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Note that $x^5$ can be expressed as either $x \times x \times x \times x^2$ or $x^2 \times x^2 \times x$ . The first case would have each term with coefficient $1$ and the second with coefficient $-2$ , So in essence, the coefficient would be $\binom{4}{3} \times 1 + \binom{4}{2}\times \binom{2}{1} \times (-2) =-20$