# How do you factor 2x^2-7x+3?

May 28, 2015

I'll use a version of the AC Method...

$A = 2$, $B = 7$ and $C = 3$ are the coefficients, ignoring any signs.

Since the sign of the constant term is $+$, we look for a pair of numbers whose product is $A C = 6$ and whose sum is $B = 7$.

The pair $\left(6 , 1\right)$ works: $6 \times 1 = 6$ and $6 + 1 = 7$.

Then use the pair we found to split the middle term into two and factor by grouping:

$2 {x}^{2} - 7 x + 3 = 2 {x}^{2} - x - 6 x + 3$

$= \left(2 {x}^{2} - x\right) - \left(6 x - 3\right)$

$= x \left(2 x - 1\right) - 3 \left(2 x - 1\right)$

$= \left(x - 3\right) \left(2 x - 1\right)$