AP EAMCET2004MathematicsBinomial Theorem
If x-4 x^2-5 x+6 can be expanded in the ascending powers of x , then the coefficient of x^3 is
Options
- A-73 648
- B73 648
- C71 648
- D-71 648
Correct answer
A. -73 648
Step-by-step solution
aligned & x-4 x^2-5 x+6 = x-4 (x-2)(x-3) & = 2 (x-2) - 1 (x-3) =2(x-2)⁻¹-(x-3)⁻¹ & =2(-2)⁻¹ (1- x 2 )⁻¹-(-3)⁻¹ (1- x 3 )⁻¹ & =- [1+ ( x 2 )+ ( x 2 )^2+ ( x 2 )^3+ ] & + 1 3 [1+ ( x 3 )+ ( x 3 )^2+ ( x 3 )^3+ ] & Coefficient of x^3 in x-4 x^2-5 x+6 . aligned aligned & =- ( 1 2 )^3+ 1 3 ( 1 3 )^3 & =- 1 8 + 1 81 = -81+8 648 & =- 73 648 aligned