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It is my intent not to allow students to leave my algebra class not being able to factor trinomials. To that I end, I employ the “retest.”

Factoring comes in November and December in my (fairly traditional) algebra text. The results on our first factoring test were about as expected: 60% of the students scored 80 or better. I wish that it would be higher. I worked for it to be higher. Every year I adjust. I wonder if we start too easy. But answers don’t matter at this moment for the 40% who scored below our expectations.

They studied over vacation (no regular assignment), and took a retest when they came back. Now, this retest is tricky. I make it similar to the original test, but shorter, with the points only adding up to 90. I also pick a slightly easier subset of questions. The goal remains 80 points. So on the retest, 20% broke 80, and 20% were assigned more tutoring.

Friday we retested again (the actual questions are posted beneath the fold). And every last student qualified. Pheww.

Seems a bit cruel, I know. I combine teaching for understanding with teaching skills, but this test is all about skill. No way to reason yourself clear, just practice and repeat. But they gain a sense of accomplishment. It makes me feel good that they eventually succeeded. And now I can say “Yes [whatever new topic we are studying] is hard. But you are a student who learned to factor, which was much harder, so I know you can do it…”

Simplify. Assume that no denominator equals zero $\frac{(5xy^2z)^3}{20x^2y^6}$ $\frac{44a^5b^2c - 33a^6bc + 110a^7c}{11a^5c}$

Express each product as a polynomial
(4m – 3n)(4m + 3n) $(5g + h)^2$
(11x – 3)(x – 4)
w(3w + 1)(10w – 3)

Factor completely. If the polynomial cannot be factored, write “prime.” $25x^2 - 16y^2$ $3z^2 + 4z - 15$ $10a^2 + 20ab + 9b^2$ $2k^2 - 21k + 40$ $40p^2q - 70pq^2 + 25q^3$ $7u^2 - 28$ $15c^3 - 5c^2d + 6c - 2d$

Solve. $8p^2 + 2p - 3 = 0$
j(4j + 3)(j – 5) = 0

13 Comments leave one →
1. January 14, 2008 am31 9:13 am 9:13 am

They never gave us “prime” polynomials on factoring tests when I was in school. ;-) Is there a good way to confirm that a polynomial is prime, or do you just fail to find a factorization and eventually give up and decide it must be prime? Is one of the examples in the “factor completely” section prime?

2. January 14, 2008 am31 9:23 am 9:23 am

Delighted you asked! Watch this one: $16u^2 + 53u + 34$
We need to find two numbers with a product of 544 and a sum of 53. Concentrate on the sum (reverse what is usually taught)
3*50 =150
10*43 = 430
15*38 = 570 (go back)
13*40 = 520
14*41 = 574

Those last two products show that we will not be able to factor this. Notice, there are no numbers between 13 and 14 to try.

I semi-require this sort of work (actually, I write a nice note if they show this to me – most do)

3. January 14, 2008 am31 10:22 am 10:22 am

Ah, that’s cleaner than what I was doing. I was thinking in terms of pairs of numbers, two of which multiply to 16, and 2 of which multiply to 34, then looking at all the combinations. So ab=16 and cd=34, and ac+bd = 53. And then looking at possibilities for a,b and c,d, combining, etc. Which is ok since there weren’t *that* many possible combinations, but your method is certainly cleaner. :) (And in my notation, amounts to noting that ac * bd = acbd = abcd = 16 * 34)

Maybe I knew that when I was in HS (the part about combining the products to make finding the factors easier). We did a fair bit of factoring, though like I said, I don’t think they ever gave us primes. There were always factors, we just had to find them. It seems to me that recognizing when there aren’t factors is pretty useful!

Of course in desperation, you could always plug into the quadratic formula and see if you get back anything useful (i.e. rational) and work back from there ;-)

4. January 14, 2008 pm31 8:58 pm 8:58 pm

Not an easy test.

I find that kids get confused when presented with the different kinds of factoring all together.

Is this for your 9th grade algebra class?

5. January 15, 2008 am31 3:41 am 3:41 am

No, not easy. But consider this a sort of honors class. And we teach it with some depth.

6. TRMilne permalink
January 17, 2008 am31 7:56 am 7:56 am

To put it in another light, reduce to prime factors …
16*34 = 2*2*2*2*2*17
Work that in combination with the “sum=53” part. You’ll need some combination of the prime factors so you don’t need to try 3*40 or 14*41, both of which contain factors that can’t be gotten from the list there.

In other words, only 2*272, 4*136, 8*68, 16*34 are the only candidates that need trying. If they don’t work, nothing will.

7. TRMilne permalink
January 17, 2008 am31 7:58 am 7:58 am

This is based on an alternate method of factoring that not all may have encountered. I’ll try to webify it and post a link at some point.

8. TRMilne permalink
January 17, 2008 am31 8:28 am 8:28 am

and 32*17

9. JBL permalink
January 17, 2008 pm31 9:17 pm 9:17 pm

There are some criteria for checking factorability of polynomials, e.g. Eisenstein’s Criterion. I don’t know any general approaches that work better than “check every reasonable possibility,” though.

Trickiest factoring problem I know: factor $x^4 + 4y^4$.

10. January 18, 2008 am31 1:53 am 1:53 am

JBL,

you get a feel for those after a while. For me, the add and subtract aspect jumps out, but that’s probably just experience.

11. TRMilne permalink
January 18, 2008 am31 2:10 am 2:10 am

Here’s that factoring method I promised. Hastily put into html but here goes …
http://www.trmilne.com/math/algebra1/extras/factoring.html

12. January 18, 2008 am31 3:30 am 3:30 am

TR, you demonstrated factoring: $12x^2 - 32x - 35$

I would multiply 12 and -35 = -420, and look for 2 numbers that added to -32 and multiplied to make -420. Sounds familiar, right?

A tip: don’t bother factoring 420. Instead, look at this list:
5 * 37 = 185
6 * 38 = 228
7 * 39 = 273
8 * 40 = 320
9 * 41 = 369
10*42 = 420
We can find the numbers mindlessly and quickly (works with your method as well).

Here we diverge: $12x^2 - 32x - 35 =$ $12x^2 - 10x + 42x - 35 =$ $2x(6x - 5) + 7(6x - 5) =$ $(6x - 5)(2x + 7)$

I submit that breaking the middle and then factoring out GCFs twice rests on an existing skill set without asking students to memorize as much procedure.

Of course, that’s just my opinion.

13. January 19, 2008 am31 7:02 am 7:02 am

JBL,

I found that problem in my algebra I book… (Dolciani), albeit C level and following several suggestive “helper” problems.

Jonathan