Factoring (Grouping Quadratic Equation)

Question

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?

Answer 1

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 $x+1$.

Answer 2

It goes like this

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

Answer 3

Its the Distributive Property of Multiplication , a basic rule.

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

In your question, your last (and confusing to you) step was,

$$(6x+1)(x+1)$$ and you wanted to know from where that $1$ came into play.

Okay, so you just multiply those two factors and see what comes,

$$(6x+1)\color{red}{(x+1)}=6x\color{red}{(x+1)}+1\color{red}{(x+1)}$$

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: $$ 2\cdot \color{red}{5}+1\cdot \color{red}{5} = 5(2+1) $$