In your opinion, what is the easiest method to use to factor and solve a quadratic function? What is the most challenging?

i like diveding so the most esest way is divedeing

See below.

Step-by-See below. step explanation:

If the coefficient of x^2 is 1 then you can quite easily factor it by looking at the factors of the last (constant) term.

For example;

Factor x^2 - 2x - 3.

We need 2 numbers whose product is the last term (-3) and whose sum is the coefficient of x (-2).

Now - 3 times + 1 = - 3 and - 3 + 1 = - 2, so the 2 numbers are - 3 and + 1 (remember to include the signs),

Now we simply place the - 3 and + 1 after the x 's in the 2 parentheses:

= (x - 3) (x + 1) and these are the factors.

The most challenging is when we have a coefficient of x^2 greater than 1.

For example

f (x) = 4x^2 - 4x - 3.

I find the 'ac' method easier in these cases.

Lets factor the above:

The 'a' in ac refers to the coefficient of x^2 which is 4 in this case and the 'c' is - 3.

Now ac, that is a times c = 4*-3 = - 12.

Now we need 2 numbers whose product is - 12 and whose sum = - 4 (that is the coefficient of x in f (x).

These 2 numbers are - 6 and + 2, so we write the function as follows, replacing the middle term - 4x by + 2x - 6x:

F (x) = 4x^2 + 2x - 6x - 3

Now we should be able to factor this by grouping, taking 2 pairs and factoring each of them.

f (x) = 2x (2x + 1) - 3 (2x + 1)

(2x + 1) is common so the factors are:

f (x) = (2x - 3) (2x + 1).

Note if you had replaced the - 4x by - 6x + 2x (the other way around) you wouldn't be able to get the common factor so you would just reverse them and try again.