Mathematics 9709/12 — October/November 2020

Walkthrough

We are asked for the coefficient of $x^3$ in the product $(1+kx)(1-2x)^5$. Expanding the whole product would be slow, so we only need the terms that can produce $x^3$. Multiplying the two factors, an $x^3$ term can arise in two ways: the constant $1$ multiplies the $x^3$ term from $(1-2x)^5$, and the term $kx$ multiplies the $x^2$ term from $(1-2x)^5$. Therefore we need the coefficients of $x^2$ and $x^3$ in $(1-2x)^5$.

Using the binomial theorem, the term containing $x^r$ in $(1-2x)^5$ is:

For $r=2$:

For $r=3$:

Thus the coefficient of $x^3$ in the product is $40k - 80$. The question states this is $20$, so:

Solving gives $k = \frac{5}{2}$. This matches the mark scheme: B1 for $-80$, B1 for $40$, M1 for forming and solving the equation, and A1 for the final value.

Key Takeaways

This question tests the binomial expansion and the idea that when two polynomials are multiplied, the coefficient of a given power is obtained by summing products of coefficients whose powers add to that power. It also reinforces careful handling of negative bases, such as $(-2)^3$.

Common Mistakes

• Forgetting the negative sign in $(-2)^3$, giving $80$ instead of $-80$.
• Using only the $x^3$ coefficient of $(1-2x)^5$ and ignoring the contribution from $kx \cdot x^2$.
• Forgetting to multiply the $x^2$ coefficient by $k$.
• Not simplifying the coefficients before forming the equation.
• Solving the equation incorrectly, for example writing $40k = -60$ by mishandling the sign of $-80$.

Things to Be Careful About

• The coefficient of $x^3$ in the product is $40k - 80$, not just $-80$.
• Make sure the binomial coefficients are evaluated correctly: $\binom{5}{2} = \binom{5}{3} = 10$.
• The mark scheme allows some working to be condensed, but the equation $40k - 80 = 20$ must be seen to gain the method mark.
• Give the final answer in simplest form: $\frac{5}{2}$.