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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?

Little Black Dog
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  • Hi, I understand how factoring works, because you need two numbers ( a and b) that add to a+b and multiply to ab... but I don't understand the grouping technique. – Little Black Dog Dec 03 '14 at 06:13

3 Answers3

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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: $$dx(fx+g)+e(fx+g)$$$$=(dx+e)(fx+g)$$

Nik Tedig
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sunny-lan
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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.

Gerry Myerson
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  • For those still confused like I was as to why we multiply a x c, it is not because of any kind of meaningful relationship between those two terms. It is simply because a x c happens to be the same as "b1" x "b2" as revealed by the above formulas. – Daniel Arant Jul 30 '23 at 00:00
  • @Daniel, let's go through an example: $2x^2+7x+6$. We have $df=2$, $dg+ef=7$, $eg=6$. $df=2$ is easy, we can just take $d=1$, $f=2$. That leaves us with $g+2e=7$, $eg=6$. Now what? Instead, $dfeg=ac=12$, $dg+ef=b=7$, so we want two numbers $dg$ and $ef$ that add up to $7$ and multiply to $12$. Easy! $dg=4$ and $ef=3$. Then $e=3$, $f=1$, $d=2$, $g=2$, done! (This was in response to an earlier comment, since deleted.) – Gerry Myerson Jul 30 '23 at 00:01
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    Yes, sorry. Shortly after posting my comment I realized why I was confused. I was looking for some sort of intuitive connection between a and b that would suggest that multiplying them will give us a constraint on dg * ef. It is simply because dg * ef is a * c by definition. Thanks for your response! – Daniel Arant Jul 30 '23 at 00:17
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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 $$(x+3)(x+2).$$

recmath
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