How do you solve #6x^2 - 11x + 2 = 0# by factoring?

D = b^2 - 4ac = 121 - 48 = 73. Since 73 is not a perfect square, this trinomial can't be factored.

To solve the quadratic equation (6x^2 - 11x + 2 = 0) by factoring, follow these steps:

1. Multiply the coefficient of (x^2) (6) by the constant term (2). This gives (6 \times 2 = 12).

2. Find two numbers that multiply to 12 and add up to the coefficient of (x) term (-11). These numbers are -9 and -2.

3. Rewrite the quadratic equation using these numbers to split the middle term: (6x^2 - 9x - 2x + 2 = 0).

4. Factor by grouping: (3x(2x - 3) - 1(2x - 3) = 0).

5. Notice that (2x - 3) is a common factor, so factor it out: ((2x - 3)(3x - 1) = 0).

6. Apply the zero-product property: (2x - 3 = 0) or (3x - 1 = 0).

7. Solve for (x) in each equation: (2x = 3 \Rightarrow x = \frac{3}{2}) or (3x = 1 \Rightarrow x = \frac{1}{3}).

Therefore, the solutions to the equation (6x^2 - 11x + 2 = 0) are (x = \frac{3}{2}) and (x = \frac{1}{3}).