If both n+2 and 2n+1 are factor of an^2+2n+b then prove that a-b=0
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f(n) = a n^2 + 2n + b
if, (n + 2) and (2n + 1) are factors of f(n)... then..
f(-2) = 0. and f(-1/2) = 0...
=> f(-2) = 4a - 4 + b = 0
=> 4a + b = 4 .........(i)
and,
f(-1/2)= a/4 -1 + b =0
=> a + 4b = 4 ........(ii)
(i) - (ii) => 3a - 3b = 0
=> a - b = 0...... ans...
if, (n + 2) and (2n + 1) are factors of f(n)... then..
f(-2) = 0. and f(-1/2) = 0...
=> f(-2) = 4a - 4 + b = 0
=> 4a + b = 4 .........(i)
and,
f(-1/2)= a/4 -1 + b =0
=> a + 4b = 4 ........(ii)
(i) - (ii) => 3a - 3b = 0
=> a - b = 0...... ans...
sonam5788:
I can't understand last three steps can you explain it again please
subtracting eqn (ii) from eqn (i)... you will get ... 3a-3b=0
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