- Level: International Baccalaureate
- Subject: Maths
- Word count: 1814
Math IB HL math portfolio type I - polynomials
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Introduction
Diana Herwono D 0861 006
- 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
- There is a conclusion that states:
Middle
Therefore, m is a factor of a3.
- 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:____________ _____
- (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
- 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.
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