Let f(x) = x6 – x5 – x3 – x2 - x and g(x) = x4 – x3 - 2-1..
If a, b, c, d are the four zeroes of the polynomialg ), then
find the value of f(a) + f(b) + f(c) + f(d).
Alor:
Answers
Given : f(x) = x⁶ - x⁵ – x³ – x² – x
g(x) = x⁴ - x³ - x² - 1
a, b, c, d are the four zeroes of the polynomial g(x)
To Find : the value of f(a) + f(b) + f(c) + f(d).
Solution
g(x) = x⁴ - x³ - x² - 1
x⁴ - x³ - x² - 1 = 0
=> x⁴ - x³ - 1 = x²
f(x) = x⁶ - x⁵ – x³ – x² – x
= x (x⁵ – x⁴ – x² – x - 1)
= x ( x ( x⁴ – x³ – x - 1 ) - 1 )
= x ( x (x² - x) - 1)
= x( x³ - x² - 1)
= x⁴ - x³ - x
( x⁴ - x³ - x² - 1 = 0 => x⁴ - x³ = x²+ 1
= x²+ 1 - x
= x² - x + 1
f(x)= x² - x + 1
f(x)= x² - x + 1 can be achieved using long division of f(x)/g(x) and getting remainder as g(x) = 0 so only remainder will give value
f(a) + f(b) + f(c) + f(d).
= (a² + b² + c² + d²) - (a + b + c + d ) + (1 +1 + 1 +1 )
= (a² + b² + c² + d²) - (a + b + c + d ) + (4 )
a , b , c and d are zeroes of g(x) = x⁴ - x³ - x² - 1
a + b + c + d = - (-1)/(1) = 1
a² + b² + c² + d² = (a + b + c + d) ² - 2∑ab
=> a² + b² + c² + d² = (1) ² - 2 ( -1)/1
=> a² + b² + c² + d² = 1 + 2 = 3
Hence
f(a) + f(b) + f(c) + f(d). = 3 - 1 + 4 = 6
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