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
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.
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A. 0
B. 2
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When x³-2x²+ax=b is divided by x²-2x-3, the remainder is x-6. The value of a and b respectively
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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