Consider the polynomial P(x) = (x + x^2-k )^5 + (x - x^2-k )^5 , where k is a positive constant. If the coefficient of x^3 in the simplified expansion of P(x) is -120 , then the coefficient of x in P(x) is :

Step-by-step solution

Using the binomial expansion for (a+b)^n + (a-b)^n , we know that the terms with odd powers of b cancel out. (a+b)^5 + (a-b)^5 = 2 [ ⁵C₀ a^5 + ⁵C₂ a^3 b^2 + ⁵C₄ a b^4 ] Substitute a = x and b = x^2-k : P(x) = 2 [ x^5 + 10 x^3 (x^2-k) + 5 x (x^2-k)^2 ] Expanding the terms inside the bracket: P(x) = 2 [ x^5 + 10x^5 - 10kx^3 + 5x(x^4 - 2kx^2 + k^2) ] P(x) = 2 [ 11x^5 - 10kx^3 + 5x^5 - 10kx^3 + 5k^2x ] P(x) = 2 [ 16x^5 - 20kx^3 + 5k^2x ] P(x) = 32x^5 - 40kx^3 + 10k^2x We are given that the coefficient of x^3 is -120 .