How do you factor #4x^2+4x-3#?

2 Answers
smendyka
Jul 1, 2017

Playing with the factors for #4# and #3# gives:

#(2x - 1)(2x + 3)#

Truong-Son N.
Jul 1, 2017

The only front factors of #4# that look convincing are #2# and #1#. We assume a factorization of the form

#(2x pm ?)(2x pm ?)#

The #3# term is negative, so we have

#(2x + ?)(2x - ?)#

The factors of #-3# are only #pm3# and #∓1#. So, we have to decide whether it's #+3# and #-1#, or #-3# and #+1#.

Since the #4x# term is positive, the terms in "#?#" are likely #+3# and #-1# rather than #-3# and #+1#, allowing the positive term to dominate the sum of the outer and inner factors.

#=> color(blue)((2x + 3)(2x - 1))#

Verify that this is correct by multiplying this out again. We could have been wrong and we could have wrongly chosen the form

#(4x pm ?)(x pm ?)#...

This would not have been correct:

#(4x - 3)(x + 1)#

The middle term would not be #4x#, even though the first and last terms would have been correct.

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