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)#