Remainder = 0
Let f ( x ) = x 2 + b x + c , where b and c are integers.
If f ( x ) is factor of both x 4 + 6 x 2 + 2 5 and 3 x 4 + 4 x 2 + 2 8 x + 5 , then find the value of f ( 1 ) .
Note: The problem is not original.
The answer is 4.
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1 solution
sir i m having no idea about the question as i have found i my book... and thanks for the solution
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No worries. You should rectify the question. x 4 + 6 x − 2 5 should be x 4 + 6 x + 2 5 .
Sir please explain your 8th line again i.e "As f ( x ) ............-2x+5) . What have you written in the beginning i am not able to understand that .Thanks
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f ( x ) ∣ g ( x ) ∧ f ( x ) ∣ h ( x ) means f ( x ) divides g ( x ) and f ( x ) divides h ( x ) .... ( a ∣ b stands for " a divides b " and ∧ stands for " and " ). As g ( x ) = ( x 2 + 2 x + 5 ) ( x 2 − 2 x + 5 ) and f ( x ) divides g ( x ) , therefore, f ( x ) must be either x 2 + 2 x + 5 or x 2 − 2 x + 5 . Now, for deciding whether f ( x ) = x 2 + 2 x + 5 or x 2 − 2 x + 5 , I used the fact that f ( x ) divides h ( x ) as well. After some synthetic division, I found x 2 − 2 x + 5 divides h ( x ) , while, x 2 + 2 x + 5 not. Hence, f ( x ) = x 2 − 2 x + 5 .
If you still have any doubt, feel free to ask.
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Thank you sir . I completely missed out that ∧ sign because i haven't seen that earlier .Btw how did you decided to factorize g ( x ) first rather h ( x ) ?
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@Chirayu Bhardwaj – It can easily be seen that g ( x ) can be reduced to a quadratic by making a simple substitution, i.e, t = x 2 which is relatively very easy to factorize as compared to h ( x ) which is a quartic equation. Furthermore, if you know Rational Root Theorem then you might be able to recognize that any rational root of h ( x ) (if exist) would be of the form β α , where α is a factor of constant term while β is a factor of x 4 . Certainly, figuring out such a root of h ( x ) to be able to factorize it would be a lengthy process.
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@Aditya Sky – Got it ! :D Thank you and my sir Deepansh for this awesome problem .
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@Chirayu Bhardwaj – chirayu sir please don't insult me by saying sir.. Everyone knows that you are my sir .... (can be seen by your brilliant stats
@Aditya Sky – By the way, I'm 16 years old so no need to call me Sir.
The question has a misprint. The first polynomial must be x 4 + 6 x 2 + 2 5 for the set answer to hold.
As far I'm able to remember, I've already encountered this question on Brilliant. The reason why I still remember this question is because of the way I factored the polynomial x 4 + 6 x 2 + 2 5 . Anyways, here's the solution :-
Say g ( x ) = x 4 + 6 x 2 + 2 5 , and, h ( x ) = 3 x 4 + 4 x 2 + 2 8 x + 5 .
Clearly, g ( x ) = x 4 + 6 x 2 + 2 5 = x 4 + 1 0 x 2 + 2 5 − 4 x 2 = ( x 2 + 5 ) 2 − ( 2 x ) 2 = ( x 2 + 2 x + 5 ) ( x 2 − 2 x + 5 ) .
As f ( x ) ∣ g ( x ) ∧ f ( x ) ∣ h ( x ) , therefore, f ( x ) = x 2 + 2 x + 5 or f ( x ) = x 2 − 2 x + 5 .
After synthetically dividing h ( x ) by x 2 + 2 x + 5 and x 2 − 2 x + 5 , it can be seen that ( x 2 − 2 x + 5 ) ∣ h ( x ) , while, ( x 2 + 2 x + 5 ) ∣ h ( x ) .
Hence, f ( x ) = x 2 − 2 x + 5 .
Therefore, f ( 1 ) = 1 2 − 2 ⋅ 1 + 5 = 4 .