How do you factor #a^2+4a+3#?

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Answer 1 · 2018-03-12T19:59:43

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

#"using the a-c method of factoring"#

#"the factors of + 3 which sum to + 4 are + 1 and + 3"#

#rArra^2+4a+3=(a+1)(a+3)#

Answer 2 · 2018-03-12T20:01:12

#(a-1)(a+4) = 0#

We need to find two numbers that add to #4# and multiply to #3#

#color(white)(.) + 4#
#color(white)(.) xx 3#
. . . . . . .
#1 xx 4# #=> -1 + 4 = 3#

So our factors are #(a-1)(a+4) = 0#

#a = 1, -4#