IB Mathematics HL
Portfolio
(Type I)
Zeros of Polynomials
By Diana Herwono
IB Candidate No: D 0861 006
May 2003
.
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
2. There is a conclusion that states:
If an integer k is a zero of a polynomial with integral coefficients, then k must be a factor of the constant term of the polynomial.
To understand this conclusion, study the function P(x) = a3x3 + a2x2 + a1x + a0
and suppose that P(k) = 0. Can you see that k must be a factor of a0?
Solution:
P(x) = a3x3 + a2x2 + a1x + a0
P(k) = a3k3 + a2k2 + a1k + a0
Suppose P(k) = 0 then
a3k3 + a2k2 + a1k + a0 = 0
a0 = -(a3k3 + a2k2 + a1k)
= k (-a3k2 - a2k - a1)
Therefore, k must be a factor of a0.
3. There is a conclusion that states:
If a0 = 0, then P(x) = anxn + an-1xn-1 + ... + a1x + a0 has integral coefficients, and the rational number k/m in lowest terms is a zero of P(x), then k must be a factor of a0, and m must be a factor of an.
To understand this conclusion, study the function P(x)= a3x3+a2x2+a1x+a0, and suppose that P(k/m) = 0. ...
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a0 = -(a3k3 + a2k2 + a1k)
= k (-a3k2 - a2k - a1)
Therefore, k must be a factor of a0.
3. There is a conclusion that states:
If a0 = 0, then P(x) = anxn + an-1xn-1 + ... + a1x + a0 has integral coefficients, and the rational number k/m in lowest terms is a zero of P(x), then k must be a factor of a0, and m must be a factor of an.
To understand this conclusion, study the function P(x)= a3x3+a2x2+a1x+a0, and suppose that P(k/m) = 0. Can you see that k must be a factor of a0, and m must be a factor of a3?
Solution:
P(x) = a3x3 + a2x2 + a1x + a0
P(k/m) = a3(k/m)3 + a2(k/m)2 + a1(k/m) + a0
Since P(k/m) = 0
P(k/m) = a3(k/m)3 + a2(k/m)2 + a1(k/m) + a0 = 0
= a3k3/m3 + a2k2/m2 + a1k/m + a0 = 0
a0 = - (a3k3/m3 + a2k2/m2 + a1k/m)
= k (a3k2/m3 - a2k/m2 - a1/m)
Therefore, k is a factor of a0.
P(k/m) = a3k3/m3 + a2k2/m2 + a1k/m + a0 = 0
a3 = - (a2k2/m2 + a1k/m + a0)/ (k3/m3)
= m (-a2k2/m3 - a1k/m2 - a0/m) / (k3/m3)
Therefore, m is a factor of a3.
4. 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) + (3a2bi + b3i) = a3 - 3a2bi - 3ab2 + b3i
a3 - 3ab2 - 3a2bi + b3i = a3 - 3ab2 - 3a2bi + b3i
_ _____
Therefore, if a is a real number, a = a + 0i = a - 0i = a
_
Because a is real and a = a, 0 is real
Therefore, 0 = 0
5. There is a conclusion that states:
If a + bi is a zero of P(x)= a3x3 + a2x2 + a1x + a0, where a, b, a3 , a2 , a1, and a0 are real numbers, then its conjugate, a - bi, is also a zero of P(x).
Use the results found in question 4 to prove this conclusion. And then, state the general conclusion if P(x)= anxn + an-1xn-1 + ... + a1x + a0.
Solution:
_
From the results in question 4, we know a = a.
So P(x)= a3x3 + a2x2 + a1x + a0 _ = 0
= a3x3 + a2x2 + a1x + a0 = 0 = 0
________________________________
P(a + bi) = a3(a + bi)3 + a2(a + bi)2 + a1(a + bi) + a0__ = 0 _______ ____
= a3 (a + bi)3 + a2 (a + bi)2 + a1 (a + bi) + a0 = 0 (a + bi)3 = (a + bi)3
= a3 (a - bi)3 + a2 (a - bi)2 + a1 (a - bi) + a0 = 0 (a + bi) = a - bi
Therefore a - bi is a zero of P(x)
Conclusion:
If a complex number, a + bi, is a zero of a polynomial function
P(x)= anxn + an-1xn-1 + ... + a1x + a0 , with real coefficients, then its conjugate,
a - bi, is also a zero of the polynomial.
6. Let us call a - bVr the conjugate radical of a + bVr, where a, b and r are rational numbers and Vr is irrational. And let a + bVr denote a - bVr, that is, a + bVr = a - bVr.
Prove that:____________ ________ ________
(1) (a + bVr) + (c + dVr) = a + bVr + c + dVr
(2) (a + bVr) - (c + dVr) = a + bVr - c + dVr
(3) (a + bVr) (c + dVr) = (a + bVr) (c + dVr)
(4) (a + bVr)3 = (a + bVr)3
_ _
If a is a rational number, a = ? 0 = ?
Solution:
_________________ ______ _______
(1) (a + bVr) + (c + dVr) = a + bVr + c + dVr
(a + c) + (bVr + dVr ) = (a - bVr) + (c - dVr)
(a + c) - (bVr + dVr ) = (a + c ) - (b Vr + dVr)
__________________ _______ _______
(2) (a + bVr) - (c + dVr) = a + bVr - c + dVr
a + bVr - c - dVr = (a - bVr) - (c - dVr)
(a - c) + (b - d) Vr = (a - c) - (bVr - dVr)
(a - c) - (b - d) Vr = (a - c) - (b - d) Vr
________________ _______ ________
(3) (a + bVr) (c + dVr) = (a + bVr) (c + dVr)
ac + adVr + bcVr + bdr__ = (a - bVr) (c - dVr)
(ac + bdr) + (adVr + bcVr) = ac - adVr - bcVr + bdr
(ac + bdr) - (adVr + bcVr) = ac + bdr - adVr - bcVr
ac - adVr - bcVr + bdr = ac - adVr - bcVr + bdr
_________ _______
(4) (a + bVr)3 = (a + bVr)3
(a + bVr) (a2 + 2abVr + b2r) = (a - bVr)3
a3 + 3a2bVr + 3ab2r + b3rVr = (a - bVr) (a2 - 2abVr + b2r)
(a3 + 3ab2r) + (3a2bVr + b3rVr) = a3 - 3a2bVr + 3ab2r - b3rVr
(a3 + 3ab2r) - (3a2bVr + b3rVr) = (a3 + 3ab2r) - (3a2bVr + b3rVr)
_
Because a is real and a = a, 0 is real
Therefore, 0 = 0
7. There is a conclusion that states:
If a + bVr is a zero of P(x) = a3x3 + a2x2 + a1x + a0, where a, b, r, a3, a2, a1, and a0 are rational numbers, but Vr is irrational, then its conjugate radical, a - bVr is also a zero of P(x).
Use the results found in question 6 and prove this conclusion.
And then, state what general conclusion should we deal with if
P(x) = anxn + an-1xn-1 + ... + a1x + a0.
Solution:
_
From the results in question 6, we know a = a
Therefore
P(x) = a3x3 + a2x2 + a1x + a0 = 0
= a3x3 + a2x2 + a1x + a0 = 0 =0
_______________________________________
P(a + bVr) = a3 (a + bVr)3 + a2 (a + bVr)2 + a1 (a +bVr) + a0 = 0
= a3 (a + bVr)3 + a2 (a + bVr)2 + a1 (a +bVr) + a0 = 0
___________ _______
(a + bVr)3 = (a + bVr)3
= a3 (a - bVr)3 + a2 (a - bVr)2 + a1 (a - bVr) + a0 = 0
_______
a + bVr = a - bVr
Therefore, a - bVr is a zero of P(x).
Conclusion:
If a + bVr is a zero of a polynomial function P(x) = anxn + an-1xn-1 + ... + a1x + a0, with rational coefficients, where a and b are rational, but Vr is irrational, then the conjugate radical, a - bVr, is also a zero of the polynomial.
8. 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.
Diana Herwono D 0861 006
