<s>[INST] How to find the value of $m$ which is a power of a polynomial, when divided by a linear polinomial gives some remainder? 
Q) The value of $m$ if $2x^m+x^3-3x^2-26$ leaves remainder of 226 when divided by $x-2$.
(1) 0
(2) 7
(3) 10
(4) all of these

How i solved it
let $p(x)=2x^m+x^3-3x^2-26$ and $g(x)=x-2=0$
therefore $x=2$
substituting the value of $x$ in $p(x)$
then: 
$2(2)^m+(2)^3-3(2)^2-26=226$
$= ((2)^2)^m+8-12-26=226$
$= ((2)^2)^m = 226+30$
$= ((2)^2)^m =(16)^2 => ((2)^2)^m=((2)^4)^2 $
so on comparing both sides, since the base are same on both sides, taking the exponents
therefore,   $ 2m=8 => m=4$!!
but there is not any option there....
so how to solve this? 
thanks
 [/INST] It is $2\cdot 2^m=2^ {m+1}$ $(\ne (2^2)^m,$ which is the wrong point in your solution). So your equation should read
$$2^{m+1}=2^8$$ from where you get $m=7.$
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