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I need help with this problem if anyone can contribute to solving this I would much appreciate it:

Find all polynomials $P$ that satisfy $P(1)=210$ and for every $x\in \mathbb{R}$: $$(x+10)P(2x)=(8x-32)P(x+6).$$

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    $\begingroup$ Welcome to MSE. What have you tried? $\endgroup$
    – jojobo
    Dec 24, 2020 at 20:13
  • $\begingroup$ Plug in $x = \frac{1}{2}$, what do you get? $\endgroup$ Dec 24, 2020 at 20:24
  • $\begingroup$ Also try $x= -5$ $\endgroup$
    – IanJ
    Dec 24, 2020 at 20:29
  • $\begingroup$ Ok thank you all for your contribution $\endgroup$
    – ssuag
    Dec 24, 2020 at 20:32

2 Answers 2

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small hint

With $ x=-10 $, we get $$P(-10+6)=P(-4)=0$$ and for $ x=4 $, it gives $$P(8)=0$$

thus

$$P(x)=(x-8)(x+4)R(x)$$ with $$R(1)=\frac{210}{-35}=-6$$

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  • $\begingroup$ Subbing $x=2$ gives yet another root. $\endgroup$
    – cosmo5
    Dec 24, 2020 at 20:38
  • $\begingroup$ Thank you very much $\endgroup$
    – ssuag
    Dec 24, 2020 at 20:38
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$$P(x) = c (-8 + x) (-4 + x) (4 + x)$$

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    $\begingroup$ My solution is right. Who down voted it? $\endgroup$
    – yugikaiba
    Dec 24, 2020 at 20:27
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    $\begingroup$ I didn't down vote it $\endgroup$
    – ssuag
    Dec 24, 2020 at 20:33
  • $\begingroup$ upvote if you find it useful $\endgroup$
    – yugikaiba
    Dec 24, 2020 at 20:33
  • $\begingroup$ at least will even out $\endgroup$
    – yugikaiba
    Dec 24, 2020 at 20:34
  • $\begingroup$ True i forgot to sum +6 f(x-6), now it is updated $\endgroup$
    – yugikaiba
    Dec 24, 2020 at 20:43

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