Question

How do you factor `21x^2 + 55x + 14`?

Answer 1

It is `21x^2+55x+14=21x^2+49x+6x+14=7x(3x+7)+2*(3x+7)=(7x+2)*(3x+7)`

Answer 2

Factor `y = 21x^2 + 55x + 14`

Ans: y = (7x + 2)(3x + 7)

Explanation:

`y = 21x^2 + 55x + 14 =` 21(x + p)(x + q)
I use the new AC Method to factor trinomial (Socratic Search)
Converted trinomial `y'= x^2 + 55x + 294 = (x + p')(x + q')`.
p' and q' have same sign. Factor pairs of 294 --> (2, 147)(3, 98)(6, 49).
This sum is (6 + 49) = 55 = b. Then, p' = 6 and q' = 49.
Therefor, `p = (p')/a = 6/21 = 2/7` and `q' = 49/21 = 7/3.`

Factored form: `y = 21(x + 2/7)(x + 7/3) = (7x + 2)(3x + 7)`