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Algebra Level 3

If x = 1 + 2 3 x=1+\sqrt[3]{2} Find A= x 5 − 2 x 4 + x 3 − 3 x 2 + 1942 x^5-2x^4+x^3-3x^2+1942


The answer is 1945.

Son Nguyen by Son Nguyen , 20, Vietnam

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1 solution

Son Nguyen
Oct 20, 2015

If A= x 5 − 2 x 4 + x 3 − 3 x 2 + 1942 x^5-2x^4+x^3-3x^2+1942 Let A B \frac{A}{B} B= x 3 − 3 x 2 + 3 x − 3 x^3-3x^2+3x-3 ======>A= x 5 − 2 x 4 + x 3 − 3 x 2 + 1942 x^5-2x^4+x^3-3x^2+1942 = ( x 3 − 3 x 2 + 3 x − 3 ) ( x 2 + x + 1 ) + 1945 (x^3-3x^2+3x-3)(x^2+x+1)+1945 =====>A=1945

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Here's the answer .We've got x 5 − 2 x 4 + x 3 − 3 x 2 + 1942 x^5-2x^4+x^3-3x^2+1942 = x 2 ( x 3 − 3 x 2 + 3 x − 3 ) + x 4 − 2 x 3 + 1942 ( 1 ) =x^2(x^3-3x^2+3x-3)+x^4-2x^3+1942(1)

*From the condition we have x 3 − 3 x 2 + 3 x − 3 = 0 ( 2 ) x^3-3x^2+3x-3=0 (2) x 2 − 2 x = ( 2 3 ) 2 − 1 ( 3 ) x^2-2x=(\sqrt[3]{2})^2-1(3) x 2 = ( 2 3 + 1 ) 2 ( 4 ) x^2=(\sqrt[3]{2}+1)^2 (4) Replace (2) to (1) we get x 4 − 2 x 3 + 1942 x^4-2x^3+1942 = x 2 ( x 2 − 2 x ) + 1942 ( 5 ) =x^2(x^2-2x) +1942 (5) Replace (3) and (4) to (5) we get ( 2 3 + 1 ) 2 [ ( 2 3 ) 2 − 1 ) ] + 1942 ( 6 ) (\sqrt[3]{2}+1)^2[(\sqrt[3]{2})^2-1)]+1942 (6) Solving (6) and we get the answer 1945 1945

P C - 5 years, 7 months ago

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pls onl facebook and see my answer

Son Nguyen - 5 years, 7 months ago

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