To find one coefficient in a product such as (1-2x)2(1+x)7, you do not need the whole product. Expand each bracket only as far as the power you want, then add the products of every pair of terms whose powers of x add up to that power. Why: each term of the product is one term of the first bracket times one term of the second, and their powers of x add. A coefficient is complete only when every such pair has been counted. Method - Expand each bracket up to the required power. Expand a short bracket such as (1-2x)2 in full. - List the pairs: for xk, take x0 with xk, x1 with xk-1, and so on, until a bracket has no term of that power. - Multiply the coefficients in each pair, signs included, and add. Example Find the coefficient of x3 in the expansion of (1-2x)2(1+x)7. (1-2x)2=1-4x+4x2 and (1+x)7=1+7x+21x2+35x3+. s From (1-2x)2 | From (1+x)7 | Product 1 | 35x3 | 35x3 -4x | 21x2 | -84x3 4x2 | 7x | 28x3 Coefficient of x3: 35-84+28=-21 Replacing x by an expression A known expansion stays true when x is replaced by any expression. Write the new bracket in the form of the one you have expanded, then substitute and keep only the powers you need. Given (1+2x)6=1+12x+60x2+. s, find the first three terms of (1+2x+4x2)6. 1+2x+4x2=1+2(x+2x2), so replace x by x+2x2: (1+2x+4x2)6=1+12(x+2x2)+60(x+2x2)2+. s=1+12x+24x2+60x2+. s=1+12x+84x2+. s Another way gives the same result: treat the bracket as a binomial, [(1+2x)+4x2]6=(1+2x)6+6(1+2x)5(4x2)+. s, whose x2 terms are 60x2+24x2=84x2. If the bracket does not start with 1, take the constant out first: (3+6x)5=35(1+2x)5. **Exam check** Count every pair: the constant of one bracket with the top power of the other is the pair most often missed. Keep the sign of a negative term in every product. After replacing x, bracket the whole expression before squaring it.

Coefficients in products of expansions

Secondary 3

To find one coefficient in a product such as $(1-2x)^2(1+x)^7$, you do not need the whole product. Expand each bracket only as far as the power you want, then add the products of every pair of terms whose powers of $x$ add up to that power.

Why: each term of the product is one term of the first bracket times one term of the second, and their powers of $x$ add. A coefficient is complete only when every such pair has been counted.

Method

Expand each bracket up to the required power. Expand a short bracket such as $(1-2x)^2$ in full.
List the pairs: for $x^k$, take $x^0$ with $x^k$, $x^1$ with $x^{k-1}$, and so on, until a bracket has no term of that power.
Multiply the coefficients in each pair, signs included, and add.

Example

Find the coefficient of $x^3$ in the expansion of $(1-2x)^2(1+x)^7$.

$(1-2x)^2=1-4x+4x^2$ and $(1+x)^7=1+7x+21x^2+35x^3+\cdots$

From $(1-2x)^2$	From $(1+x)^7$	Product
$1$	$35x^3$	$35x^3$
$-4x$	$21x^2$	$-84x^3$
$4x^2$	$7x$	$28x^3$

Coefficient of $x^3$: $35-84+28=-21$

Replacing $x$ by an expression

A known expansion stays true when $x$ is replaced by any expression. Write the new bracket in the form of the one you have expanded, then substitute and keep only the powers you need.

Given $(1+2x)^6=1+12x+60x^2+\cdots$, find the first three terms of $(1+2x+4x^2)^6$.

$1+2x+4x^2=1+2(x+2x^2)$, so replace $x$ by $x+2x^2$:

$(1+2x+4x^2)^6=1+12(x+2x^2)+60(x+2x^2)^2+\cdots=1+12x+24x^2+60x^2+\cdots=1+12x+84x^2+\cdots$

Another way gives the same result: treat the bracket as a binomial, $\left[(1+2x)+4x^2\right]^6=(1+2x)^6+6(1+2x)^5(4x^2)+\cdots$, whose $x^2$ terms are $60x^2+24x^2=84x^2$.

If the bracket does not start with $1$, take the constant out first: $(3+6x)^5=3^5(1+2x)^5$.

Exam check Count every pair: the constant of one bracket with the top power of the other is the pair most often missed. Keep the sign of a negative term in every product. After replacing $x$, bracket the whole expression before squaring it.

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From \((1-2x)^2\)	From \((1+x)^7\)	Product
\(1\)	\(35x^3\)	\(35x^3\)
\(-4x\)	\(21x^2\)	\(-84x^3\)
\(4x^2\)	\(7x\)	\(28x^3\)