Coefficient of x in the expansion of $(2x^2+(x-3))^8$

Question

$(2x^2+(x-3))^8$

How do I find the coefficient of x in this expansion?

Answer 1

$$(2x^2+(x-3))^8=a_0+a_1x+a_2x^2+a_3x^3\cdots$$

$$\frac{d}{dx}(2x^2+(x-3))^8=0+a_1+a_22x+a_33x^2\cdots$$

$$\lim_{x\to 0}\frac{d}{dx}(2x^2+(x-3))^8=a_1$$

$$-17496=a_1$$

Answer 2

$$(2x^2+x-3)^8=(2x^2+x-3)\cdot(2x^2+x-3) \cdots (2x^2+x-3)$$

Note that something like $ax$ you get only in one way: if you choose $x$ from one of eight parentheses and $-3$ from the other, so this coefficient is $\displaystyle 8 \cdot (-3)^7 $.

Answer 3

By the binomial theorem $$(2x^2+(x-3))^8=\sum_{k=0}^8{8\choose k}2^{8-k}x^{16-2k}(x-3)^{k}$$ for $k<8$ term power of $x$ is bigger than $1$,so $k=8$ than it simplifies to find coefficient of $x$ in $(x-3)^8$ which is $8\cdot(-3)^7=-17496$