Factor by grouping the trinomial: $6x^2+x −1$
a)
$6x^2+ 3x −2x −1$
$3x(2x + 1) + 1(−2x − 1)$
$(3x + 1) (−2x− 1)$
b)
$3x(2x+1)−1(2x+1)$
$(3x − 1)(2x+1)$
Beginner in Algebra and I have a question on factoring by grouping trinomials, especially concerning the last parentheses, the subtraction/negative signs especially. I am getting things wrong on a regular basis and would like some advice. I'll try and state as clearly as I can. Also, and not really part of the question I guess but can't you group these in any way and still get the same answer? Or no?
But for the question. For the part labeled a), specifically $+ 1(−2x − 1)$ If done this way it gives me the opposite of the answer. Or does it? Or am I getting the signs mixed up and not remembering certain math rules in order to get this right?
For example, in part b, I can get the correct answer if I switch the signs and make the 1 a $-1$ and the other two values, $2x + 1$. But am I allowed to do that? It feels like I shouldn't have the freedom to just change those signs w/o somehow factoring them out in some way. Or am I making it too hard and this is possible?
Remember the distributive property, $ab+ac=a(b+c)$. For part b, $-1(2x+1)=-1(2x)+-1(1)=-2x-1$. Now that each addend has a common factor of $2x+1$, you can combine like terms, which is again using the distributive property.$a=2x+1,b=3x$, and $c=-1$.
For part a, does $c=2x+1$ or $-2x-1$? You cannot combine like terms as written. You could do $-3x(-2x-1)+1(-2x-1)$ in which case you have $c=-2x-1$. Combining like terms then yields $(-3x+1)(-2x-1).$ The third line in part a does not follow from the second. Try substituting $x=1$.