Find without actual division,the remainder if ................................. ?Answer pls
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hey guy your answer is -1
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Roja45:
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Sol : Let p ( x ) = 4x^5 - 7x^3 - x^3 + 8,
and f ( x ) = 2x + 3.
Zero of f ( x ) , 2x + 3 = 0
2x = -3
So, x = -3 / 2.
By Remainder theorem ,
p ( x ) = 4 x^5 - 7x^3 - x^2 + 8
p ( -3/2 ) = 4 ( -3 / 2 )^5 - 7 ( -3 / 2 )^3 - ( -3 / 2 )^2 + 8
p ( -3 / 2 ) = 4 ( - 243 / 32 ) - 7 ( - 27 / 8 ) - ( 9 / 4 ) + 8
p ( -3/2 ) = ( -243 / 8 ) + ( 189 / 8 ) - ( 9 / 4 ) + 8
-243 + 189 - 18 + 64
p (-3/2) = ---------------------------------------
8
-261 + 253
p ( -3/2) = -----------------------
8
p ( -3/2) = -8 / 8 = -1.
So, when ( 4x^5 - 7x^3 - x^2 + 8 ) is divided by ( 2x + 3 ) , then remainder is ( -1 ).
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and f ( x ) = 2x + 3.
Zero of f ( x ) , 2x + 3 = 0
2x = -3
So, x = -3 / 2.
By Remainder theorem ,
p ( x ) = 4 x^5 - 7x^3 - x^2 + 8
p ( -3/2 ) = 4 ( -3 / 2 )^5 - 7 ( -3 / 2 )^3 - ( -3 / 2 )^2 + 8
p ( -3 / 2 ) = 4 ( - 243 / 32 ) - 7 ( - 27 / 8 ) - ( 9 / 4 ) + 8
p ( -3/2 ) = ( -243 / 8 ) + ( 189 / 8 ) - ( 9 / 4 ) + 8
-243 + 189 - 18 + 64
p (-3/2) = ---------------------------------------
8
-261 + 253
p ( -3/2) = -----------------------
8
p ( -3/2) = -8 / 8 = -1.
So, when ( 4x^5 - 7x^3 - x^2 + 8 ) is divided by ( 2x + 3 ) , then remainder is ( -1 ).
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