, we have transformations that are trivial.

Attempt 3:  1 + A*x + B*x^2 → (1 + C*x^3) / (1 + D*x)

1 + A*x + B*x^2 = (1 + C*x^3) / (1 + D*x)

(1 + A*x + B*x^2) * (1 + D*x) = 1 + C*x^3

(A + D)*x + (B + A*D)*x^2 + B*D*x^3 = 1 + C*x^3

Comparing the powers of x:

constant:  0 = 0

x:  A + D = 0

x^2:  B + A*D = 0

x^3:  B*D = C

This implies that:

A + D = 0

D = -A

B + A*D = 0

B+ A*-A = 0

B = A^2   (this restricts A and B)

B * D = C

(A^2)*(-A) = C

C = -A^3

We can conclude that B = A^2, C = -A^3, D = -A

The relationship between A, B, C, and D are all connected in this case.

Examples:

A = 2 ⇒ B = 4, C = -8, D = -2 and

1 + 2*x + 4*x^2 = (1 - 8*x^3) / (1 - 2*x)

A = -2 ⇒ B = 4, C = 8, D = 2 and

1 - 2*x + 4*x^2 = (1 + 8*x^3) / (1 + 2*x)

Note:  Casio fx-991EX Week - September 5, 2022 to September 9, 2022

Eddie

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