In the expansion $(1 +px)(1 +qx)^4$ in ascending powers of $x$ , the coefficient of the $x$ term is $-5$ and there is no $x^2$ term. Find the value of $p$ and $q$.
My attempted answer
1) I expanded the second bracket $(1+qx)^4 $
2) I multiplied my answer with the second bracket
3) I grouped the common $x$ terms together
The problem is that the question tells you that there is no $x^2$ term, how do I work with what I am given in the question ?
HINT: Using the binomial theorem and expanding the product, the coefficient of $x$ is $p+{4\choose 1} q$ and the coefficient of $x^2$ is ${4\choose 1} qp+{4\choose 2}q^2$.
Now you just need to compare coefficients.
Remark: "there is no $x^2$ term" is another way to say "the coefficient of $x^2$ is $0$"