What are the factors of 3x squared minus x plus 4?

Answer 1

That doesn't factor neatly. Applying the quadratic equation, we find two imaginary solutions: (1 plus or minus the square root of -47) divided by 6.

x = 0.16666 + 1.142609100066840i = 1st root = r1

x = 0.16666 - 1.142609100066840i = 2nd root = r2

where i is the square root of negative one. Then factor as (x - r1)(x - r2)

Or you can complete the square to transform the trinomial into the difference of the squares

3x2 - x + 4 factor out 3

= 3(x2 -(1/3)x + 4/3) add and subtract (1/6)2

= 3[x2 -(1/3)x + (1/6)2 - 1/36 + 4/3] simplify

= 3[(x - 1/6)2 + 47/36] write 47/36 = - (-47/36) = - (√-1√47/6)2 = - (i√47/6)2

= 3[(x - 1/6)2 - (i√47/6)2] write the difference of the squares

= 3[(x - 1/6) - (i√47/6)][(x - 1/6) + (i√47/6)]

= 3[(x - (1 + i√47)/6)][(x - (1 - i√47)/6].