a*****g
发帖数: 1320 | 1 Determine the number of positive, negative, and complex roots of the
polynomial. Assume n is a natural number. P(x)=3x^（2n+1）+4x^（2n+2）-5x^（
2n+3）-7x
--Explain why every polynomial of odd degree with real coefficients must
have at least one real root ? Another possible answer is that | a*****g
发帖数: 1320 | 2 甜甜是个优秀的好姑娘, 百忙之中给我回站内信了, 特此感谢.
1. complex conjugate roots are always pairs, so for odd degree of
polynomials, it must have at least 1 real root.
2. P(x)=3x^（2n+1）+4x^（2n+2）-5x^（2n+3）-7x
let p(x)=0, -5x^（2n+3）+4x^（2n+2）+3x^（2n+1）- 7x =0
x=0, so only 2n+2 roots to discuss. we only concern
q(x)= -5x^（2n+2）+4x^（2n+1）+3x^（2n）- 7 =0
n=1, there are 2 signs changes, so either 2 positive zeros or 0.
take q(-x)= -5x^（2n+2）- 4x^（2n+1）+3x^（2n）- 7=0
there are 2 signs changes, so either 2 negative zeros or 0.
no matter how much n is, this does not change. so there must be at
least (2n-2) complex roots.
let n=1, q(x)=-5x^4+4x^3+3x^（2）-7=0
your child can solve this one, to get how many positive roots and
negatives roots, then you can add or deduct from 2n+2. that is final
solution.
【在 a*****g 的大作中提到】
: Determine the number of positive, negative, and complex roots of the
: polynomial. Assume n is a natural number. P(x)=3x^（2n+1）+4x^（2n+2）-5x^（
: 2n+3）-7x
: --Explain why every polynomial of odd degree with real coefficients must
: have at least one real root ? Another possible answer is that
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