Answer:

Since x

2

−1 is a factor of ax

4

+bx

3

+cx

2

+dx+e.

Then (x−1) and x+1 are also factors of ax

4

+bx

3

+cx

2

+dx+e.

Let, f(x)=ax

4

+bx

3

+cx

2

+dx+e

Since (x−1) is a factor of f(x).

Then f(1)=0. [Using Remainder theorem]

or, a+b+c+d+e=0........(1).

Again since (x+1) is a factor of f(x).

Then f(−1)=0. [Using Remainder theorem]

or, a−b+c−d+e=0........(2).

Now adding (1) and (2) we get,

2(a+c+e)=0

or, a+c+e=0

Using this from (2) we get,

b+c=0

So a+c+e=b+d=0

Step-by-step explanation:

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Step-by-step explanation:

Since x2−1 is a factor of ax4+bx3+cx2+dx+e.

Then (x−1) and x+1 are also factors of ax4+bx3+cx2+dx+e.

Let, f(x)=ax4+bx3+cx2+dx+e

Since (x−1) is a factor of f(x).

Then f(1)=0. [Using Remainder theorem]

or, a+b+c+d+e=0........(1).

Again since (x+1) is a factor of f(x).

Then f(−1)=0. [Using Remainder theorem]

or, a−b+c−d+e=0........(2).

Now adding (1) and (2) we get,

2(a+c+e)=0

or, a+c+e=0

Using this from (2) we get,

b+c=0

So a+c+e=b+d=0