So I have a question that confuses me, hopefully, you guys could clear it up for me.

$$6x^2 + 7x + 1$$ Step 1: $$6x^2 + 6x + x + 1$$ Step 2: $$6x(x+1)+(x+1)$$ Now here comes my question, we have to factor it out so: $$(6x+1)+(x+1)$$ What mathematical property allows us to write $(6x+1$)?

I'm talking about the $1$ here.

I see that $6x(x+1)+(x+1)$ is the same as $6x(x+1)1<<(x+1)$ but this somehow puzzles me because I'm used to it being okay for multiplication since $1 * anything$ does not change anything. But later on we actually add them? As in $(6x+1 <<)(x+1)$

Maybe it's one too many for today but I'm really confused about this. Can someone see what I mean here?

• Did you leave out a plus sign in a couple spots? Nowhere is there a $6x(x+1)(x+1)$. – Mike Jan 3 '18 at 7:43
• Yeah I did, thanks. – user472288 Jan 3 '18 at 7:45

When you factor out terms, you are using the distributive property, $ab+ac=a(b+c)$. In step $2$, $6x$ was factored from each of the first $2$ terms $(a=6x)$. When $6x$ was factored from $6x$, what was left was $1$. If you understand that part of it, for the final step, factor out $x+1$ from the $2$ remaining terms. Just as you got $1$ when factoring $6x$ from $6x$, you also get $1$ when factoring $x+1$ from $x+1$.

It goes like this

$$6x^2 + 6x + x + 1 = 6x(x+1) + 1(x+1) = (6x+1)(x+1)$$

• yeah its just a mistake, will correct it but my question stands with the +1 that we're adding.. – user472288 Jan 3 '18 at 6:36
• That's just how factoring works. $ac + bc = (a+b)c$. Here we have $a = 6x$, $b = 1$ and $c = x+1$. Notice both terms have a common factor of $x+1$ – Dylan Jan 3 '18 at 6:37
• So anytime we don't have a $b$ other than 1 it's going to be 1? – user472288 Jan 3 '18 at 6:39
• The $1$ is implicit. $ac + c = (a+1)c$ for example – Dylan Jan 3 '18 at 6:40
• Argh not sure why I don't quite get it. You see, I thought the correct answer would be $(6x)(x+1)$ but we added the 1 here to make it $(6x+1)(x+1)$. Maybe I should just look at it tomorrow. – user472288 Jan 3 '18 at 6:43

Its the Distributive Property of Multiplication , a basic rule.

That is, $$a(b+c) = ab +ac$$

$$(6x+1)(x+1)$$ and you wanted to know from where that $1$ came into play.
$$(6x+1)\color{red}{(x+1)}=6x\color{red}{(x+1)}+1\color{red}{(x+1)}$$
So , you have distributed $(6x+1)$ over $(x+1)$ so what is its converse (or you can say,reverse)?
It is, $$6x\color{red}{(x+1)}+1\color{red}{(x+1)} = (6x+1)\color{red}{(x+1)}$$ Or we have just taken $(x+1)$ out as an common term.
Example: $$2\cdot \color{red}{5}+1\cdot \color{red}{5} = 5(2+1)$$