How do you factor and solve #3x^2+6x+3=0#?

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Answer 1 · 2015-05-23T09:25:12

#3x^2+6x+3=0#

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like #ax^2 + bx + c#, we need to think of 2 numbers such that:

#N_1*N_2 = a*c = 3*3 = 9#
and
#N_1 +N_2 = b = 6#

After trying out a few numbers we get #N_1 = 3# and #N_2 =3#
#3*3 = 9#, and #3+3= 6#

#3x^2+6x+3= 3x^2+3x +3x+3#

# = 3x(x+1) +3(x+1)#

#color(green)((3x+3)(x+1)# is the factorised form.

Answer 2 · 2015-12-17T00:31:51

#x=-1#

Factor a #3# out.

#3(x^2+2x+1)=0#

Realize that the inner portion is a perfect binomial square, equal to #(x+1)^2#.

#3(x+1)^2=0#

Now, to solve for #0#, realize that #(x+1)^2# must equal to #0# in order for the entire expression to be equal to #0#, since #3xx0=0#.

#(x+1)^2=0#
#x+1=0#
#x=-1#

You can check this by graphing the equation. The graph should touch the #x#-axis at #x=-1#.

graph{3x^2+6x+3 [-10.41, 12.09, -2.56, 8.7]}