Question

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

Answer 1

Since x2 - 1 = (x - 1) is a factor of

p(x) = ax4 + bx3  + cx2 + dx + e  

∴ p(x) is divisible by (x+1) and (x-1) separately  

⇒ p(1) = 0 and p(-1) = 0

p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0

⇒  a + b + c + d + e = 0 -----------(i)  

Similarly, p(-1) = a (-1)4 + b (-1)3  + c (-1)2 + d (-1) + e = 0

⇒ a - b + c - d + e = 0  

⇒ a + c + e = b + d -----------------(ii)

Putting the value of a + c + e in eqn , we get  

a + b + c + d + e = 0

⇒ a + c + e + b + d = 0

⇒ b + d + b + d = 0  

⇒ 2(b+d) = 0

⇒ b + d = 0 ---------------------------(iii

comparing equations (ii) and (iii) , we get....

 

a + c + e = b + d = 0

HOPE IT HELPS !

THANK YOU !

MARK ME BRAINLIEST ^_^