1

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

anni
  • 349

1 Answers1

1

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 $m=7.$

mfl
  • 29,399
  • but how, 2 is multiplying with m? – anni Oct 13 '14 at 13:48
  • If you have in mind that $2^m$ means to multiply $2$ by $2$ $m$ times, when you multiply again by $2$ you are multiplying $2$ by $2$ $m+1$ times. – mfl Oct 13 '14 at 13:50