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# Math IB HL math portfolio type I - polynomials

Extracts from this document...

Introduction

Diana Herwono D 0861 006

1. Let P(x) = a5x5 + a4x4 + a3x3 + a2x2 + a1x + a0.

Using the sums of (a5 + a3 + a1) and (a4 + a2 + a0), determine whether

P(1) = 0?  and P(-1) = 0?

Examine these examples:

(1) P(x) = x5 - 3x4 + 2x3 + 4x2 + 6x - 10

(2) P(x) = x5 - 3x4 + 2x3 - 4x2 + 6x + 10

(3) P(x) = x5 + 3x4 + 2x3 - 4x2 + 6x + 10

(4) P(x) = x5 + 3x4 + 2x3 - 4x2 + 6x - 10

## What is your conclusion for the general case when

P(x) = anxn + an-1xn-1 + … + a2x2 + a1x1 + a0?

Solution:

(1)         P(x) = x5 - 3x4 + 2x3 + 4x2 + 6x - 10

(a5 + a3 + a1) = 9

(a4 + a2 + a0) = -9

P(1) = 0

P(-1) = -18

(2)         P(x) = x5 - 3x4 + 2x3 - 4x2 + 6x + 10

(a5 + a3 + a1) = 9

(a4 + a2 + a0) = 3

P(1) = 12

P(-1) = -6

(3)        P(x) = x5 + 3x4 + 2x3 - 4x2 + 6x + 10

(a5 + a3 + a1) = 9

(a4 + a2 + a0) = 9

P(1) = 18

P(-1) = 0

(4)        P(x) = x5 + 3x4 + 2x3 - 4x2 + 6x - 10

(a5 + a3 + a1) = 9

(a4 + a2 + a0) = -11

P(1) = -2

P(-1) = -20

Conclusion:

From the above examples, we see that when:

a5 + a3 + a1 = -(a4 + a2 + a0)                 P(1)= 0

a5 + a3 + a1 = a4 + a2 + a0                 P(-1)= 0

Therefore for the general case,

if an-1 + an-3 + … + a1 = -(an + … + a2 + a0)         then P(1) = 0

if an-1 + an-3 + … + a1 = an + … + a2 + a0         then P(-1) = 0

1. There is a conclusion that states:

Middle

2k2/m3 - a1k/m2 - a0/m) / (k3/m3)

Therefore, m is a factor of a3.

1. a-bi is called the conjugate of a+bi, where a and b are real numbers and

i = (-1).  Let a+bi denote a-bi. In other words, a+bi = a-bi.

Prove that:____________ _____

1. (a + bi) + (c + di) = a + bi + c + di

(2) (a + bi) - (c + di) =_a + bi – c + di

(3) (a + bi) (c + di) = (a+ bi) (c + di)

(4) (a + bi)3 = (a + bi)3

_        _

If a is a real number, a = ?  0 = ?

Solution:

_____

Let a + bi = a – bi

______________    _____    _____

(1)        (a + bi) + (c + di) = a + bi + c + di

(a + c) + (bi + di) = a – bi + c – di

(a + c) + (b + d)i  = (a + c) – (bi + di)

(a + c) - (b + d)i   = (a + c) - (b + d)i

______________    _____   _____

(2)        (a + bi) - (c + di) = a + bi – c + di

(a - c) + (bi + di) = (a – bi) – (c – di)

(a – c) - (b + d)i  = (a - c) - (bi - di)

(a - c) - (b + d)i   = (a - c) - (b - d)i

____________       _____   ______

(3)                (a + bi) (c + di)   = (a + bi) (c + di)

ac + adi + bci – bd   = (a – bi) (c – di)

(ac – bd) + (ad + bc)i = ac – adi – bci - bd

(ac - bd) - (ad + bc)i   = (ac - bd) - (ad + bd)i

_______    _____

(4)                                     (a + bi)3= (a + bi)3

(a + bi) (a2 + 2abi – b2)   = (a – bi)3

a3 + 3a2bi – 3ab2 – b3i     = (a – bi) (a2 – 2abi – b2)

(a3 - 3ab2)

Conclusion

nxn + an-1xn-1 + … + a1x + a0,  with rational coefficients, where a and b are rational, but √r is irrational, then the conjugate radical, a - b√r, is also a zero of the polynomial.
1. There is a conclusion that states:

The sum of the zeros of the polynomial P(x) = anxn + an-1xn-1 + … + a1x + a0, with an≠0, is equal to –an-1∕an, and the product of the zeros is equal to a0∕an if n is even and - a0∕an if n is odd.

To understand this conclusion, study the function

P(x) = a3x3 + a2x2 + a1x + a0, with a3≠0, and suppose that x1, x2, and x3 are its three zeros.  Can you see that x1 + x2 + x3 = -a2∕a3, and x1x2x3 = -a0∕a3?

Solution:

If P(x) = a3x3 + a2x2 + a1x + a0 and a3≠0, its zeros are x1 , x2 , x3.

P(x) =x3 + (a2∕a3)x2 + ( a1∕a3)x + (a0∕a3) = 0.                                  (divide by a3)

x3 + (a2∕a3)x2 + ( a1∕a3)x + (a0∕a3) = (x - x1) (x - x2) (x - x3)

=x3 - (x1 + x2 + x3)x2 + (x1x2 + x2x3 + x3x1)x - x1x2x3

Therefore,  x1 + x2 + x3 = -a2∕a3  and  x1x2x3 = -a0∕a3.

Conclusion:

In the polynomial P(x) = anxn + an-1xn-1 + … + a1x + a0,with an≠0, the sum of the zeros is equal to –an-1∕an, and the product of the zeros is equal to a0∕an if n is even and - a0∕an if n is odd.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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