SOLUTION
[tex]\begin{gathered} \frac{3x^2+2x-21}{-2x^2-2x\text{ +12}}\times\frac{2x^2+25x+63}{6x^2+7x-49} \\ \\ \frac{3x^2+9x-7x-21}{-2x^2-6x+4x+12}\times\frac{2x^2+18x+7x+63}{6x^2+21x-14x-49} \\ \\ \frac{3x\mleft(x+3\mright)-7\mleft(x+3\mright)}{-2x(x+3)+4(x+3)}\times\frac{2x(x+9)+7(x+9)}{3x(2x+7)-7(2x+7)} \\ \\ \frac{(3x-7)(x+3)}{(-2x+4)(x+3)}\times\frac{(2x+7)(x+9)}{(3x-7)(2x+7)} \end{gathered}[/tex]
Cancelling out the common terms we have
[tex]\begin{gathered} \frac{(3x-7)}{(-2x+4)_{}}\times\frac{(x+9)}{3x-7)} \\ \text{cancelling out (3x-7) we have } \\ \frac{(x+9)}{(-2x+4)} \end{gathered}[/tex]
Therefore, from
[tex]\begin{gathered} \frac{(x+9)}{(-2x+4)} \\ \text{if a =1, then b = 9, c = -2 and d = 4} \end{gathered}[/tex]
So, b = 9,
c = -2
d = 4
