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.
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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:_______ _____ _____
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(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
- 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.
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Let us call a - b√r the conjugate radical of a + b√r, where a, b and r are rational numbers and √r is irrational. And let a + b√r denote a - b√r, that is, a + b√r = a - b√r.
Prove that:____________ ________ ________
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(a + b√r) + (c + d√r) = a + b√r + c + d√r
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(a + b√r) - (c + d√r) = a + b√r – c + d√r
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(a + b√r) (c + d√r) = (a + b√r) (c + d√r)
-
(a + b√r)3 = (a + b√r)3
_ _
If a is a rational number, a = ? 0 = ?
Solution:
_________________ ______ _______
(1) (a + b√r) + (c + d√r) = a + b√r + c + d√r
(a + c) + (b√r + d√r ) = (a - b√r) + (c - d√r)
(a + c) - (b√r + d√r ) = (a + c ) - (b √r + d√r)
__________________ _______ _______
(2) (a + b√r) – (c + d√r) = a + b√r – c + d√r
a + b√r – c - d√r = (a - b√r) – (c - d√r)
(a – c) + (b – d) √r = (a – c) – (b√r - d√r)
(a – c) – (b – d) √r = (a – c) – (b – d) √r
________________ _______ ________
(3) (a + b√r) (c + d√r) = (a + b√r) (c + d√r)
ac + ad√r + bc√r + bdr__ = (a - b√r) (c - d√r)
(ac + bdr) + (ad√r + bc√r) = ac - ad√r - bc√r + bdr
(ac + bdr) - (ad√r + bc√r) = ac + bdr - ad√r - bc√r
ac - ad√r - bc√r + bdr = ac - ad√r - bc√r + bdr
_________ _______
(4) (a + b√r)3 = (a + b√r)3
(a + b√r) (a2 + 2ab√r + b2r) = (a - b√r)3
a3 + 3a2b√r + 3ab2r + b3r√r = (a - b√r) (a2 - 2ab√r + b2r)
(a3 + 3ab2r) + (3a2b√r + b3r√r) = a3 - 3a2b√r + 3ab2r - b3r√r
(a3 + 3ab2r) - (3a2b√r + b3r√r) = (a3 + 3ab2r) - (3a2b√r + b3r√r)
_
Because a is real and a = a, 0 is real
Therefore, 0 = 0
- There is a conclusion that states:
If a + b√r is a zero of P(x) = a3x3 + a2x2 + a1x + a0, where a, b, r, a3, a2, a1, and a0 are rational numbers, but √r is irrational, then its conjugate radical, a - b√r 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 + b√r) = a3 (a + b√r)3 + a2 (a + b√r)2 + a1 (a +b√r) + a0 = 0
= a3 (a + b√r)3 + a2 (a + b√r)2 + a1 (a +b√r) + a0 = 0
___________ _______
(a + b√r)3 = (a + b√r)3
= a3 (a - b√r)3 + a2 (a - b√r)2 + a1 (a - b√r) + a0 = 0
_______
a + b√r = a - b√r
Therefore, a - b√r is a zero of P(x).
Conclusion:
If a + b√r 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 √r is irrational, then the conjugate radical, a - b√r, is also a zero of the polynomial.
- 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.
