# Everything I Learned In Eighth-Grade Math

My title is an exaggeration. In eighth grade Prealgebra I learned many things, but I confess that I didn’t learn well from that particular teacher that particular year. What I most clearly remember learning I picked up from a substitute who filled in a few weeks. It’s a method for factoring quadratic expressions into binomial expressions, and I must admit, it’s not very good. It’s cumbersome and totally useless once one knows the quadratic equation. But it’s fun to do, and I liked it a lot, and I’ve never seen it described as a way to factor quadratic expressions. So let me put it on the web and do what I can to preserve its legacy, and get hundreds of people telling me what it actually is and how everybody but the people I know went through a phase of using it.

It’s a method which looks at first like it’s going to be a magic square, but it’s not, and I’m at a loss what to call it. I don’t remember the substitute teacher’s name, so I can’t use that. I do remember the regular teacher’s name, but it wasn’t, as far as I know, part of his lesson plan, and it’d not be fair to him to let his legacy be defined by one student who just didn’t get him.

The first important thing we need is a quadratic polynomial, and I’ll take a specific one, here, $10 x^2 - 7x - 12$. The method was taught when we’d use just integer coefficients, and would want to factor into binomials with integer coefficients. In principle it would work fine with decimals, but if you’re doing that there’s really no reason to use this in preference to the quadratic formula. It does go all right if you have roots or radicals as coefficients, though.

The second thing needed is a three-by-three grid. It will look like a magic box, but don’t worry about things like sums across rows being equal. I’ve added letters to identify rows and columns, which aren’t part of the scheme, but make it easier to describe what I’m doing.

 i j k A B C

Into the first row go the first and last terms of the polynomial. That is, in row A, column i, goes the $10 x^2$ term. In row A, column j, goes the $-12$ term. And that’s all that we use from the polynomial for right now. We’ll come back to the $-7 x$.

One rule of this pseudo magic square is that, within reach row, the thing in the first column times the thing in the second column equals the thing in the third column. That is, into row A, column k, will go the product of row A column i and row A column j: $10 x^2 \times -12 = -120 x^2$. So here’s the table now:

 i j k A $10 x^2$ -12 $-120 x^2$ B C

Now for the second rule of this pseduo magic table: within each column, the third row and the second row multiply together to make the first row. That is, row C column k times row B column k will equal row A column k.

The next step in this is going to be factoring $-120 x^2$. There are, in principle, an infinite number of possible factorings. This number is cut down a little bit if we assume we’re working on integer coefficients, which is why this is a good method in prealgebra and rubbish once you know the quadratic formula. (And it again works fine if you have a coefficient-times-a-radical, and are reasonably sure the radical is part of the binomial factorization.)

More specifically, though, we’re looking for a factorization of $-120 x^2$ into exactly two terms which added together give us the middle term in the polynomials: they have to add together to make $-7 x$. So the two terms are going to be some number times x and some (possibly other) number times x. We have to find those numbers.

This is the part that’s fun, by the way, at least if you like factoring numbers. You need two integers which multiplied together are -120 and which added together are -7. That’s a word problem in itself, but if you’re not asked to do word problems you can just start reeling off factors and see if any promising combinations turn up: 1 and -120 (adds to -119, no good); 2 and -60 (adds to -58, closer but still bad); 3 and -40; 4 and -30; 5 and -22; 6 and -20; 8 and -15 and oh look at that. Isn’t that wonderful? (If we had started off on -1 and 120, we’d probably notice that -8 and 15 gives us positive 7 and then we can just swap the minus from one to another.)

This step requires a little awareness of positive and negative products, by the way. If row A, column k is positive, then it’s possible the factors are both positive or are both negative. If row A, column k is negative, then one factor has to be positive and one negative.

But we’ve got it now: $-120 x^2 = (8x) \times (-15x)$, and $8x + (-15x) = -7x$. So into column k, in rows B and C, go $8 x$ and $-15 x$. It doesn’t matter which one you put where. I tend to put the larger number (in magnitude) up top, but you don’t have to if you don’t want to. This makes the grid now:

 i j k A $10 x^2$ -12 $-120 x^2$ B $-15 x$ C $8 x$

Next comes filling in the two rows, and this is another fun little puzzle. The elements in row B, column i and row B, column j have to multiply together to make row B, column k. Simultaneously, the elements in row C, column i, and row B, column i, have to multiply together to make row A, column i. And if that’s not enough, the elements in row C, column i, and row C, column j, have to multiply together to make row C, column k, while at the same time the element in row C, column j times the element in row B, column j, has to equal row A, column j. I realize you’re now all lost. Let me actually do the work; that’s easier.

Let me take row B, first. The thing in column i times the thing in column j has to equal $-15 x$. Well, the x goes in column i; the x always goes in column i, or else the products of column i terms couldn’t have an $x^2$ in it. The minus sign is easy for this problem; it goes into column j, or else the products in column j wouldn’t get to -12. As for factoring 15 … well, if we’re sticking to integers, that’s either 1 and 15 or 3 and 5. 15 won’t multiply by any other integers to give us 10 or 12, so we can rule that out. Either row B, column i gets the 3 or the 5. 3 times any integer isn’t going to give us the 10 that’s in row A, column i; and 5 times any integer isn’t going to give us the 1 that’s in row A, column j. So we must mean to put into row B, column i, the $5 x$, and into row B, column j, the -3. My grid is now:

 i j k A $10 x^2$ -12 $-120 x^2$ B $5 x$ -3 $-15 x$ C $8 x$

Now to finish off row C. Again, the element in row C, column i multiplied by the element in row C, column j, must be $8 x$; and the x goes into row C, column i. 8 can be factored into 1 times 8, or 2 times 4, and that’s about it if we’re limited to pairs, which we always are in this method.

Looking up the rows tells us whether the 2 or the 4 goes into row C, column i: two times five is ten, so $2x \times 5x = 10x^2$ and so into row C, column i, goes $2x$ while into row C, column j, goes 4. And as promised, -4 times 3 is -12. So now the table is:

 i j k A $10 x^2$ -12 $-120 x^2$ B $5 x$ -3 $-15 x$ C $2 x$ 4 $8 x$

And now for the final step. The first term in the binomial factoring of the polynomial is going to be the element in row B, column i, plus the element in row C, column j: $5x + 4$. The second term in the binomial factoring is going to be the element in row C, column i, plus the element in row B, column j: $2x - 3$. (In class we drew diagonal bubbles around these; I don’t know any practical way to do that in HTML or LaTeX.) That’s our factorization.

And sure enough, if we check, $(5x + 4) \times (2x - 3) = 10x^2 - 7x - 12$. It’s also the answer we get using the quadratic formula and then fiddling around to make sure we have the scaling right.

Obviously, the quadratic formula is superior on all counts — it’s quicker, it’s easier to remember all the steps involved, it doesn’t get awkward if you have to work with decimals, it takes less space — except that this lets you make pretty pictures and fill them with numbers. And maybe it helps ease you into the quadratic formula.

(And obviously it works if we have something that’s a quadratic in form, even if it has higher order powers: $10x^4 - 7x^2 - 12$ uses exactly the same routine, except we would write $x^2$ instead of x in rows B and C.) ## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

## 10 thoughts on “Everything I Learned In Eighth-Grade Math”

1. educationrealist says:

Is there any reason why you wouldn’t use the generic rectangle method instead? I find this method far more convoluted, and it covers the same material.

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1. Joseph Nebus says:

Well, I don’t know the rectangle method, at least not by that name.

I can’t defend this as a factoring method on any grounds, really, except that it’s how I first learned to do factoring systematically (or semi-systematically), and I hadn’t wanted it to be completely lost to time.

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1. educationrealist says:

No need to defend it. I think a method is essential for lower ability kids, and wasn’t criticizing you at all. I know teachers who use this method in preference to the rectangle (also known as box and diamond) and have just never been sure why, as it seems to have a bit more complexity with no added benefit. I thought maybe you could shed some insight.

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1. Joseph Nebus says:

I hadn’t encountered the box-and-diamond method, as best as I know, before, but now that I know what to search for it does look like a more straightforward method. Why that wasn’t what we got back in middle school I couldn’t guess; maybe it was just how my substitute was first taught factoring.

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1. educationrealist says:

If you’re interested, I have a long and short doc that takes students through the procedure. It’s a step by step document you might find helpful.

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1. Joseph Nebus says:

I am interested, yes, and would appreciate it. Thank you.

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2. Blinky the Wonder Wombat says:

After determining the two x terms, I’ve found it easier to just insert them into the original equation thus:

10x^2 -15x+8x-12

Look for a common factor in the first term and one of the two middle terms, in this case, 5x:

5x(2x-3) + 8x-12

Now find a common factor in the other two terms, in this case, 4:

5x(2x-3) +4(2x-3)

Hey look, (2x-3) is common to both sides! Factor it out and get:

(2x-3) * (5x+4)

I found that this method seemed a little more logical to my children.

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1. Joseph Nebus says:

Mm, yes, looking for the common factor between the first term and either of the intermediate ones does work out. That might be easier for children to learn. It does avoid another two rounds of guess-the-factoring for the horizontal rows.

‘Course, it does leave the rest of the little boxes un-filled, and that seems like a shame or an invitation to tic-tac-toe. But there’s always something lost in adapting methods, isn’t there?

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