Help with binomial expansion exercise

Question

I have missed a few lessons in school and I had to learn this topic pretty much on my own. And now I am stuck on this question while preparing for the test: "Find c given the expansion $(1+cx)(1+x)^4$ includes the term $22x^3$" I am pretty stumped because I have no idea on where to start

Answer 1

We are interested in the coefficient on $x^3$. Think about how you would expand $(1+cx)(1 + x)^4$ to get an $x^3$ term.

Let's first expand $(1+x)^4$. From this, there are $\binom{4}{3}$ ways to get an $x^3$ term and $\binom{4}{2}$ ways to get an $x^2$ term. Therefore, the $x^3$ term and $x^2$ term will have coefficients $\binom{4}{3} = 4$ and $\binom{4}{2} = 6$ respectively.

Now we can multiply these terms by either the $1$ or the $cx$ terms from $(1+cx)$ to yield the final $x^3$ term in the complete expansion. First, we pick up a coefficient of $4$ by multiplying $1$ with $4x^3$. Next, we pick up a coefficient of $6c$ by multiplying $cx$ with $6x^2$. Combining, our final result is $(4 + 6c)x^3$, and the coefficient must equal $22$...

Answer 2

$$(1+cx)(1+x)^4=(1+cx)(1+4x+6x^2+4x^3+x^4)=$$ $$=1+4x+6x^2+4x^3+x^4+cx+4cx^2+6cx^3+4cx^4+cx^5=$$ $$=1+(4+c)x+(6+4c)x^2+(4+6c)x^3+(1+4c)x^4+cx^5=$$

$$(4+6c)x^3=22x^3\Rightarrow 4+6c=22\Rightarrow c=3$$