@Nishi Agarwal
You are going wrong somewhere.
Yes, $x^{3} + x^{2}$ is very much able to detect any single bit error, even without MSB as 1.
The thing is that, if we look at it, this also contains 1 as the MSB, as shown;
$x^{3} + x^{2} = x^{2}(x + 1)$
So, all in all, doubt still stands na... if all the two terms are for sure to have 1 as the MSB after taking out common, what's the point of specifically requirng 1 as MSB.