plzzz solve question 15
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zero of x-1 =1
p(x)=2x^2 + ax^2 + 3x -5
p(1) =2×1^2 +a×1^2 + 3×1 -5
=2+a+3-5
=5-5+a
=a
in second p(x)=x^3 + 2x^2 -5x -a
then p(1)= 1^3 +2×1^2-5×1-a
= 1+2-5-a
=3-5-a
=-2-a
according to condition both reminder are same. .
then,,,
a= -2-a (transform -a to LHS)
a+a= -2
2a= -2
a=-2/2
a=-1
hope this solution helps you....
p(x)=2x^2 + ax^2 + 3x -5
p(1) =2×1^2 +a×1^2 + 3×1 -5
=2+a+3-5
=5-5+a
=a
in second p(x)=x^3 + 2x^2 -5x -a
then p(1)= 1^3 +2×1^2-5×1-a
= 1+2-5-a
=3-5-a
=-2-a
according to condition both reminder are same. .
then,,,
a= -2-a (transform -a to LHS)
a+a= -2
2a= -2
a=-2/2
a=-1
hope this solution helps you....
eminent:
heyy
plzzz anyone solve question number 18 tooo
plzzzz I will mark it as brainliest answer
yes dude.....
i will send it wait .....
thanks
but now I don't need
I have done it
thank you for helping help
thats good ....
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