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

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Answer 1 · 2015-05-28T20:27:12

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 $AC=6$ and whose sum is $B=7$ .

The pair $(6, 1)$ works: $6 xx 1 = 6$ and $6 + 1 = 7$ .

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

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

$=(2x^2-x)-(6x-3)$

$=x(2x-1)-3(2x-1)$

$=(x-3)(2x-1)$

How do you factor 2x^2-7x+32x^2-7x+3?