余定理–多项式|第9类数学

多项式是由变量和系数组成的代数表达式。变量有时也称为不确定。我们可以使用多项式执行任何运算，无论是乘法，除法，减法还是加法。具有一个变量的多项式示例为x ² + x – 8，y ³ + y ² – 52，z ² +64

多项式一词源自希腊语“ poly ”(意为“许多”)和“ nominal ”(意为“项”)，因此统称为“许多项”。多项式不能具有无限项。

剩余定理

令g(x)为1或大于1的多项式，令b为任何实数。如果将g(x)除以线性多项式x – b，则余数为p(b) 。

证明

令g(x)为1或大于1的多项式。假设g(x)除以(x – b)，则商为q(x)，余数为r(x)，即，

g(x) = (x – b) q(x) + r(x)    ……(1)

由于x – b的次数为1，而r(x)的次数小于x – b的次数，因此r(x)的次数为0。这意味着r(x)是恒定的。因此，我们可以将等式(1)写为

 g(x) = (x – b) q(x) + r     ……(2)

特别地，如果x = b，则方程式(2)变为

p(b) = (b – b) q(b) + r  =>  r

剩余定理的样本问题

问题1：当g(x)= x ⁴ – x ³ + x ² – 2x + 1除以x – 2时，求出余数。

解决方案：

Zero of x – 2 is 2, so as per the remainder theorem 

Remainder, in this case, will be g(2). 

So, 

g(2) = (2)⁴ – (2)³ + (2)² + 2(2) + 1 = 17

问题2：找到多项式x ² – 5x + 4的根

解决方案：

The approach of solving such questions include to choose a number in such a manner, that putting it to use, will yield a zero remainder.

f(x) = x² – 5x + 4 

f(4) = 4² – 5(4) + 4 

f(4) = 20 – 20 = 0

So, (x – 4) must be a factor of x² – 5x + 4 

问题3：将t ³ – 2t ² + 4t + 5除以t – 1时，求出余数。

解决方案：

Since, here it is already given that we need to find the remainder when the given quotient is divided by t – 1. So, accordingly we will put 1 in place of x, to solve and get the remainder.

Here, p(t) = t³ – 2t² + 4t + 5, and the zero of t – 1 is 1

∴ g(1) = (1)³ – 2(1)² + 4 + 5 = 8

By the Remainder Theorem, 8 is the remainder when t³ – 2t² + 4t + 5 is divided by t – 1

问题4：找出x ³时的余数– x ² + 2除以x – 2。

解决方案：

Here, we will put x = 2 in the given quotient to find the remainder

 2^3 – 2^2 + 2

= 8 – 4 + 2

= 6

By the Remainder Theorem, 6 is the remainder when x³ – x² + 2 is divided by x – 2.

问题5：通过x ³ – x ² – 4的除法运算得到0作为余数。

解决方案：

We will use Hit & Trial method to find the answer,

Clearly putting x = 2, will give zero as remainder.

So, this will be the answer to yield a zero remainder on division by x – 2.