Question

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

Answer 1

Answer:

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.

Answer 2

i like diveding so the most esest way is divedeing