Question 542451
<pre>
We want the numerical coefficient of the term in x<sup>27</sup>
of the complete expansion of (x²+5x+2)<sup>5</sup>(x²-7x+3)<sup>9</sup>.

(x²+5x+2)<sup>5</sup>(x²-7x+3)<sup>9</sup> equals this product of 14 trinomials:
<font size = 1>
(x²+5x+2)(x²+5x+2)(x²+5x+2)(x²+5x+2)(x²+5x+2)*(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)(x²-7x+3)</font>

The complete expansion can be considered as the sum of all products of 14
factors where one factor is taken from each of the 14 trinomials.

since 27 = 13·2+1 The only way to get a term in x<sup>27</sup> is to take the
x² term from 13 of the trinomials and the x-term from the remaining one. 

There will be 14 terms in x<sup>27</sup> to combine, of which 

5 will be of this form  

(5x)*(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²) = 5x<sup>27</sup>

with the (5x) in the 5 positions 1-5

and the other 9 will be of this form:

(x²)(x²)(x²)(x²)(x²)*(-7x)*(x²)(x²)(x²)(x²)(x²)(x²)(x²)(x²) = -7x<sup>27</sup>

with the (-7x) in the 9 positions 6-14

Therefore the term in x<sup>27</sup> is

5(5x<sup>27</sup>) + 9(-7x<sup>27</sup>) = 25x<sup>27</sup> - 63x<sup>27</sup> = -38x<sup>27</sup>

The coefficient of that term is -38.
 
The answer is -38.

Edwin</pre>