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Q.
In expansion of $\left(x^2-x-2\right)^5$, the coefficient of $x^2$ is
Binomial Theorem
Solution:
$( xC -2)^5(1+ x )^5=-(2- x )^5(1+ x )^5 $
$=-\left[{ }^5 C _0 2^5-{ }^5 C _1 2^4 x +{ }^5 C _2 2^3 x ^2+\ldots \ldots\right]\left[{ }^5 C _0+{ }^5 C _1 x +{ }^5 C _2 x ^2+\ldots \ldots\right]$
$\therefore \text { Coefficient of } x ^2 \text { is }=-{ }^5 C _0 2^5 \cdot{ }^5 C _2+{ }^5 C _1 \cdot 2^4 \cdot{ }^5 C _1-{ }^5 C _2 2^3=0$