# How do you factor 12f² +35f +8?

May 20, 2015

$12 {f}^{2} + 35 f + 8 = \left(3 f + 8\right) \left(4 f + 1\right)$

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

The determinant of this quadratic is given by the formula:

$\Delta = {b}^{2} - 4 a c = {35}^{2} - \left(4 \times 12 \times 8\right) = 1225 - 384$

$= 841 = {29}^{2}$

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

$f = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 35 \pm 29}{24}$

That is $f = - \frac{8}{3}$ or $f = - \frac{1}{4}$.

These values correspond to the factors $\left(3 f + 8\right)$ and $\left(4 f + 1\right)$.

May 21, 2015

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 . $y = {x}^{2} + b x + c$ = (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 . $f \left(x\right) = a {x}^{2} + b x + c$ = (x - p)(x - q). Convert f(x) to $f ' \left(x\right) = {x}^{2} + b x + \left(a . c\right)$ = (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.