Math, asked by gayathrigayihoney, 8 months ago

if (x²-1)is a factor of ax⁴+bx³+cx²+dx+e,show that a+c+e=b+d=0​

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Answered by krishpambhar777
5

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

Hope this helps . Please mark as brainliest.

Answered by rajvivek2580
5

Answer:

Brainly.in

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AjayT7805

13.07.2018

Math

Secondary School

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Answered

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.

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