Mathematics 9709/11 — May/June 2017

Walkthrough

We need the coefficients of $x^2$ and $x^3$ in the expansion of $(3 - 2x)^6$. By the binomial theorem, each term has the form

For $x^2$, choose $k = 2$. This means we pick 2 of the 6 factors to contribute $-2x$, and the remaining 4 factors contribute $3$. The coefficient is

For $x^3$, choose $k = 3$. Now 3 factors contribute $-2x$ and the remaining 3 contribute $3$, giving

The coefficient of $x^3$ is negative because $(-2)^3 = -8$. Finally, divide the two coefficients:

Key Takeaways

• The binomial theorem lets us find any single coefficient without writing out the whole expansion.
• The coefficient of $x^k$ in $(p + qx)^n$ is $\binom{n}{k} p^{n-k} q^k$.
• Remember to include the binomial coefficient, the power of the constant term, and the power of the coefficient of $x$.
• A negative base raised to an odd power gives a negative coefficient, which affects the sign of the final ratio.

Common Mistakes

• Forgetting the binomial coefficient $\binom{6}{2}$ or $\binom{6}{3}$.
• Using the wrong powers: for $x^2$, the power of $3$ must be $4$, not $2$.
• Losing the sign: $(-2)^3 = -8$, so the coefficient of $x^3$ is negative.
• Omitting the negative sign in the final answer. The mark scheme requires the negative sign to appear before the fraction or in the numerator.
• Writing the coefficient with an $x$ attached; the coefficient is the number alone.

Things to Be Careful About

• The general term is $\binom{6}{k} 3^{6-k} (-2)^k x^k$, so the coefficient of $x^k$ is $\binom{6}{k} 3^{6-k} (-2)^k$.
• Check the sign of each coefficient before forming the ratio.
• Simplify $\frac{4860}{-4320}$ fully. Dividing numerator and denominator by $540$ gives $\frac{9}{-8} = -\frac{9}{8}$.
• The mark scheme accepts unsimplified coefficient forms, but the powers must be correct and the final ratio must be simplified with the correct sign.