Question

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

Answer 1

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.

: )

Answer 2

Hi

solution:
if
${x}^{2} - 1$
is a factor of given polynomial,then
${x}^{2} = 1 \\ x = + - \sqrt{1} \\ x = - 1 \\ x = + 1$
satisfies the polynomial.
1) put x= 1 into the polynomial
$p(x) = a {x}^{4} + b {x}^{3} + c {x}^{2} + dx + e \\ p(1) = 0 \\ a( {1)}^{4} + b( {1)}^{3} + c( {1)}^{2} + d(1) + e = 0 \\ a + b + c + d + e = 0$
2) put x = -1
$p( - 1) = 0 \\ a( { - 1)}^{4} + b( { - 1)}^{3} + c( { - 1)}^{2} + d( - 1) + e = 0 \\ a - b + c - d + e = 0 \\ a + c + e = b + d$
hope it helps you.