Simplify the integrand by factoring the numerator.

Alright ~LIZZY💞🥺 — let's do this. Step 1: Simplify the integrand by factoring the numerator. The integrand is ((x-2)(6x^2 - x - 2))/((2x+1)). First, factor the quadratic term 6x^2 - x - 2. We look for two numbers that multiply to 6 × (-2) = -12 and add to -1. These numbers are -4 and 3. 6x^2 - x - 2 = 6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (2x+1)(3x-2) Now substitute this back into the integrand: ((x-2)(6x^2 - x - 2))/((2x+1)) = ((x-2)(2x+1)(3x-2))/((2x+1)) Assuming 2x+1 ≠ 0, we can cancel the (2x+1) terms: (x-2)(3x-2) Expand this product: (x-2)(3x-2) = x(3x-2) - 2(3x-2) = 3x^2 - 2x - 6x + 4 = 3x^2 - 8x + 4 So the integral becomes: _-2^3 (3x^2 - 8x + 4) dx Step 2: Find the antiderivative of the simplified expression. The antiderivative of 3x^2 - 8x + 4 is: F(x) = 3x^2+12+1 - 8x^1+11+1 + 4x F(x) = (3x^3)/(3) - (8x^2)/(2) + 4x F(x) = x^3 - 4x^2 + 4x Step 3: Evaluate the definite integral using the Fundamental Theorem of Calculus. _-2^3 (3x^2 - 8x + 4) dx = [x^3 - 4x^2 + 4x]_-2^3 = (3^3 - 4(3)^2 + 4(3)) - ((-2)^3 - 4(-2)^2 + 4(-2)) Calculate the value at the upper limit (x=3): 3^3 - 4(3)^2 + 4(3) = 27 - 4(9) + 12 = 27 - 36 + 12 = -9 + 12 = 3 Calculate the value at the lower limit (x=-2): (-2)^3 - 4(-2)^2 + 4(-2) = -8 - 4(4) - 8 = -8 - 16 - 8 = -32 Subtract the lower limit value from the upper limit value: 3 - (-32) = 3 + 32 = 35 The final answer is 35. Send me the next one 📸