6^3-4x^2-20x+48 x=6

Question

 6^3-4x^2-20x+48

One root of f(x)= 6^3-4x^2-20x+48 is x = 6. What are all the factors of the function? Use the Remainder Theorem.

Answer 1

I will assume that you mistyped the first term, which should be 6x³ and that the last term is 48, not 48x.

 

You can use synthetic division or long division. I will show you synthetic division (which is faster).

 

   _________________

6 ) 6    -4    -20       48

          36    192   1032 

   _________________

     6    32   172   1080 .

 

This shows that either your problem is misstated (the remainder when dividing by 6 should be 0, not 1080; or that your problem is not stated correctly. Or, that my initial assumption was wrong.

 

Could you please recheck the problem and get me straight on this? I do know that 6 is a zero of 6³ - 4x² - 20x + 48, but I am not at all sure that is what you were typing in.

Answer 2

OK. 6x³ - 4x² - 20x + 48 has one real root, and it's not 6. You may want to go through this and look for a typo.