Polynomial – Exercise 2.3 – Class IX

  1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by

(i) x + 1

(ii) x – 1/2

(iii) x

(iv) x + π

(v) 5 + 2x

Solution:

(i) x + 1

p(x) = x3 + 3x2 + 3x + 1 and the zero of x + 1 is – 1

So, p(1) = (-1)3 + 3(-1)2 + 3(-1) + 1

= -1 + 3 – 3 + 1

= 0

So, by the remainder theorem, remainder obtained on dividing p(x) by x + 1 is 0.

So, p(x)  is the multiple of x + 1

 

(ii) x – 1/2

p(x) = x3 + 3x2 + 3x + 1 and zero of x – 1/2 is 1/2

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

= 1/8  + 3*1/4 +  3*1/2 + 1

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

= 1+6+12+8/8 = 27/8

So, by the remainder theorem, remainder obtained on dividing p(x) by x – 1/2 is 27/8.

 

(iii) x

p(x) = x3 + 3x2 + 3x + 1 and zero of x  is 0

p(0) = (0)3 + 3(0)2 + 3(0) + 1

= 0  + 3*0 +  3*0 + 1

= 1

So, by the remainder theorem, remainder obtained on dividing p(x) by x  is 1.

 

(iv) x + π

p(x) = x3 + 3x2 + 3x + 1 and zero of x + π is -π

p(-π) = (-π)3 + 3(-π)2 + 3(-π) + 1

= -π3  + 3π2 –  3π + 1

So, by the remainder theorem, remainder obtained on dividing p(x) by x + π  is = -π3  + 3π2 –  3π + 1.

 

(v) 5 + 2x

p(x) = x3 + 3x2 + 3x + 1 and zero of 5 + 2x  is –5/2

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

= –125/8 +3*25/4 – 3*5/2 + 1

= -125+150-60+8/8

= ‑27/8

So, by the remainder theorem, remainder obtained on dividing p(x) by 5 + 2x  is –27/8.


  1. Find the remainder when x3 – ax2 + 6x – a is divided by x – a

Solution:

p(x) = x3 – ax2  + 6x – a and zero of x –  a is a

p(1) = a3 – a(a)2 + 6(a) – a

= a3 – a3 + 6a – a

= 5a

So, by the remainder theorem, remainder obtained on dividing p(x) by x – a  is 5a.


  1. Check whether 7 + 3x is a factor of 3x3 + 7x

Solution:

7 + 3x = 0

7 = – 3x

x = –7/3

Zero of 7 + 3x is –7/3

p(-7/3) =  3x3 + 7x

= 3(-7/3)3 + 7(-7/3)

= 3(-343/27) – 7(7/3)

= – 343/949/3

= –490/9

So, remainder = – 490/9 which is different from 0.

Therefore, (3x + 7) is not a factor of the polynomial 3x3 + 7x


 

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