Finding the value of the constant $p$

Question

NOTE: Binomial expansion of $(1-2x)^{5}$ -

$1-10x+40x^{2}-80x^{3}+80x^{4}-32x^{5}$

The question -

It is given that, when $(1+px)(1-2x)^{5}$ is expanded, there is no term in $x^{5}$. Find the value of the constant $p$.

How do I go about solving this? I know it's something to do with canceling out the powers (or is it?).

Any help would be highly appreciated.

Answer 1

Coefficient of $x^5$ in the expansion of $(1+px)(1-2x)^5$ $$=1.{5 \choose 5}(-2)^5+p.{5 \choose 4}(-2)^4=0$$ $$2^4(5p-2)=0$$ $$\therefore p=\frac{2}{5}$$

Answer 2

So, the coefficient of $x^5$ is

$$-32+p\cdot80$$ which needs to be zero, right?