If(X^{2}-1) is a factor of ax^{4}+bx^{3}+cx^{2}+dx+e,show that a+c+e=b+d=0

**Given :** *x* ^{2} – 1 is a factor of *f*(*x*) = *ax* ^{4} + *bx* ^{3} + *cx* ^{2} + *dx* + *e*

Now (*x* ^{2} – 1) can be written as (*x* + 1) (*x* – 1)

⇒ (*x* – 1) and (*x* + 1) are factors of *f*(*x*)

⇒ *f*(–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 *..... (1)

and *f *(1) = 0

⇒ *a*(1)^{4} + *b*(1)^{3} + *c*(1)^{2} + *d*(1) + *e *= 0

⇒ *a *+ *b* + *c *+ *d *+ *e *= 0

⇒ *b + d + b + d *= 0* *( from (1) )

⇒ 2 (*b + d*) = 0

⇒ *b + d *= 0 ..... (2)

from (1) and (2) we get

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

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