1. Find the remainder when xº + 3×2 + 3x + 1 is divided by1(i) x + 1(ii) x(ii) x2(iv) x + TT About the author Arya
Answer: (i) x + 1 Apply remainder theorem =>x + 1 =0 => x = – 1 Replace x by – 1 we get =>x3+3×2 + 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+3×2 + 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+3×2 + 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+3×2 + 3x + 1 =>(- π)3 + 3(-π)2 + 3(-π) + 1 => – π3 + 3π2 – 3π + 1 Remainder is – π3 + 3π2 – 3π + 1 Reply
Answer: (i) x + 1 Apply remainder theorem =>x + 1 =0 => x = – 1 Replace x by – 1 we get =>x3+3×2 + 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+3×2 + 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+3×2 + 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+3×2 + 3x + 1 =>(- π)3 + 3(-π)2 + 3(-π) + 1 => – π3 + 3π2 – 3π + 1 Remainder is – π3 + 3π2 – 3π + 1 Step-by-step explanation: I hope it will be helpful mark me as brain list Reply
Answer:
(i) x + 1
Apply remainder theorem
=>x + 1 =0
=> x = – 1
Replace x by – 1 we get
=>x3+3×2 + 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+3×2 + 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+3×2 + 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+3×2 + 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+3×2 + 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+3×2 + 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+3×2 + 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+3×2 + 3x + 1
=>(- π)3 + 3(-π)2 + 3(-π) + 1
=> – π3 + 3π2 – 3π + 1
Remainder is – π3 + 3π2 – 3π + 1
Step-by-step explanation:
I hope it will be helpful
mark me as brain list