Solution:
Given that,
[tex](x^2 + 6x - 16)(x - 2) = ax^3 + bx^2 + cx + d[/tex]
We have to find the value of [tex]\frac{c}{d}[/tex]
Let us simplify the given expression
Simplify the left hand side of equation
[tex](x^2 + 6x - 16)(x - 2)\\\\Distribute\:parentheses\\\\x^2x+x^2\left(-2\right)+6xx+6x\left(-2\right)+\left(-16\right)x+\left(-16\right)\left(-2\right)\\\\Simplify\\\\x^2x-2x^2+6xx-6\cdot \:2x-16x+16\cdot \:2\\\\x^3+4x^2-12x-16x+32\\\\\mathrm{Add\:similar\:elements:}\\\\x^3+4x^2-28x+32[/tex]
Therefore,
[tex]x^3 + 4x^2 - 28x + 32 = ax^3 + bx^2 + cx + d[/tex]
On comparing both sides we get,
c = -28
d = 32
Therefore,
[tex]\frac{c}{d} = \frac{-28}{32} = \frac{-7}{8}[/tex]
Thus value of c/d is -7/8
