How do you solve #3x^2+5x+2=0 #?

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Answer 1 · 2016-03-13T21:20:13

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

3 and 2 are both prime. Thus the only factors of them is 1 and 3 for 3; 1 and 2 for 2.

Assuming there are integer solutions for #x# then it has to be true that one of the following is true. If not, then it is down to the formula.

#(3x+-2)(x+-1) # as the first option
#(3x+-1)(x+-2)# as the second option

In the given question:
The constant 2 is positive so the signs in the brackets are the same.
The #5x# is positive so both signs are +

#color(blue)("Test 1")#

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

# =" " 3x^2+3x+2x+2#

#color(blue)( =" " 3x^2+5x+2)#