Products of Binomial Expansions | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Include every pair that makes the power

Multiplying two terms adds their powers: $x^p x^q=x^{p+q}$ . In a product of two polynomial expansions, an $x^2$ term can come from three power pairs: $0+2$ , $1+1$ , or $2+0$ .

First factor × second factor

constant × x²

→ term in x²

x × x

→ term in x²

x² × constant

→ term in x²

Multiply within each row, then add the three contributions.

Multiply the coefficients within each pair, then add all contributions. A missing term has coefficient zero; do not invent a contribution for it.

Expand only as far as needed

Worked example

Find the coefficient of $x^2$ in $(1+2x)^4(3-x)$ .

The second factor has only a constant and an $x$ term. We need only the first three terms of the first factor:

\begin{aligned}(1+2x)^4={}&1+8x+24x^2\\&+\cdots\end{aligned}

The dots retain the unlisted higher-power terms; this is not equality to a three-term polynomial.

3(24)+(-1)(8)=64

There is no third contribution because $3-x$ has no $x^2$ term.

Worked example

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

The needed starts are $1+3x+3x^2+\cdots$ and $1-4x+4x^2$ . All three pairs contribute:

1(4)+3(-4)+3(1)=-5

For an $x^m$ coefficient in polynomial factors, list all pairs whose powers add to $m$ . If a factor includes negative powers, a higher positive power may also contribute; check the actual powers before truncating.

Now you try

Your turnIndependent practice

(a) Find the coefficient of $x^3$ in $(1+2x)^4(2-x)$ .

(b) Find the coefficient of $x^2$ in $(1+x)^4(1-x)^3$ .

Show worked answer

(a) The first expansion starts $1+8x+24x^2+32x^3+\cdots$ . The coefficient is $2(32)-24=40$ .

(b) Use $1+4x+6x^2+\cdots$ and $1-3x+3x^2-\cdots$ . The coefficient is $3-12+6=-3$ .

An independent check for (b) is $(1+x)^4(1-x)^3=(1+x)(1-x^2)^3$ . Only $-3x^2$ contributes to the target power.

Include every pair that makes the power

Multiplying two terms adds their powers: $x^p x^q=x^{p+q}$ . In a product of two polynomial expansions, an $x^2$ term can come from three power pairs: $0+2$ , $1+1$ , or $2+0$ .

First factor × second factor

constant × x²

→ term in x²

x × x

→ term in x²

x² × constant

→ term in x²

Multiply within each row, then add the three contributions.

Multiply the coefficients within each pair, then add all contributions. A missing term has coefficient zero; do not invent a contribution for it.

Expand only as far as needed

Worked example

Find the coefficient of $x^2$ in $(1+2x)^4(3-x)$ .

The second factor has only a constant and an $x$ term. We need only the first three terms of the first factor:

\begin{aligned}(1+2x)^4={}&1+8x+24x^2\\&+\cdots\end{aligned}

The dots retain the unlisted higher-power terms; this is not equality to a three-term polynomial.

3(24)+(-1)(8)=64

There is no third contribution because $3-x$ has no $x^2$ term.

Worked example

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

The needed starts are $1+3x+3x^2+\cdots$ and $1-4x+4x^2$ . All three pairs contribute:

1(4)+3(-4)+3(1)=-5

Now you try

Your turnIndependent practice

(a) Find the coefficient of $x^3$ in $(1+2x)^4(2-x)$ .

(b) Find the coefficient of $x^2$ in $(1+x)^4(1-x)^3$ .

Show worked answer

(a) The first expansion starts $1+8x+24x^2+32x^3+\cdots$ . The coefficient is $2(32)-24=40$ .

(b) Use $1+4x+6x^2+\cdots$ and $1-3x+3x^2-\cdots$ . The coefficient is $3-12+6=-3$ .

An independent check for (b) is $(1+x)^4(1-x)^3=(1+x)(1-x^2)^3$ . Only $-3x^2$ contributes to the target power.