Recurring roots
Let f ( x ) = x 3 + x 2 − 2 x − 1 . If f ( a ) = 0 , find f ( a 2 − 2 ) .
The answer is 0.
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1 solution
nice question :) and nice solution too :)
You might anticipate that a 6 − 5 a 4 + 6 a 2 − 1 is a factor of a 3 + a 2 − 2 a − 1 by using polynomial division. Indeed it is and the other factor is a 3 − a 2 − 2 a + 1 . So f ( a 2 − 2 ) = ( a 3 − a 2 − 2 a + 1 ) ∗ f ( a ) = ( a 3 − a 2 − 2 a + 1 ) ∗ 0 = 0 . Love your questions, sir!
Sorry but i am unable to understand how u took a^2 common from (a^4-3a^2...) step 3 to 4 and 4 to 5 i cannot comprehend.
From f ( a ) = 0 we get 1 = x 3 + x 2 − 2 x . Through a series of substituting for the constant and subsequent factoring we get:
f ( a 2 − 2 ) = ( a 2 − 2 ) 3 + ( a 2 − 2 ) 2 − 2 ( a 2 − 2 ) − 1 = a 6 − 5 a 4 + 6 a 2 − 1 = a 6 − 5 a 4 + 6 a 2 − a 3 − a 2 + 2 a
= a ( a 5 − 5 a 3 − a 2 + 5 a + 2 )
= a 2 ( a 4 − 3 a 2 + a + 1 )
= a 3 ( a 3 + a 2 − 2 a − 1 )
= a 3 f ( a ) = 0 .