Question

if(x^2-1) is a factor of ax^4+bx^3+cx^2+dx+e.prove that a+c+e=b+d=0​

Answer 1

Step-by-step explanation:

We are given,

(x² - 1) if a factor of ax⁴ + bx³ + cx² + dx + e

Thus, if it is a factor

x² - 1 = 0

x² = 1

x = √1

x = +1 or -1

Let's take x = -1

Let p(x) = ax⁴ + bx³ + cx² + dx + e

We know that,

If x = -1 is a solution of p(x), then p(-1) = 0

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

Now, all the powers with even numbers will be positive and odd numbered power will be negative.

For ex:- (-1)⁴ = (-1) × (-1) × (-1) × (-1)

= 1 × (-1) × (-1)

= (-1) × (-1)

= 1

(-1)³ = (-1) × (-1) × (-1)

= 1 × (-1)

= (-1)

Similarly,

0 = a - b + c - d + e

Taking all the negative terms to the Left we get,

a + c + e = b + d

Hence proved

Hope it helped and you understood it........All the best

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