Question

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

Answer 1

Answer:

if (x² - 1) is a factor , x = 1 or x = -1

Substitute x = -1 in eq.

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...

Answer 2

Given:

x² -1 is a factor of P(x) = ax⁴ +bx³ + cx²+dx +e

if x² -1 is a factor of P(x) then P(±1) = 0

x² -1 = 0

or, x² = 1

or, x = ± 1

substituting value of x in P(X)

P(x) = ax⁴ +bx³ + cx²+dx +e

or, P(1) : a(1)⁴ + b(1)³ + c (1)²+dx + e = 0

or, P(1) : a + b +c+d+e = 0 ------- equ(1)

Simlarly,

P(-1) : a(-1)⁴ + b(-1)³ + c (-1)²+d(-1) + e = 0

or, P(-1): a -b +c-d + e = 0

or, P(-1): a + c+ e = b +d ------- equ(2)

From equ (1) & (2) , We get

a + c +e = b + d = 0