Find the coefficient of x in the expansion of $(2x^2+x-3)^8$. This is a question from IB past papers.
I factorized the equation to (2x+3)(x-1) and I tried finding the coefficients, but I got a wrong answer. Maybe I have forgotten how to solve it. Can someone tell me the way to solve it, but not the whole answer?
 A: The term $2x^{2}$ makes no contribution to coefficient of $x$. Just expand $(x-3)^{8}$ by Binomial theorem. The answer is $-3^{7}(8)$
A: $$=\sum_{r=0}\binom8r(x-3)^{8-r}(2x^2)^r$$
So, we are interested in $r=0$
What is the coefficient of $x$ in $\binom80(x-3)^8?$
A: $P(x)=(2x^2+x-3)^8$ is  a polynomial in $x$.
$P'(x)=8(2x^2+x-3)^7(4x +1);$
$P'(0)=8(-3)^7(1), $
The coefficient of $x$ is:
$P(0)= 8(-3)^7(1).$
A: Just do $${(x-3)^8}$$ and then use your pascal triangle to find the coefficent
                  1
               1     1
             1     2     1
           1     3     3     1
          1     4     6     4     1
        1     5    10    10     5     1
     1     6    15    20    15     6     1
  1     7    21    35    35    21     7     1
1     8    28    56    70    56    28     8     1

Thus equal to $${(-1)(3^7)(8)} = -17496$$
A: Your method also works. We have
$$
(2x+3)^{8}(x-1)^{8} = \left(\sum_{i=0}^{8}\binom{8}{i}(2x)^{i}3^{8-i}\right)\left(\sum_{j=0}^{8} \binom{8}{j}x^{j}(-1)^{8-j}\right)
$$
and the coefficient of $x^{1}$ is
$$
\binom{8}{1}\times 2^{1} \times 3^{7} \times\binom{8}{0}\times (-1)^{8}+ \binom{8}{0}\times 3^{8}\times \binom{8}{1}\times (-1)^{7} = 16\times 3^{7} - 8\times 3^{8} = -8\times 3^{7
}
$$
