Solve for $x$: $$\sum_{r=0}^5 {5\choose r} (-1)^rx^{5-r}3^r = 32$$
Looks like binomial theorem. So this would simplify to $(-x+3)^5=32$, and solving gives $x=2$. Is this correct?
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$\begingroup$ close but it's not $(-x+3)^5$ but $(x-3)^5$, can you tell me why? $\endgroup$– Sentinel135Commented Mar 21, 2017 at 0:51
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$\begingroup$ Oh oops, because it is to the same power as the 3... $\endgroup$– suomynonACommented Mar 21, 2017 at 0:52
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$\begingroup$ There you go. we can merge the $(-1)^r(3)^r$ into $(-3)^r$. So what does that equal instead? $\endgroup$– Sentinel135Commented Mar 21, 2017 at 0:54
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$\begingroup$ $(x-3)^5\!\!\!$ $\endgroup$– suomynonACommented Mar 21, 2017 at 0:57
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1$\begingroup$ $x=5\!\!\!\;\;$ $\endgroup$– suomynonACommented Mar 21, 2017 at 1:02
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