How do you factor the trinomial #6a^2 +5a+1#?

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Answer 1 · 2015-11-21T17:56:34

#6a^2+2a+3a+1=(2a+1)(3a+1)#

#6a^2+5a+1# is a quadratic equation in the form of #ax^2+bx+c#, where #a=6, b=5, and c=1#.

Use the AC method of factoring.

Multiply #a# times #c#.

#6xx1=6#

Determine two numbers that when multiplied equal #6# and when added equal #5#. The numbers #2# and #3# meet the criteria.

Rewrite the equation with #2a and 3a# in place of #5a#.

#6a^2+2a+3a+1#

Group each pair of terms.

#(6a^2+2a)+(3a+1)#

Factor out #2a# from the first pair of terms.

#2a(3a+1)+1(3a+1)#
I'm showing the #1# in front of #(3a+1)# so that it is easier to see where #(2a+1)# comes from.

Factor out #(3a+1)# and rewrite.

#(2a+1)(3a+1)#