Question

If (x^2-1) is a factor of ax^4+bx^3+cx^2+dx+e, show that a+c+e=b+d=0.give me step by step explaination.
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Answer 1

Answer:

a+c+e = b+d = 0

Step-by-step explanation:

x²-1 is a factor

So, x²-1 = (x-1)(x+1)

So, x-1 , x+1 are two factors of a given biquadratic polynomial f(X) = ax⁴+bx³+cx²+dx+e

Now, We can write ,

f(x) = f(1) = f(-1) = 0

f(1) = 0

(substituting x=1 in th biquadratic polynomial f(x))

a(1)⁴+b(1)³+c(1)²+d(1)+e=0

a+b+c+d+e = 0 ------> (1)

f(-1) = 0

(substituting x= -1 in the biquadratic polynomial f(x))

a(-1)⁴+b(-1)+c(-1)²+d(-1)+e = 0

a-b+c-d+e = 0 ------> (2)

From equations (1)&(2),

Subtracting them,

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

2(b+d) = 0

b+d = 0 ------>(3)

Adding them,

ing them,(a+b+c+d+e)+(a-b+c-d+e) = 0

(a-b+c-d+e) = 02(a+c+e) = 0

a+c+e = 0 -------> (4)

Hence from equations (3) & (4),

we proved,

(a+c+e) = (b+d) = 0