How do you solve #x^3 -3x^2 +16x -48 = 0#?

Question

Isaiah Anthony · Accepted Answer

#3 and +-i4#. See the Socratic graph of the cubic, making x-intercept 3..

Based on the coefficients' sign changes, the equation has a maximum of three positive roots, the sign remains unchanged, when x is changed to #-x#. And so, there are no negative roots. At x = 0, the cubic is zero, so it becomes #(x-3)(x^2+16)# The other solutions are from by #x^2+16=0#, giving #x =+-i4# [2, 4, -500, 500]} 0, -500, 500]} graph{(x-3)(x^2+16) For approximating the solution, it is large y vs small x, not to scale.

Nathan Axel · Answer

undefined To solve the equation $ x^3 - 3x^2 + 16x - 48 = 0 $, you can use a method called factoring by grouping. First, group the terms:

$ (x^3 - 3x^2) + (16x - 48) = 0 $

Then factor out the greatest common factor from each group:

$ x^2(x - 3) + 16(x - 3) = 0 $

Now, you have a common factor of $ (x - 3) $, so you can factor that out:

$ (x - 3)(x^2 + 16) = 0 $

Now, set each factor equal to zero and solve for $ x $:

1. $ x - 3 = 0 $ 
   $ x = 3 $

2. $ x^2 + 16 = 0 $ 
   $ x^2 = -16 $ 
   $ x = \sqrt{-16} $
   Since the square root of a negative number is imaginary, there are no real solutions.

So, the real solution to the equation is $ x = 3 $.