Question

1. Find the remainder when x + 3x + 3x + 1 is divided by the following
Lincar polynomials
1.x+1
2.x
3.x+π
4.5+2x
​

Answer 1

Answer:

Question 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

Apply remainder theorem

=>x + 1 =0

=> x = - 1

Replace x by – 1 we get

=>x3+3x2 + 3x + 1

=>(-1)3 + 3(-1)2 + 3(-1) + 1

=> -1 + 3 - 3 + 1

=> 0

Remainder is 0

(ii) x –1/2

Apply remainder theorem

=>x – 1/2 =0

=> x = 1/2

Replace x by 1/2 we get

=>x3+3x2 + 3x + 1

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

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

Add the fraction taking LCM of denominator we get

=>(1 + 6 + 12 + 8)/8

=>27/8

Remainder is 27/8

(iii) x

Apply remainder theorem

=>x =0

Replace x by 0 we get

=>x3+3x2 + 3x + 1

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

=> 0+0 +0 + 1

=> 1

Remainder is 1

(iv) x + π

Apply remainder theorem

=>x + π =0

=> x = - π

Replace x by – π we get

=>x3+3x2 + 3x + 1

=>(- π)3 + 3(-π)2 + 3(-π) + 1

=> - π3 + 3π2 - 3π + 1

Remainder is - π3 + 3π2 - 3π + 1

(v) 5 + 2x

Apply remainder theorem

=>5+2x =0

=> 2x = - 5

=> x = - 5/2

Replace x by – 5/2 we get

=>x3+3x2 + 3x + 1

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

=> -125/8 + 75/4 – 15/2 + 1

Add the fraction taking LCM of denominator

=>(-125 + 150 - 60 + 8 )/125

=> -27/8

Remainder is -27/8