Math, asked by Laur5318, 8 months ago

If x²-1 is a factor of ax4+bx³ +cx²+dx+e, then
A. a + c + e = b + d
B. a + b + e = c + d
C. a + b + c = d + e
D. b + c + d = a + e

Answers

Answered by nikitasingh79
5

Given: x² - 1 is a factor of ax⁴ + bx³ + cx² + dx + e,

 

Solution :  

Let f (x) = ax⁴ + bx³ + cx² + dx + e be the given polynomial. Then x² - 1 is a factor of f(x)

⇒ (x - 1) (x + 1) is a factor of f(x)

⇒ (x - 1) and (x + 1) is a factor of f(x)

⇒ f(1) = 0 and f(-1) = 0  

 

On Substituting x = 1 in f (x) :  

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

⇒ a × 1 + b × 1 + c × 1 +  d + e = 0

⇒ a + b + c +  d + e = 0 …………(1)

 

 

On Substituting x = - 1 in f (x) :  

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

⇒ a × 1 + b × -1 + c × 1 - d + e = 0

⇒ a - b + c - d + e = 0 …………(2)

Adding and subtracting these two equations, we get :  

Adding :  

a + b + c +  d + e = 0

a - b + c - d + e = 0

------------------------------

2a + 2c + 2e = 0

-------------------------------

2(a + c + e) = 0 …………(3)

 

Subtracting :  

a + b + c +  d + e = 0

a - b + c - d + e = 0

(-) (+) (-) (+) (-)  

------------------------------

2b + 2d = 0

--------------------------

2(b + d) = 0………….(4)

From eq 3 & 4 :  

⇒ a + c + e = b + d  

Among the given options option (A) a + c + e = b + d is correct.

HOPE THIS ANSWER WILL HELP YOU…..

 

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Answered by Anonymous
2

Answer:

Step-by-step explanation:

On Substituting x = 1 in f (x) :  

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

⇒ a × 1 + b × 1 + c × 1 +  d + e = 0

⇒ a + b + c +  d + e = 0 …………(1)

 

 

On Substituting x = - 1 in f (x) :  

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

⇒ a × 1 + b × -1 + c × 1 - d + e = 0

⇒ a - b + c - d + e = 0 …………(2)

Adding and subtracting these two equations, we get :  

Adding :  

a + b + c +  d + e = 0

a - b + c - d + e = 0

------------------------------

2a + 2c + 2e = 0

-------------------------------

2(a + c + e) = 0 …………(3)

 

Subtracting :  

a + b + c +  d + e = 0

a - b + c - d + e = 0

(-) (+) (-) (+) (-)  

------------------------------

2b + 2d = 0

--------------------------

2(b + d) = 0………….(4)

From eq 3 & 4 :  

⇒ a + c + e = b + d  

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