Factors of $$48x^{3} - 8x^{2} - 93x - 45$$ are

(A) : (4x + 3)(4x ­- 3)(3x ­- 5)

= $$(16x^2 - 12x + 12x - 9)(3x - 5)$$

= $$(16x^2 - 9)(3x - 5)$$

= $$48x^3 - 80x^2 - 27x + 45$$

(B) : (4x - 3)(4x ­- 3)(3x -­ 5) 

= $$(16x^2 - 24x + 9)(3x - 5)$$

= $$48x^3 - 80x^2 - 72x^2 + 120x + 27x - 45$$

= $$48x^3 - 152x^2 + 147x - 45$$

(C) : (4x + 3)(4x + 3)(3x -­ 5) 

= $$(16x^2 + 24x + 9)(3x - 5)$$

= $$48x^3 - 80x^2 + 72x^2 - 120x + 27x - 45$$

= $$48x^3 - 8x^2 - 93x - 45$$

=> Ans - (C)