Step-by-step explanation:FACTORING a cubic polynomial is not very easy. I suggest trying out a combination of p/q approach for polynomials and factor and/or remainder theorem. p/q, where p are the possible integer factors of the constant (in your case that is 6), and q are the possible integer factors of the coefficient of the leading term (in your case that is 2). Thus p = 1, 2, 3, 6 q = 1, 2 p/q = 1, 1/2, 2, 3, 3/2, 6 which may be either positive or NEGATIVE. p/q LISTS down the possible factors of your polynomial, so this will help you greatly factor out any polynomial. Using factor and/or remainder theorem, you can determine whether the numbers you have are roots (possible factors) of a polynomial. You apply it by substituting a number for the x's, and if the answer comes out a 0, then that number is a root. Let's try that now. 2(1)^3 + 3(1)^2 - 5(1) - 6 = -6 2(-1)^3 + 3(-1)^2 - 5(-1) - 6 = 0 Since it zeroes at -1, then -1 is a root, which makes (x - (-1)) or (x+1) a factor of the cubic. Thus, we just need to figure out what would be the remaining polynomial, after we take out (x+1) from it. In short, 2x^3 + 3x^2 - 5x - 6 = (x + 1)(...) To determine this, we need to use either synthetic division or long division. I prefer synthetic division. 2     3       -5     -6      |   -1 (because this is the root)      -2       -1     6 2     1       -6      0 Thus, the remaining polynomial's coefficients are 2, 1 and -6. The remaining polynomial is a polynomial which is a DEGREE lower than the starting polynomial. Because the starting polynomial is cubic (degree 3), the remaining polynomial is quadratic (degree 2) 2x^3 + 3x^2 - 5x - 6 = (x + 1)(2x^2 + x - 6) Since the remaining polynomial is quadratic, we can just simply factor out using traditional factoring. 2x^2 + x - 6 = (2x - 3)(x + 2) Thus, the factored form of the cubic is 2x^3 + 3x^2 - 5x - 6 = (x + 1)(2x - 3)(x + 2) You may expand the factors by multiplying to check. Hope this helps!!!!!