Can anyone solve 1 iv
Please help me
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Answered by
105
=>x+1=0
=>x=-1
Sub,x=-1 in
i)p(x)= x^3+x^2+x+1
=>p(-1)=(-1)^3+(-1)^2+(-1)+1
=>-1+1-1+1
=>0
ii)p(x)= x^4+x^3+x^2+x+1
=>p(-1)=(-1)^4+(-1)^3+(-1)^2+1
=>1-1+1+1
=>3-1
=>2
iii) p(x)=x^4+3x^3+3x^2+x+1
=>p(-1)=(-1)^4+3(-1)^3+3(-1)^2+(-1)+1
=>1-3+3-1+1
=>1
iv)p(x)=x^3-x^2-(2+sqrt2)x+sqrt2
=>p(-1)=(-1)^3-(-1)^2-(2+sqrt2)(-1)+sqrt2
=>-1-1+2-sqrt2+sqrt2
=>-2+2-sqrt2+sqrt2
=>0
So, equation (i) and (iv) are the factors of (x+1)
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Answered by
2
Step-by-step explanation:
1)x^3+x^2+x+1=(-1)^3+1^2-1+1
=-1+1-1+1
=0
2)x^4+x^3+x^2+x+1
=(-1)^4+(-1)^3+(-1)^2-1+1
=1-1+1-1+1
=-2+3
=1
3)x^4+3x^2+3x^2+x+1
=1-3+3-1+1
=5-4
=1
4)x^3-x^2-(2+√2)x+√2
=-1-1+2-√2+√2
=-2+2-√2+√2
=0
therefore eq. 1 and 4 is factor of (x+1)
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