9类NCERT解决方案–第2章多项式–练习2.3

问题1.将x ³ + 3x ² + 3x + 1除以除法求余数

(i)x + 1

Solution:

x + 1 = 0

x  = −1

Therefore remainder will be f(x):

f(−1) = (−1)³ + 3(−1)² + 3(−1) + 1

= −1 + 3 − 3 + 1

= 0

(ii)x – 1/2

Solution:

x – 1/2 = 0

x = 1/2

Therefore remainder will be f(x):

f(1/2) = (1/2)³ + 3(1/2)² + 3(1/2) + 1

= (1/8) + (3/4) + (3/2) + 1

= 27/8

(iii)x

Solution:

x = 0

Therefore remainder will be f(x):

f(0) = (0)³ + 3(0)² + 3(0) + 1

= 1

(iv)x +圆周率

Solution:

x + pi = 0

x = −pi

Therefore remainder will be f(x):

f(−pi) = (−pi)³ + 3(−pi)² + 3(−pi) + 1

= −pi³ + 3pi² − 3pi + 1

(v)5 + 2x

Solution:

5 + 2x = 0

2x = −5

 x = -5/2

Therefore remainder will be f(x) :

f(-5/2) = (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1 

= (-125/8) + (75/4) – (15/2) + 1

= -27/8

^{问题2.当x 3} − ax ² + 6x − a除以x – a时，求出余数。

Solution:

Let f(x) = x³ − ax² + 6x − a

x − a = 0

∴ x = a

Therefore remainder will be f(x):

f(a) = (a)³ − a(a²) + 6(a) − a

= a³ − a³ + 6a − a 

= 5a

问题3.检查7 + 3x是否是3x ³ + 7x的因数。

Solution:

7 + 3x = 0

3x = −7

 x = -7/3

Therefore remainder will be f(x):

f(-7/3) = 3(-7/3)³ + 7(-7/3) 

= – (343/9) + (-49/3)

= (-343- (49) * 3)/9

= (-343 – 147)/9

= – 490/9 ≠ 0

∴ 7 + 3x is not a factor of 3x³ + 7x