Question

How do you factor 12f² +35f +8?

Answer 1

`12f^2+35f+8 = (3f+8)(4f+1)`

To find this, notice that `12f^2+35f+8` is in the form `af^2+bf+c`, with `a=12`, `b=35` and `c=8`.

The determinant of this quadratic is given by the formula:

`Delta = b^2-4ac = 35^2-(4xx12xx8) = 1225-384`

`= 841 = 29^2`

Being a perfect square, the quadratic has zeros for rational values of f given by the formula:

`f = (-b+-sqrt(Delta))/(2a) = (-35+-29)/24`

That is `f = -8/3` or `f = -1/4`.

These values correspond to the factors `(3f+8)` and `(4f+1)`.

Answer 2

I use the new AC Method (Google, Yahoo Search) to factor trinomials.

y = 12x^2 + 35x + 8 = (x - p)(x - q)
Converted trinomial: (a.c = 96)
y' = x^2 + 35x + 96 = (x - p')(x - q'). Find 2 numbers p' and q' that product = ac = 96 and sum = b = 35.
Compose factor pairs of 96 ->(2, 48)(3, 32). This sum is 35 = b.
Then, p' = 3 and q' = 32
Then, p = p'/a = 3/12 = 1/4, and q = q'/a = 32/12 = 8/3

Factored form: f(x) = (x - p)(x - q) = (x + 1/4)(x + 8/3) = (4x + 1)(3x + 8)

Summary of the new AC Method .
CASE 1 . `y = x^2 + bx + c` = (x - p)(x - q). Find p and q knowing product = c and sum = b. Compose factor pairs of c. Find the pair whose sum equals to b.
CASE 2 . `f(x) = ax^2 + bx + c` = (x - p)(x - q). Convert f(x) to `f'(x) = x^2 + bx + (a.c)` = (x - p')(x - q'). Find p' and q' exactly like Case 1 . Compose factor pairs of (a.c) and find the pair whose sum is (b), or -b. Next find p = (p')/a and q = (q')/a.

This new AC Method is fast, systematic, no factoring by grouping, and no solving binomials.