if (x²-1)is a factor of ax⁴+bx³+cx²+dx+e,show that a+c+e=b+d=0
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Answer:
Step-by-step explanation:
x² - 1 = 0
(x + 1)(x - 1) = 0
x = -1 or x = 1
p(x) = ax⁴ + bx³ + cx² + dx + e
As x + 1 = 0 is factor , p(x) = 0
a(-1)⁴ + b(-1)³ + c(-1)² + d(-1) + e = 0
a - b + c - d + e = 0
a + c + e = b + d + 0
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AjayT7805
13.07.2018
Math
Secondary School
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If (x²-1)is a factor of ax⁴+bx³+cx²+dx+e,show that at a+c+e=b+d=0.
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mysticd
mysticd Genius
Hi,
Let p(x) = ax^4+bx^3+cx^2+dx+e
It is given that x^2-1 is a factor of p(x).
x^2-1 = ( x + 1 )( x -1 )
Therefore , ( x + 1 ) , ( x - 1 ) are factors of
p( x ).
1 ) p( 1 ) = 0 [ By factor theorem ]
a + b + c + d + e = 0
rearranging the terms , we get
( a + c + d ) + ( b + e ) = 0 ----( 1 )
2 ) p( - 1 ) = 0
a - b + c - d + e = 0
a + c + e = b + d -----( 2 )
put ( 2 ) in equation ( 1 ) , we get
b + d + b + d = 0
2( b + d ) = 0
b + d = 0---( 3 )
put ( 3 ) in equation ( 1 ) , we get
a + c + e = b + d = 0
Hence proved.
I hope this helps you.