Question

If there is no term in `x` in the expression `(1−2x)(1+ax)^5`, what is the value of `a`?

Answer 1

'There is no term in `x`' means that all terms in the expansion are either constants (numbers) or powers of `x` greater than 1 (e.g. `x^2` or bigger). See the working below, and the solution is `a=2/5`.

Explanation:

We need to expand and find a value of `a` such that there are no terms in `x`:

`(1-2x)(1+ax)^5`

Rather than manually work through the expansion of the fifth power, we can use the binomial theorem:

`(1+ax)^5 = 1 + 5ax+10(ax)^2+10(ax)^3+5(ax)^4+(ax)^5`

We can multiply each of these terms by `(1-2x)`, and then we need to choose a value of `a` so that any terms in `x` cancel out.

`(1-2x)(1+ax)^5 = (1-2x) + 5ax(1-2x) +10(ax)^2(1-2x)+10(ax)^3(1-2x)+5(ax)^4(1-2x)+(ax)^5(1-2x)`
`=1-2x+5ax-10ax^2+10a^2x^2-20a^2x^3+10a^3x^3-20a^3x^4+5a^4x^4-20a^4x^5+a^5x^5-2a^5x^6`

There are only two terms in `x`, and when we collect like terms we want them to add to `0`, so:

`-2x+5ax=0`

`5ax=2x`

Divide both sides by `5x` to make `a` the subject:

`a=(2cancel(x))/(5cancel(x)) = 2/5`