Question

The coefficient of $x^5$ in the expansion of $(x^2-x-2)^3$ is

• A. 6
• B. -3
• C. -6
• D. 3

Answer 1

Answer: (B)

-3

Answer 2

To find the coefficient of $x^5$ in $(x^2-x-2)^3$, we can use the multinomial theorem. The general term in the expansion of $(x^2+(-x)+(-2))^3$ is $\frac{3!}{p!q!r!}(x^2)^p(-x)^q(-2)^r$, where $p+q+r=3$. This simplifies to $\frac{3!}{p!q!r!}(-1)^q(-2)^r x^{2p+q}$. We need $2p+q=5$.

Possible non-negative integer solutions for $(p,q,r)$ satisfying $p+q+r=3$ and $2p+q=5$:

• If $p=0$, $q=5$. Then $r=-2$ (not valid).

• If $p=1$, $q=3$. Then $r=-1$ (not valid).

• If $p=2$, $q=1$. Then $r=0$ (valid). The coefficient for this term is $\frac{3!}{2!1!0!}(-1)^1(-2)^0 = 3 \times (-1) \times 1 = -3$.

• If $p=3$, $q=-1$ (not valid).

Thus, the only term contributing to $x^5$ is when $(p,q,r)=(2,1,0)$, and its coefficient is -3.