Which of the following CANNOT be a root of a polynomial in x

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Which of the following CANNOT be a root of a polynomial in x of the form 9x^5+ax^3+b,where a and b are integers?
(a)-9 (b) -5 (c)1/4 (d)1/3 (e)9

把root代入原公式会得到0,所以如果
9(-9)^5+a(-9)^3+b
9(-5)^5+a(-5)^3+b
9(1/4)^5+a(1/4)^3+b
9(1/3)^5+a(1/3)^3+b
9(9)^5+a(9)^3+b
当中的一个式子结果不是0的话,那么这个数字就不是这个polynomial的root.
注意因为a和b是整数,所以-9,-5,9 都有可能是root.
那么就在1/4和1/3两个分数中选择了.
9(1/4)^5=9/1024,a(1/4)^3=a/64=16a/1024
9(1/3)^5=1/27,a(1/3)^3=a/27
如果a=-1的话,1/27+a/27=0,如果a=26的话,1/27+a/27=1
总之,9(1/3)^5+a(1/3)^3+b也可以得到整数,即1/3也有可能做root.
按排除法算,答案就是1/4,C了.
如果你想确认一下的话,可以再检查一下1/4.
9(1/4)^5=9/1024,a(1/4)^3=a/64=16a/1024
如果1/4是root的话,9(1/4)^5+a(1/4)^3=9/1024+16a/1024必须等于一个整数.
也就是说,(9+16a)/1024=整数
1024的倍数肯定是偶数,16a (a是整数)也一定是偶数.
但是16a+9是奇数,所以不会有a可以让(9+16a)/1024=整数这个条件成立.
所以,1/4不可能是root.

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