We are learning factoring by grouping - The teacher explained the process but didn't explain the logic behind it. You need to multiply the coefficient on the x-squared term by the constant to get a number. You then need to find two numbers which multiply to this number, and add to the co-efficient on the x-term. Then you split the x-term between these two numbers, group them, and factor them.
Why does this work?
On
The grouping technique you are referring to is the "reverse" of the distribution property.
Recall that the distribution property is as follows: $$a(x+y)=ax+ay.$$ we may make use of this the following way as well: $$(a+b)(x+y)=(a+b)x+(a+b)y,$$ Where we have treated (a+b) as one number, namely the sum.
Now it's only a matter of reversing this. Consider the following example: $$x^2+5x+6.$$ To do this we find two numbers that add up to 5 and multiply to be 6. Those numbers are 2 and 3. So we have: \begin{align*} x^2+5x+6&=x^2+2x+3x+6\\ &=x(x+2)+3(x+2), \end{align*} which, from above, we recognize to be $$(x+3)(x+2).$$
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Short answer: You have to remember, grouping is simply splitting the middle term so that it can be grouped:
Take the expression $ 3x^2+10x+8 $ The terms have no common factors, but if you just split the middle term right: $$3x^2+4x+6x+8$$ You can suddenly "group" the expression like such: $$x(3x+4)+2(3x+4)$$ Woah - Magic! Now we have $(x+2)(3x+4)$!
Long answer: Continuing off Gerry Myerson's answer:
Given the expression: $ ax^2+bx+c $ You could rewrite it as such: $(dx+e)(fx+g)$
Which proves the following is true (multiply it out): $$ ax^2+bx+c=dfx^2+(dg+ef)x+eg $$ What you want to do first is to find $dg$ and $ef$. Since you already know that $df = a$, and $eg =c$, all you have to do is factor $dfeg$, or $ac$, into $dg$ and $ef$ (remember, $dg+ef=b$). Once you have found $dg$ and $ef$, you can rewrite the expression as such: $$dfx^2+dgx+efx+eg $$ Now, you can "group" the equation into two factorable terms like so: $$dfx^2+dgx = dx(fx+g)$$$$ efx+eg=e(fx+g)$$ Yay! Now your expression is simply: $$dx(fx+g)+e(fx+g)$$$$=(dx+e)(fx+g)$$
So, you're given $ax^2+bx+c$, and you want to find $d,e,f,g$ such that $$ax^2+bx+c=(dx+e)(fx+g)$$ Multiply out the right side, and what we want is $$a=df,\quad b=dg+ef,\quad c=eg$$ The first step in the process gets you the number $ac$, which is $dfeg$. So you are trying to split $dfeg$ into two factors that add up to $dg+ef$. Those two factors are going to be $dg$ and $ef$. I'm not sure what you mean by "group them", but you factor them as $d\times g$ and $e\times f$, and then you have the numbers $d,e,f,g$ that you're looking for.