1. Find the remainder when xº + 3x2 + 3x + 1 is divided by
1
(i) x + 1
(ii) x
(ii) x
2
(iv) x + TT
Answers
Answer:
(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
Answer:
(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
Step-by-step explanation:
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