Please Help with Algebra Problem :)

I have one last question to do on my review and I just can not figure this one out.

19. One solution of the equation given below is x=8. Find the other two solutions of that equation algebraically.

x^3 - 7x^2 - 14x + 48 = 0
I went to math lab and I have the answers at x= -3, 2, 8 but I have no idea how to solve this problem to get those answers.

Thanks so much for your time!

Shelby

Have you tried using Synthetic division ( Horners Method ) ...

By the way x=8 is a root of the equation, therefore x-8 is a factor.

19. One solution of the equation given below is x=8. Find the other two solutions of that equation algebraically.

x^3 - 7x^2 - 14x + 48 = 0
I went to math lab and I have the answers at x= -3, 2, 8 but I have no idea how to solve this problem to get those answers.

Click to expand...

Since x = 8 is a solution, that means that (x - 8) is a factor of x^3 - 7x^2 - 14x + 48. Therefore, if x^3 - 7x^2 - 14x + 48 is divided by (x - 8), using either synthetic or long division methods, another factor will be found -- one that is a quadratic. Once you have that quadratic, set it equal to zero and solve for x, using the quadratic formula or any other method you know.