# explain factoring quadratic equations by grouping to me like I'm an idiot

So somehow I tested into intermediate algebra and I’m doing pretty well except for this factoring quadratic equations by grouping thing. Could someone either explain it to me like I’m an idiot or point to an idiot level explanation?

Here’s were I’m at. Here’s a couple of examples. One I managed to solve, just not the right way, and one that left me pretty freaken confused.

The one I managed to solve on my own:

1/10R² - 5/2= 0

x10 to make everything whole. This left me with:

R² - 25 = 0

Looking at it now it was pretty obvious R = |5|, but algebra doesn’t have an intuition operator so I worked out this set of steps to show R = |5|:

+25 on both sides

R² = 25

Now I just have to squareroot it:
sqrt(R²) = sqrt(25)

The squareroote of R² is R, and 25 is 5 and -5.

Therefor R= -5, or 5

However the factoring by grouping way I scribbled down when someone asked in class:

1/10R² - 5/2= 0

x10 again, makes sense

R² - 25 = 0

Now here is where it gets crazy somehow we end up with:

(R+5)(R-5) = 0

Well obviously one of those is 0 since you can only get 0 from multiplication if 0 is one of the items being multiplied. Therfore R = -5, or +5, but where in the fire caves did (R+5)(R-5) = 0 come from?

3y² + 11y = 0

The steps were:
3y² + 11y = 0
(y+4)(3y-1) = 0
Y = -4, 3y-1= 0

3y = 1
Divide by 3
Leaving us with:
Y= 1/3, and y = -4

I understand everything but the parentheses sets. Where do they come from? My attempts to figure this out say I need to find the greatest common factor, but what do I do with greatest common factor when I find it and where did the – 4 come from? It doesn’t seem to be a factor of anything. I’ve tried googling, reading, no go. Help?

freaken edit window.

That second equation should be:

3y² + 11y -4 = 0

not

3y² + 11y = 0

I misread it (my handwriting is terrible).

Just to make sure we’re clear where we’re starting from, do you see why

(R + 5)(R - 5) = 0

is equivalent to

R[sup]2[/sup] - 25 = 0

?

(R[sup]2[/sup] - 25) is equivalent to (R - 5)(R + 5) because if you multiply it out you get (R[sup]2[/sup] - 25)

Watch:
(R - 5)(R +5)
=R[sup]2[/sup] + 5R -5R -25
=R[sup]2[/sup] - 25

Finding the factors is often a job for intuition or trial-and-error, but once you have them, you can check them easily. It’s worth remembering, though, the formula for the difference of two squares, since that one shows up often: A^2 - B^2 = (A+B)(A-B). This is the one used in your R^2 - 25 example.

You’re just concerned with finding a factor, not necessarily a greatest factor. As for the -4, in a previous step you had converted your equation to (y+4)(3y-1) = 0. At least one of those two factors must be zero, since the product is zero, so you either have y + 4 = 0 or 3y - 1 = 0. If it’s 3y - 1 = 0, then you get y = 1/3, and if it’s y + 4 = 0, then you get y = -4.

When you have the difference of two squares (say x[sup]2[/sup] - a[sup]2[/sup]), you can always factor it into two first-order binomials, meaning (x+a)(x-a). To see that this works, you can undo it by applying FOIL:

(x+a)(x-a) = xx - ax + ax - aa = x[sup]2[/sup] - a[sup]2[/sup].

Does that make sense? Note that the +ax and -ax add to zero so they go away.

So, since R[sup]2[/sup] - 25 is the difference of two squares, you get (R+5)(R-5) as the factors. If you’re ever unsure, just apply FOIL to make sure it’s right.

As you note, if you have ab = 0, then either a or b must be zero. So if we assume (R+5) is zero, then R = -5. If (R-5) is zero, then R = 5. And those are our two solutions.

That’s an ugly one because it wants to trick you. 3 and 11 are prime numbers, so it looks like it doesn’t want to be factored.

Remember the general pattern: when you foil (a+b)(c+d), you end up with ac + ad + bc + bd. If the expression you’re factoring is in the standard order, then a and c are your variable terms and b and d are constants. So ac will be your second-order term (variable squared) and ad and bc will be your first-order terms (variable not squared.) The trick is that you have to make ad and bc add up to the original first-order term.

The factoring above works because when we multiply a and d, we get -1y, and when we multiply b and c, we get +12y. The sum of those is 11y.

So here’s how you get there: Remember that factors a and c have to multiply to get our second-order term. Since you can’t factor 3 (it’s prime) the only thing you can factor out of it is one of the y’s. So that leaves you with
(y )(3y )

The next step is to pick two integers that will cause us to get our middle ad and bc terms to add up to 11y. We have to multiply one of them by 3y. As luck would have it, 4 * 3y gets us to 12y, which is close:

(y + 4)(3y )

Now we need our ad term to subtract one y. That’s easy since a is already one y, so using a -1 flips the sign:

(y + 4)(3y - 1)

So that’s the process. I don’t know quite how to explain it as a generalized process other than to say you have to pick and choose stuff to make the FOIL work. Sometimes this involves brute force for me, but if you practice it you get an intuition for it.

Since if ab = 0 either a or b must be zero, then our answers to the equation are y + 4 = 0, so y = -4, and 3y - 1 = 0, so 3y = 1 and y = 1/3.

On second thought, that results in an extra -4 hanging around. I don’t think that answer makes sense.

As friedo says,

3y² + 11y = y * (3y + 11)

3y[sup]2[/sup] + 11y - 4 = (y + 4) * (3y - 1)

It’s unclear where the mistake in writing this came from, but most likely, the OP meant the second line, and simply forgot the “- 4” in the original problem.

Huzzah. My world is whole once more.

It’s a standard form for stating a quadratic equation as a set of factors. The zero should be self explanatory as it is the right hand side unaltered from the previous equation. As for the left hand side, any quadratic should have two factors of the form

(r[sub]1[/sub] - x[sub]1[/sub])(r[sub]0[/sub] - x[sub]0[/sub]) = 0.

and in general for an equation of degree n you have:

(r[sub]n[/sub] - x[sub]n[/sub])(r[sub]n-1[/sub] - x[sub]n-1[/sub])…(r[sub]1[/sub] - x[sub]1[/sub])(r[sub]0[/sub] - x[sub]0[/sub]) = 0.

There will be n solutions which may or may not be distinct; and obviously those numbers will be the various values of r, because they are the ones that will result in a zero product. I don’t know if this is still the case, but in older algebra texts it was common to speak of the “zeros” of an equation interchangeably with its “solutions”.

Ah, right. Sorry, didn’t see this before. :smack:

Let me work through an example the way I was taught. Let’s try

x²-17x+72=0

We want to factor that into (x-k)(x-m)=0. So k times m must equal 72. What pairs are potential factors of 72 (and thus are potential values for k and m)? The choices are (1, 72), (2, 36), (3, 24), (4,18), (6,12), and (8, 9). Which choice is right? Well, we also know that k+m must equal 17, and the only pair that fits that is (8,9). So

x²-17x+72=0 is equivalent to (x-8)(x-9)=0, and so x is 8 or 9. Testing that out we get

64-136+72=0 which is true and 81-153+72= 0 which is also true. Success. So for x²-ax+b=0 the method is to find all the integer factor pairs that equal b when multiplied together and equal a when added together. It takes a little time to work through the possibilities, but not that long.

I get it!

That is basically an abstract version of how we all did multiplication by hand in the third grade. It’s not the vulcan mental calisthenics it appeared to be. It finally clicked.

Via the distributive property we can prove 4²-2² = (4-2)(4+2)

(4-2)(4+2)

4² + 42 + -24 + -(2²)

Which simplifies to
4²-2²
So to surmise:

(x-y)(x+y) = x² + xy + -xy + -(y²) = x²-y²

I think I’m getting it!

Thanks everyone! All your posts really are helping. My head feels about like a square peg after a journey through a round hole, but that’ll pass. I had the same feeling after learning subnetting.

Also friedo, sorry about the reality altering typo.

I’m gonna get some sleep and come back and reread things. Thanks again!

The better you understand multiplication (including the famous FOIL method), the better you’ll understand factoring. Most algebra classes cover multiplcation (of polynomials) first, then factoring polynomials, then solving quadratic equations by means of factoring.

By the way, I’m trying to decide whether or not to nitpick the phrase “factoring quadratic equations.” Technically, what you’re factoring is not the equation per se, but the polynomial that makes up one side of the equation. But I don’t know whether this is an important distinction or not.

Yes, exactly. That’s a great insight to have. Third grade multiplication is just multiplication of polynomials along with treating the variable as 10 in the end; what you’re doing now is just the same thing, only in some ways actually cleaner and easier (for not having that last bit). And, as you saw for yourself, all it is is the distributive property. Seems like you’ve got it down quite well now.

I’ve recently learned about algebra tiles, which are used to (among other things) show how quadratic equations relate to the area of rectangles and squares. For example, consider 3x[sup]2[/sup]+13x+4 (which factors to (3x+1)(x+4) ). You would have:
[ul]
[li]Three square tiles with side length x (i.e., 3 x-by-x tiles)[/li][li]Thirteen rectangular tiles with side lengths 1 and x (i.e., 13 1-by-x tiles)[/li][li]Four square tiles with side length 1 (i.e., 4 1-by-1 tiles)[/li][/ul]
You can then shift these tiles around to form a rectangle which is (3x+1)-by-(x+4) – the act of forming a rectangle is the same as factoring.

You can read some about algebra tiles here and here. There’s an online applet that lets you practice factoring here

Wow! These are a fantastic tool. I bet I could get a lot better at factoring by visualizing these things instead of brute-forcing it.

And very easy to make yourself. My geometry students would make them with straightedge and compass for my algebra students. Hint: make the length of x = sqrt(6)

Although algebra is pretty timeless, I wouldn’t get too excited about 5 year old zombies.