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ANS. Q 12,14,15&16
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hope it helps you........................
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hukam0685:
question no 15 have some problem,it can't be solved
you can just cross check by putting values of x from answer
ok
i will check
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p-1=0
p=1
p^10+p^8+p^6-p^4-p^2-1
=(1)^10+(1)^8+(1)^6-(1)^4-(1)^2-1
=1+1+1-1-1-1=0
hence p-1 is a factor of given polynomial
13.4a^2-9b^2-2a-3b
=using identity a^2-b^2=(a+b)(a-b)
=(2a)^2-(3b)^2-(2a+3b)
(2a+3b)(2a-3b) -(2a+3b)
=(2a+3b)[2a-3b-1]
14. given let p(x)=3x^2-mx-na
T.P. a=(m+n)/3
(x-a) is a factor of given equation
x-a=0
x=a
putting x=a in the equation
p(a)=3(a)^2-m×a-na=0
3a^2-a(m+n)=0
3a^2=a(m+n)
3a=(m+n)
a=(m+n)/3
ans.hence proved
p=1
p^10+p^8+p^6-p^4-p^2-1
=(1)^10+(1)^8+(1)^6-(1)^4-(1)^2-1
=1+1+1-1-1-1=0
hence p-1 is a factor of given polynomial
13.4a^2-9b^2-2a-3b
=using identity a^2-b^2=(a+b)(a-b)
=(2a)^2-(3b)^2-(2a+3b)
(2a+3b)(2a-3b) -(2a+3b)
=(2a+3b)[2a-3b-1]
14. given let p(x)=3x^2-mx-na
T.P. a=(m+n)/3
(x-a) is a factor of given equation
x-a=0
x=a
putting x=a in the equation
p(a)=3(a)^2-m×a-na=0
3a^2-a(m+n)=0
3a^2=a(m+n)
3a=(m+n)
a=(m+n)/3
ans.hence proved
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