X cube +3x square +3x+1/x+1
Answers
Answer:
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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
x+1=0
x=-1
by substituting it
f(x)=x^3+3x^2+3x+1
f(-1)=(-1)^3+3(-1)^2 +3(-1)+1
f(-1)=-1+3-3+1
f(-1)=0