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Find real solutions for $x$, in$ \,\,\,f(x)=2x^{98}+5x^{97}+5x^{96}+...5x+3=0$, It is also given that $x+1$ is a factor

Since $x+1$ is a factor, we can write $f(x) =(x+1)(2x^{97}+3x^{96}+2x^{95}...+3)$

can someone give a hint what to do next? Thanks.

Kenta S
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emil
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1 Answers1

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$f(x)=5x(x^{97}+x^{96}+...+x+1)+3-3x^{98}=5x \frac{1-x^{98}}{1-x}+3(1-x^{98})$ for $x \ne 1$.

Can you proceed ?

Fred
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  • Thanks for your answer. And if I complete it like this, is it ok? $\frac{(1-x^{98})(2x+3)}{(1-x)}=0$ So $x=\frac{-3}{2}$ or $x=-1$ ( here we are neglecting x=1 answer since it doesn't satisfy f(x) – emil Jul 05 '18 at 11:35
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    Yes, we have $f(x)=0 \iff x \in {-\frac{3}{2}, -1}$ – Fred Jul 05 '18 at 11:40