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(Here’s the older posts: 1 2 3) $6b^2 - 9bk + 10bc - 15ck$ Now we’ve spent a day factoring sort of thing
3b(2b – 3k) + 5c(2b – 3k) Group the terms, pull a common factor from each.
(2b – 3k)[3b + 5c] Pull the binomial common factor forward.

(I use the square brackets for the visual effect, and clearly tell the kids that they are exactly the same as regular parentheses.) $12y^2 + 3y + 200y + 50$ factor that!
3y(4y + 1) + 50(4y + 1)
(4y + 1)[3y + 50] $12a^2 + 203a + 50$ How hard is this? Not very. Notice that 203a can be ‘broken’ into 3a + 200a. But how do we know to use 3 and 200, and not, for example, 13 and 190? Well, we just saw it done. (With different letters. Lots of kids don’t immediately see that they are the same problem. If you survive on x alone, consider jazzing it up…) But consider another: $88y^2 + 247y + 120$ We’re not running 124 trials (with errors) to see what works. We need to step back for some theory. (below the fold —>)

Consider (ax + by)(cx + dy) = $acx^2 + adxy + bcxy + bdy^2 =$ $acx^2 + (ad + bc)xy + bdy^2 =$

Here’s the trick. We have ad + bc in the middle. We’d like to find ad and bc separately. If we knew their product, we already know their sum, we could do a quick search. But we do know their product. Multiply the first and last coefficients: ac*bd = abcd, the same as the product of ad and bc. You’ll reread this paragraph, since it’s never clear until you’ve done a few. Here’s some with numbers: $10g^2 + 33g + 9$ So how do we break 33 up? We find two numbers that add to 33 and multiply to make 10*9 = 90. No thinking here, they are 30 and 3.

For consistency we’ll name the two numbers we are considering. “p” and “s” are widely used for the product and sum of the roots of a quadratic, so we’ll use as alternates the Greek letters pi and sigma, “Π” and “Σ” (I’m not wildly in love with this, but I got it from Steve Konrad, who gave me the rest, and the kids like the odd symbols, so I’ve stuck with it).

Now let’s look at $12a^2 + 203a + 50$. Π = 12*50 = 600 and Σ= 203. Can we find two numbers that add to 203 and multiply to make 600? Sure, 200 and 3. Does it matter which order? No. Watch: $12a^2 + 3a + 200a + 50$
3a(4a + 1) + 50(4a + 1)
(4a + 1)[3a + 50], or… $12a^2 + 200a + 3a + 50$
4a(3a + 50) + 1(3a + 50)
(3a + 50)[4a + 1], (the 1 helps to see…)

Only one more for this post. $88y^2 + 247y + 120$ Π = 88*120 = 10,560 and Σ= 247.

Here I differ from Steve. Instead of searching the factors of 5280, we can quick search for sums of 247, moving apart if the product is too high, moving together if it is too low

147*100 = 14,700 too high
197*50= 9,850 too low
187*60 = 11,220 too high
190*57 = 10,830 too high
192*55 = 10,560 good, so: $88y^2 + 247y + 120$ = $88y^2 + 55y + 192y + 120$ =
11y(8y + 5) + __(8y + 5) (it’s got to be 8y + 5)
11y(8y + 5) + 24(8y + 5) =
(8y + 5)(11y + 24)

Ask questions, if you like. The next post will have some minus signs, more examples, and a link to a web site where they explain 90% of this nicer than I could. And with good visuals.

5 Comments leave one →
1. Jackie permalink
September 16, 2007 pm30 5:54 pm 5:54 pm

Love the pi & sigma “trick”. I try to avoid using “s” as it looks like a 5 to too many kids. Great idea – thanks!

2. September 17, 2007 pm30 6:46 pm 6:46 pm

Thanks for the trick. I like it!
You have a typo, I think, in the “Here I differ…” paragraph. Don’t you mean, “sums of 247”?

3. September 17, 2007 pm30 8:22 pm 8:22 pm

Right you are. I started with another example, and was sloppy when I changed it.

4. Anonymous permalink
March 23, 2013 am31 8:40 am 8:40 am

Great to see this blog. I have been teaching factoring using this method (I call it by grouping) for a few years and have begun to wonder if I should go back to the guess and check method many people learned and what the book outlines. Nice to see some other support for what I have been doing.

I like how you teach finding the GCF that is a binomial before the factoring. I had not been doing that and will add it this year.

Thanks for your posts.