It is often difficult to factor a trinomial in the form ax^2 + bx + c when lead coefficient is not one. It is especially difficult to factor when the lead coefficient is not prime. The following is a method that makes this much easier.
We multiply ax^2 + bx + c by a to get ax^2 + bx + c and try to factor this realizing that after we are done we must divide out the factor of a we put in originally. Suppose this can be factored into
(ax + r)(ax + s). This new expression multiplies out as a2x2 + axs + axr + rs or a^2x^2 + x(as + ar) + rs. Thus, we only need to find an r and s such that r + s = b (as + ar = ab) and rs = ac. We only need to find r and r.
Example: Factor: 6x^2 + 11x – 10
1. Multiply 6 by the -10 to create the trinomial x^2 + 11x – 60 (I call this the dummy trinomial. The silly name helps the kids remember that it is only temporary.)
2. (6x + )(6x - ) and then only need to find the numbers that multiply to – 60 and add to 11.
3. (6x + 15)(6x – 4) (still the dummy)
4. And finally: (6x + 15) /3(6x - 4)/2 = (2x + 5)(3x - 2)
We multiply ax^2 + bx + c by a to get ax^2 + bx + c and try to factor this realizing that after we are done we must divide out the factor of a we put in originally. Suppose this can be factored into
(ax + r)(ax + s). This new expression multiplies out as a2x2 + axs + axr + rs or a^2x^2 + x(as + ar) + rs. Thus, we only need to find an r and s such that r + s = b (as + ar = ab) and rs = ac. We only need to find r and r.
Example: Factor: 6x^2 + 11x – 10
1. Multiply 6 by the -10 to create the trinomial x^2 + 11x – 60 (I call this the dummy trinomial. The silly name helps the kids remember that it is only temporary.)
2. (6x + )(6x - ) and then only need to find the numbers that multiply to – 60 and add to 11.
3. (6x + 15)(6x – 4) (still the dummy)
4. And finally: (6x + 15) /3(6x - 4)/2 = (2x + 5)(3x - 2)
