Cool roots.#Pythagorean
given that P ( x ) = x 3 + a x 2 + b x − 1 4 7 6 0 has integral roots α , β , γ such that α 2 + β 2 = γ 2 find P ( 1 0 ) m o d 8 8 . d e t a i l s & a s s u m p t i o n 0 = ∣ α ∣ < ∣ β ∣ < ∣ γ ∣ , α < β < γ work with the maximum value of α , β , γ
All roots are positive.
The answer is 50.
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1 solution
You could just make P ( x ) = ( x − 9 ) ( x − 4 0 ) ( x − 4 1 ) .
But I think you might want to state that α , β , and γ are positive integers. Because P ( x ) can also be ( x + 9 ) ( x + 4 0 ) ( x − 4 1 ) or ( x + 9 ) ( x − 4 0 ) ( x + 4 1 ) . Try to be specific.
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the question asked the maximum of α . β , γ
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No need to find coef of a and b because, p(x)=(x-9)(x-41)(x-40) because it is monic polynomial and 9,40,41 are roots of p(x) which implies p(x)=(x-9)(x-40)(x-41).Therefore p(10)=1* -31 * -30=930. Hard part is finding that Pythagorean triplet!
I think you meant Vieta's, not Newton's
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sir, are you sure, i partially know about these cause i'm a 7th grader.
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Yes, here's a link: http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas still a cool problem, nonetheless!
theyre similar
first, lets review veita 's sums α + β + γ = − a α β + β γ + γ α = b α β γ = − c = 1 4 7 6 0 notice that 1 4 7 6 0 = 9 × 4 0 × 4 1 which is the α , β , γ in P ( x ) satisfying the Pythagorean theorem , so α + β + γ = 4 0 + 4 1 + 9 = 9 0 a = − 9 0 α β + β γ + γ α = 4 1 × 4 0 + 4 0 × 9 + 4 1 × 9 = 2 3 6 9 b = 2 3 6 9 the function becomes P ( x ) = x 3 − 9 0 x 2 + 2 3 6 9 x − 1 4 7 6 0 insert ten P ( 1 0 ) = 1 0 3 − 9 0 × 1 0 2 + 2 3 6 9 0 − 1 4 7 6 0 P ( 1 0 ) = 1 0 0 0 − 9 0 0 0 + 2 3 6 9 0 − 1 4 7 6 0 p ( 1 0 ) = 9 3 0 9 3 0 = 5 0 (mod 88)