FACTORING PRACTICE: X-Method, Factoring by Grouping, Difference of Squares, Cross-Cancelling.

[@vespa2860]

As I have said, the factor by grouping was new to me. I now understand it, but still tend to use my old method. Not sure if any is quicker or easier. I simply remove the a coefficient First use it to multiply with the c ; as you do. And use the X method (again, as you do). Then simply reintroduce the a again as the denominator tin the two factors). eg: 2x^2 + 3x +1 becomes x^2 + 3x +2 Use the X method to find (x + 2) (x + 1) then reintroduce the a term (in this case 2) as the denominator in the pair i.e.: (x + 2/2) (x + 1/2) Simplify as far as possible (x + 1) or, if no simplification is possible, use the denominator as the coefficient of x: (2x +1)

[@helpwithmathing]

This is a "pre-coffee" reply, but I wonder whether this method would get tripped up when there is more than one possible way to split up the "a". Can you play with how your method would work with 4x^2 -4x -3 and report back?

[@vespa2860]

@@helpwithmathing First remove 4 and muliply it by -3 Leaves: x^2-4x-12 X method reveals -6 and 2 so (x-6) (x+2) then reintroduce the 4 as denominator in both, gives: (x-6/4) (x+2/4) simplify both as far as possible, gives (x-3/2) (x+1/2) When cannot reduce any further, use the denominators as their respective coefficients of x (2x-3) (2x+1) I think either method will work provided the X method is actually possible.

[@helpwithmathing]

@@vespa2860 Ah! I now understand something I didn't understand before. When you talk about "using the denominator as their respective coefficients of x" you mean to multiply the factor by the current reduced denominator of the factor, not by the original "a" you pulled out and then reintroduced as the denominator. I feel like I now want to make a video in a, b and c, showing why this works! Thanks for the alternative method!