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

1 Answer
May 5, 2018

The answer is #2x^2+7x+3=(2x+1)(x+3)#

Explanation:

We notice that #x^2# has 2 in front, so we, therefore, want an expression of the form
#(2x+a)(x+b)#
If we multiply, we get #2x^2+(a+2b)x+ab#

The two should be equal term for term, i.e.
#a+2b=7#
#ab=3#

Now if #ab=3# and both are to be whole numbers (no fraqtions), then #a=1, b=3# or #a=3, b=1#
#a=1, b=3# gives #a+2b=1+2*3=7#, which fits
#a=3, b=1# gives #a+2b=3+2*1=5#, which doesn't fit.

Therefore
#2x^2+7x+3=(2x+1)(x+3)#