If $x=2\sin\frac{3\pi}{10}$, find the value of the expression $$ W =2x^5+x^4-4x^3-4x^2-x+7 $$ After a bit of thought, I got : $$ x = \frac{\sqrt{5}+1}{2} $$ But I don't think my teacher expected me to substitute the value of $x$ into the expression. Is there anything I'm missing ?
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1Absolutely, you won't substitute that value into the expression – Saksham Jun 11 '17 at 18:24
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Even without know the value of $2\sin\dfrac{3\pi}{10}$ using https://math.stackexchange.com/questions/827540/proving-trigonometric-equation-cos36-circ-cos72-circ-1-2, $$\dfrac x2=2\left(\dfrac x2\right)^2-1+\dfrac12\iff x^2-x-1=0$$ So, $$2\sin\dfrac\pi{10}$$ will be another root – lab bhattacharjee Jun 18 '17 at 11:43
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Use $x^2-x-1=0$. It must help: $$2x^5+x^4-4x^3-4x^2+x+7=(x^2-x-1)(2x^3+3x^2+x)+7=7.$$
Michael Rozenberg
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you just made an edit, you gave a factor $(x^2 - x + 1) $ which differs form $x^2 - x - 1.$ I just revised the question to what I think was intended by the people/book asking – Will Jagy Jun 11 '17 at 18:29
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