How do you solve 3x cubed plus 2x squared minus 12x minus 8 equals 0?

Answer 1

Let f(x) = 3x^2 + 2x^2 - 12x - 8

Although there are systematic ways for factorising cubics, the simplest probably involves finding one root by trial and error, and the other by solving a quadratic equation.

Take a factor of + or - 8 (the last coefficient) and substitute for x in f(x).

Try x = 4. f(4) = 168 so x = 4 is not a root.

Try x = 2. f(2) = 0 so x = 2 is a root and

by the remainder theorem, (x - 2) is a factor of f(x).

The other factor must be a quadratic. Suppose it is ax^2 + bx + c where a, b and c are constants to be determined.

Now f(x) = 3x^2 + 2x^2 - 12x - 8 = (x - 2)*(ax^2 + bx + c)

= ax^3 + bx^2 - 2ax^2 + cx - 2bx - 2c

= ax^3 + (b - 2a)x^2 + (c - 2b)x - 2c

Comparing coefficients of

x^3: 3 = a

x^2: 2 = b - 2a = b - 6 so b = 8

x^1: -12 = c - 2b = c - 16 so c = 4

x^0: -8 = -2c so c = 4 (this last acts as a check).

Thus the quadratic factor of f(x) is 3x^2 + 8x + 4

and using the quadratic formula to find its roots, x = -2 or x = -2/3.

Therefore, adding in the earlier solution, the three roots are x = (-2, -2/3 and 2).