Coefficient in a Product with a Binomial Expansion

To find one coefficient in a product, you don’t need the full expansion. Expand $(1-x)^6$ only up to $x^2$, then collect the pairs of terms whose powers multiply to $x^2$.

Find the coefficient of $x^2$ in the expansion of $(1 + 2x)(1 - x)^6$. [4]

The working

Step 1, expand $(1-x)^6$ as far as the $x^2$ term using $\binom{6}{r}$: $(1-x)^6 = 1 - \binom{6}{1}x + \binom{6}{2}x^2 - \cdots = 1 - 6x + 15x^2 - \cdots \quad \text{(M1, A1)}$

Step 2, decide which products give $x^2$. Multiplying by $(1 + 2x)$:

• $1 \times (15x^2) = 15x^2$
• $2x \times (-6x) = -12x^2$

(The $2x \times 1$ gives $x^1$, and $1\times 1$ gives the constant, so neither contributes.) (M1)

Step 3, add the contributions: $15x^2 - 12x^2 = 3x^2$

The coefficient of $x^2$ is $\boxed{3}$. (A1)

Where the marks are won and lost

• You need both contributions: constant $\times\ x^2$ term and $2x \times x$ term. Missing the second is the classic error.
• Get the signs right in $(1-x)^6$: the $x$ term is $-6x$, so $2x \times (-6x) = -12x^2$ is negative.
• Expanding further than $x^2$ wastes time; stop at the term you need.

Common mistakes

• Only taking $1 \times 15x^2$ and forgetting the cross term.
• Sign slip: treating the $x$ term of $(1-x)^6$ as $+6x$.
• Using $\binom{6}{2} = 30$ instead of $15$ (it is $\frac{6\times5}{2}$).

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