Originally Posted by

**Archie Meade** Hi fishkeeper,

Higher powers of x will not be involved in computing the $\displaystyle x^1$ term, since multiplications of those will only contribute to higher powers of x.

For instance

$\displaystyle (3+x+x^2)(2)=6+2x+2x^2$

The x term results from the multiplication of x by the constant.

Or..

$\displaystyle (3+x+x^2)(2+x+x^2)=6+3x+3x^2+2x+x^2+x^3+2x^2+x^3+x ^4$

$\displaystyle =6+(3x+2x)+(3x^2+x^2)+(x^3+x^3)+x^4$

The resultant x term is the sum of the constant from one factor multiplied by the x term in the other factor from both sets of factors.

When CaptainBlack wrote out the pair of square brackets multiplied together,

he only wrote out the first few terms of the binomial expansions, as higher powers of x will not contribute to the $\displaystyle x^1$ term.

That's all the question wanted you to discover.

The question wanted you to see that only a pair of terms from both expansions contribute to the resultant x term.

CaptainBlack placed the first few terms of the binomial expansions of both $\displaystyle (1-2x)^4$ and $\displaystyle (2+x)^9$ into the square brackets, as it is unnecessary to write the rest of them to obtain the coefficient of the $\displaystyle x^1$ term only.

if you needed the entire expansion, then all the terms of both would be required.