How to factor this quadratic?

Question

 3x^2 + 7x + 31 = 0?  

Answer 1

3x² + 7x + 31 = 0

Divide each side by 3.

x² + (7/3)x + 31/3 = 0

Separate variables and constants aside by subtracting 31/3 from each side.

x² + (7/3)x = - 31/3.

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

Here x coefficient = 7/3. So, (half the x coefficient)² = [(7/3) / 2]² = 49/36.

Add 49/36 to each side.

x² + (7/3)x + 49/36 = - 31/3 + 49/36.

x² + (7/3)x + (7/6)² = [(- 31)(12) + 49]/36.

x² + (7/3)x + (7/6)² = - 323/36.

[x + 7/6]² = i²(323/36).

Take square root both sides.

x + 7/6 = ± (i√323)/6.

x = - 7/6 ± (i√323)/6.

x = [- 7 ± i√323]/6.

The factor form of equation is [x-(- 7 + i√323)/6][x-(- 7 - i√323)/6]=0