Question

How do I factor 3x2+4x=1 and #x2-2x-8

Answer 1

thats the same as 6+4x=1
take out a 2 : 2(3+2x)=1
divide each side by 2: 3+2x=(1/2)
take 3 from each side: 2x= (-5/2)
then divide each side by 2 to get x:
x= -5/4

Answer 2

I have an impression that the character x stands for the variable, and not a multiplication sign. Also, the equal sign is meant to be a + sign, and the # is meant to be a three (just a guess).

The question should then be:
"How do I factor 3x2+4x+1 and 3x2-2x-8?"

1.
3x²+4x+1
From 3x² look for a pattern, since the only factors of three are 1 and 3.
(3x+a)(x+b)
Since a*b=1, a=1 and b=1 or a=-1 and b=-1.
Since the second term is +4x, we conclude that a=1 and b=1.
Try
(3x+1)(x+1)

2.
3x2-2x-8
Proceed as in #1,
(3x+a)(x+b)
Now, since the constant term -8 has multiple factors (1,2,4 and 8), we have to make more trial combinations that make a+3b=-2 (coefficient of -2x) and a*b=-8 (constant term).
Make a table of a,b, a+3b, a*b and calculate, using possible values of a and b as &plusmn(1,2,4,8). Notice that since a+3b=-2, at least one of a or b has to be negative, also b=-8/a
a b a+3b a*b
1 -8 -23 -8
2 -4 -10 -8
4 -2 -2 -8 (we found the combination).
So the factors are:
(3x+4)(x-2)
check:
(3x+4)(x-2)=3x²-2x-8 OK.