今天智课小编为大家带来的就是SAT数学考点实例讲解 多项式函数的详细内容,在sat考试的数学部分,考察的知识点是比较多的,小伙伴们在备考数学的时候还是喜欢比较用例题来进行讲解!赶紧来看看吧!

  SAT数学多项式函数考点实例

  A polynomial function P has zeros -3,3/2,and 8. Which of the following polynomial functions could define P?

  多项式函数P存在 -3,3/2,8三个零点,则P的多项式函数是以下哪一个?

  A. P(x)=-3(x-3/2)(x-8) B.P(x)=-(x-3)(x+3/2)(x+8) C. P(x)=(x+3)(3x-2)(x-8) D. P(x)=(x+3)(2x-3)(x-8)

  答案:D

  解析:Recall that if K is a zero of a polynomial function defined as y=f(x), then x-k is a factor of f.

  要牢记,如果K是函数y=f(x)的零点,则x-k是函数f的一个因式

  Since the polynomial function P has the zeros -3,3/2,and 8,it follows that (x-(-3)),(x-3/2),and (x-8) must be factors of P.

  既然该多项式函数P有-3,3/2,8三个零点,则可得到(x-(-3)),(x-3/2), (x-8)都是P的因式。

  Therefore, we can define P as P(x)=a(x+3)(x-3/2)(x-8), where a is a nonzero constant.

  所以,我们可以定义函数P为P(x)=a(x+3)(x-3/2)(x-8), 其中a是非零常量。

  A constant factor, such as a, does not affect the zeros of the polynomial function. In order to rewrite the equation with integral coeffecients, let a=2.

  最后一步,用整数系数改写一下该方程。

  If a=2, it follows that

  P(x)=a(x+3)(x-3/2)(x-8)

  =2(x+3)(x-3/2)(x-8)

  =(x+3)(2x-3)(x-8).

  so the polynomial that could define P is P(x)=(x+3)(2x-3)(x-8).

  得到最终的可能结果之一为P(x)=(x+3)(2x-3)(x-8).

  以上就是智课小编为大家带来的SAT数学考点实例讲解 多项式函数的相关内容,希望对大家的备考有所帮助,更多精彩内容敬请关注智课网!


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