Find the term for $x^3$ in $(3x+4)(x-2)^4$

Question

I'm sure there's an algebraic trick to it using the binomial theorem/Pascal's triangle, but I can't think of it right now. How do I get to the solution without manually expanding everything?

Answer 1

Hint:

$$(x-2)^4 = \sum_{i=0}^4 \binom4i x^i(-2)^{4-i}$$

Don't compute everything, just extract what you need from this formula.

Answer 2

The coefficient of $x^3$ in $(3x+4)(x-2)^4 = 3$ times the coefficient of $x^2$ in $(x-2)^4 +4$ times the coefficient of $x^3$ in $(x-2)^4$

Now the $(r+1)^{th}$ term of $(a+b)^n$ is $\displaystyle\binom nra^{n-r}b^r,0\le r\le n$

Answer 3

You need the $x^3$ and $x^2$ terms of $(x-2)^4$ (why?)

$(x-2)^4 = x^4 +(4/1!) x^3(-2) + ((4\cdot 3)/2!)x^2(-2)^2+.......$

Consider:

1) $4×[(4/1!)(-2)]$ and

2)$3× [((4 \cdot 3)/2!)(-2)^2]$.

What has happened, and

what is left to do ?