¿Puedes o no utilizar el producto notable estudiado, cuando los binomios que se multiplican son: (3x+2a)(3x+5a)?

13) As they’ve given the zeros, you can use those to create factors. Since the zeros are x=-1, x=-3, and x=4, by making the right side 0 in each equation, you get the factors (x+1), (x+3), and (x-4). Then, you can multiply those factors using box method or FOIL or whichever method you prefer. By multiplying (x+1) and (x+3), I got x^2+4x+3. Then, I multiplied that by (x-4) to get x^3-13x-12, which is your final answer.

14) Using the same steps in 13, I found the factors. As the zeros are x=-6 and x=0, I got the factors (x) and (x+6). Then, I multiplied them and got x^2+6x, which is the answer.

15) x=1, x=1, x=2. These become the factors (x-1), (x-1), and (x-2). Then, by multiplying (x-1) and (x-1) you get x^2-2x+1. Multiply that by (x-2) to get your final answer, x^3-4x^2+5x-2.

16) Zeros are x=-1, x=5, and x=-2. You get the factors (x+1), (x-5), and (x+2). Multiply (x+1) and (x-5) to get x^2-4x-5. Multiply that by the other factor, (x+2), to get your final answer, x^3-2x^2-13x-10.