sat2数学关于高次多项式Rational zero (root ) theorem 是什么意思呀 还有 if p+qi is a zero ,then p-qi is aslo a zero是为什么呀谢谢啦

sat2数学关于高次多项式Rational zero (root ) theorem 是什么意思呀 还有 if p+qi is a zero ,then p-qi is aslo a zero是为什么呀谢谢啦
sat2数学
关于高次多项式
Rational zero (root ) theorem 是什么意思呀

还有 if p+qi is a zero ,then p-qi is aslo a zero
是为什么呀
谢谢啦

sat2数学关于高次多项式Rational zero (root ) theorem 是什么意思呀 还有 if p+qi is a zero ,then p-qi is aslo a zero是为什么呀谢谢啦
（1）就是ab=0,则a=0或者b=0. a,b可以使式子,也可以是数.
（2）P+Qi=0,则Q=0,P=0.则P-Qi自然为0了.

Rational root theorem
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In algebra, the rational root theorem (or rational root test) states a constraint on rational soluti...

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Rational root theorem
From Wikipedia, the free encyclopedia
Jump to: navigation, search
In algebra, the rational root theorem (or rational root test) states a constraint on rational solutions (or roots) of the polynomial equation
a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\!
with integer coefficients.
If a0 and an are nonzero, then each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., the greatest common divisor of p and q is 1) satisfies
* p is an integer factor of the constant term a0, and
* q is an integer factor of the leading coefficient an.
Thus, a list of possible rational roots of the equation can be derived using the formula x = \pm \frac{p}{q}.
For example, every rational solution of the equation
3x^3 - 5x^2 + 5x - 2 = 0\,\!
must be among the numbers symbolically indicated by
± \tfrac{1,2}{1,3}\,,
which gives the list of possible answers:
1, -1, 2, -2, \frac{1}{3}, -\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}\,.
These root candidates can be tested, for example using the Horner scheme. In this particular case there is exactly one rational root. If a root candidate does not satisfy the equation, it can be used to shorten the list of remaining candidates. For example, x = 1 does not satisfy the equation as the left hand side equals 1. This means that substituting x = 1 + t yields a polynomial in t with constant term 1, while the coefficient of t3 remains the same as the coefficient of x3. Applying the rational root theorem thus yields the following possible roots for t:
t=\pm\tfrac{1}{1,3}
Therefore,
x = 1+t = 2, 0, \frac{4}{3}, \frac{2}{3}
Root candidates that do not occur on both lists are ruled out. The list of rational root candidates has thus shrunk to just x = 2 and x = 2/3.
If a root r1 is found, the Horner scheme will also yield a polynomial of degree n − 1 whose roots, together with r1, are exactly the roots of the original polynomial.
It may also be the case that none of the candidates is a solution; in this case the equation has no rational solution. The fundamental theorem of algebra states that any polynomial with integer (or real, or even complex) coefficients must have at least one root in the set of complex numbers. Since the roots of a polynomial with real coefficients occur in complex conjugate pairs, and a real number is self-conjugate, any polynomial of odd degree (the degree being n = 3 in the example above) with real coefficients must have a root in the real numbers.
If the equation lacks a constant term a0, then 0 is one of the rational roots of the equation.
The theorem is a special case (for a single linear factor) of Gauss's lemma on the factorization of polynomials.
The integral root theorem is a special case of the rational root theorem if the leading coefficient an = 1.
第二个问题你的qi是什么啊？

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晕 楼上的英语应该给人家翻译一下啊 呵呵 楼上可能一时没注意看吧 这是带有实部虚部的表达式
表达式 p+qi p代表实数 而qi代表复数 若 p+qi=0 ，即实部虚部都等于零
即 p q 都等于零 所以 p-qi is aslo a zero