Polynomials Question involving relative prime numbers

[abhinavs]

Two integers are said to be relatively prime if their highest common factor is 1. If a and
b are relatively prime it is possible to find integers x and y such that ax + by = 1. For
example 51 and 44 are relatively prime.
Repeated use of the division identity
leads to:
51 = 44 × 1 + 7
44 = 7 × 6 + 2
7 = 3 × 2 + 1
Reversing these steps leads to:
1 = 7 − 3 × 2
= 7− 3(44 − 7 × 6)
= 19 × 7 − 3 × 44
= 19(51 − 44 × 1) − 3 × 44
= 19 × 51 − 22 × 44

(a) Use this method to find integers a and b such that 87a + 19b = 1
(b) Find polynomials A(x) and B(x) such that 1 = A(x)(x^2−x)+B(x)(x^4+4x^2−4x+4).

I was able to complete part (a), however, part (b) seemed impossible.

Here are the three equations I have come up with so far:
x^4 +4x^2-4x+4 = (x^2-x)(x^2 + x +5)+ x+4 (1)
x^2-x = (x+4)(x-5) + 20
x+4 = (20(x+3))/20 + 1
I am unable to solve them simultaneously.
Please help.

 

[mmm4444bot]

Two integers are said to be relatively prime if their highest common factor is 1. If a and
b are relatively prime it is possible to find integers x and y such that ax + by = 1.

For example 51 and 44 are relatively prime.

Repeated use of the division identity leads to:
51 = 44 × 1 + 7
44 = 7 × 6 + 2
7 = 3 × 2 + 1
Reversing these steps leads to:
1 = 7 − 3 × 2
= 7− 3(44 − 7 × 6)
= 19 × 7 − 3 × 44
= 19(51 − 44 × 1) − 3 × 44
= 19 × 51 − 22 × 44

(a) Use this method to find integers a and b such that 87a + 19b = 1

(b) Find polynomials A(x) and B(x) such that 1 = A(x)(x^2−x)+B(x)(x^4+4x^2−4x+4).

I was able to complete part (a), however, part (b) seemed impossible.

Here are the three equations I have come up with so far:

x^4 + 4x^2 - 4x + 4 = (x^2 - x)(x^2 + x + 5) + x + 4

x^2 - x = (x+4)(x-5) + 20

x + 4 = 20*(x+3)/20 + 1

I am unable to solve them simultaneously.

I got the same.

We're doing good, so far, but we're not solving these equations simultaneously. (That would be finding values of x that make all three equations true.)

Instead, we're trying to find polynomials named A(x) and B(x) such that the equation given in part (b) is true.

Reverse the steps, like they did in the example. Keeping track of each expression (that is, matching the forms part-for-part, as shown in the reversed-steps example) is a bit of a mind bender; be careful with substitutions and factoring. It's all algebraic manipulation of those equations, to arrive at:

1 = A(x)*(x^2 - x) + B(x)*(x^4 + 4x^2 - 4x + 4)

I get a 4th-degree polynomial for A(x) and a 2nd-degree polynomial for B(x).

---

Starting with the third equation, solve it for 1:

1 = (x + 4) - (x + 3)/20 * 20

Now, solve the second equation for 20, and substitute the result for 20 above.

Continue following the logic given in the example … don't multiply out or simplify everything at each step … try to follow the same form as they did, like here:

= 7− 3(44 − 7 × 6)
= 19 × 7 − 3 × 44

By the way, what did you get for your answer to part (a) ? :cool: