# How do you factor 10z^2-17z+3?

May 23, 2015

$10 {z}^{2} - 17 z + 3 = \left(5 z - 1\right) \left(2 z - 3\right)$

Given that the constant term is positive and the middle term is negative, I was looking for a factorization like $\left(5 z - a\right) \left(2 z - b\right)$ with $a > 0$, $b > 0$ and $a b = 3$. That does not give many choices to try.

Of course there might have been a solution of the form $\left(10 z - a\right) \left(z - b\right)$, but the middle term being $- 17$ - close to $- 3 \times 5$ pointed to the first factorization to try.

May 23, 2015

Factor y = 10x^2 - 17x + 3 = (x -p)(x - q). I use the new AC Method.
Converted trinomial: x^2 - 17x + 3 = (x - p')(x - q')
Compose factor pairs of a.c = 30 -> (1, 30)(2, 15)> This last sum is 17 = -b. Then p' = -2 and q' = -15.
Next: p = p'/2 = -2/10 = -1/5, and q = q'/2 = -15/10 = -3/2
Factored form: y = (x - 1/5)(x - 3/2) = (5x - 1)(2x - 3).