Help with Factoring: Trinomials

(A new question of the week)

I recently had a pleasant discussion of factoring, with the kind of student for whom I returned to teaching: one who has been away from math for a while, and with greater maturity has the determination to succeed. We’ll see several examples of the “ac-grouping” method of factoring a trinomial, within a larger problem simplifying a rational expression.

A product of rational expressions

The question came from Brittney in early February:

Good morning,

Thank you for this service! I really need it.

To begin, I should state that I was placed into College Algebra by mistake. I am to graduate with my Associate of Arts in May. So failure isn’t an option. I haven’t taken math in 19 years. I’m 35. I haven’t learned a “system” yet. I have literally taught myself and have seen a tutor once or twice a week. I’m behind!

My first question is: How do I solve a problem like the following: (I am specifically having trouble with factoring when there is a first term in the trinomial of a number near a letter. Ex: 5y2.)

Here is the problem:

y2 + 10y + 25        5y2-11y-12
------------- times ----------
5y2 + 29y + 20          y+5

I hope this makes sense, I tried to type it properly.. I should add that I know that you’re supposed to factor.

So 1*25 = 25. Then 5+5 = 10, and 5*5 = 25. So (y+5)(y+5) for the first numerator.

But when I get to the first denominator I don’t understand. I want to multiply 5*20 = 100. Then find two numbers to add to 29. That 5 in front of the y2 is throwing me off. Someone else helped me by simply showing me the answer. But I want to know why. Perhaps answering like I know nothing at all would be best, since math is super hard for me! Thank you so very much.

I need help with the rest of the problem as well please. Thank you!

Brittney has correctly factored \(y^2+10y+25\) as \((y+5)(y+5)\), by finding a pair of numbers whose product is 25 (the product of the first and last coefficients), and whose sum is 10 (the middle coefficient). In this case, where the coefficient of the squared term is 1, we can just put those two numbers, 5 and 5, into the factors, filling in the blanks in \((y+\_)(y+\_)\).

But now she is struggling to factor \(5y^2+29y+20\) using the same technique, finding a pair of numbers whose product is \(5\cdot 20=100\) and whose sum is \(29\). Nothing is wrong in the work so far, and the 5 itself isn’t the problem, except that it results in larger numbers! A big part of what Brittney needs is encouragement. The real issue is going to be the actual factoring part, which is quite different from what she did for the first numerator.

I answered, recalling that part of the reason I chose to teach at a community college was the knowledge that I would find there the adult students I’d most enjoyed helping at Ask Dr. Math:

Hi, Brittney.

I’ve taught a number of students in more or less your same position; you’re why I went into teaching! So let’s take a look.

First, I’ll write the problem in a way that’s easier to type, though a little harder to read:

(y^2+10y+25)/(5y^2+29y+20) * (5y^2-11y-12)/(y+5)

You’ve done a good job in the first factoring:

[(y+5)(y+5)]/(5y^2+29y+20) * (5y^2-11y-12)/(y+5)

The other two polynomials you have to factor are a harder type, as you observe; there are a couple different ways to do this, and your work gives me a hint that you have learned, at least partly, the method I most like to teach when there isn’t time to make you a master of factoring. I call it the “ac-grouping” method, but there are several other names I’ve heard.

The ac-grouping method of factoring

Multiplying the first and third coefficients in \(ax^2+bx+c\), to find the required product \(ac\), is the first part of the ac-grouping method; the second part, which we haven’t seen evidence of yet, is “factoring by grouping”. Many students struggle at that transition, forgetting that this is where the method diverges from what they have learned when the leading coefficient is 1. I find this method easier to explain than other methods, which is the main reason I teach it; students who come having mastered another method are free to use it!

I’m going to demonstrate on a different example so you have two chances to try it on your own in this problem. We’ve explained the method many times, a couple of which you can read here, if what I say isn’t detailed enough:

Factoring Trinomials

Factoring Trinomials in Quadratic Equations

Factorization by Decomposition and the Distributive

(We’ve also explained why it works, but you may not need that at the moment, unless you’re like me.)

Those links don’t currently work, as they were disabled soon after this discussion; we hope they will return soon. I chose to give a full explanation of the method, starting with the part Brittney had done correctly. In particular, I wanted to use a fresh example, one with a negative coefficient, so that I could demonstrate features not found in Brittney’s first denominator, but which she would face in the second numerator.

Step 1: Find two numbers

(If you read carefully, you’ll see a typo on my next line, which I am forced to leave as is, because it becomes part of our discussion later.)

Now, I’ll demonstrate how to use this method to factor 3x^2+2y-8, which I’ve chosen just to make it a little hard but not too much so:

First, we identify the three coefficients: a=3, b=2, c=-8.

We multiply ac = (3)(-8) = -24.

Now our first goal, much like the work you showed, is to find a pair of numbers whose product is -24 and whose sum is b (2). I like to write:

Product = -24
Sum = 2

As you probably know (though you may not be an expert yet), we can find the two numbers by listing factors of 24 and, because the sign of the product is negative, looking for a pair whose difference is 2:

1, 24: difference is 23
2, 12: difference is 10
3, 8: difference is 5
4, 6: difference is 2 — got it!

Now, since the product has to be negative, one of these has to be negative; and since the sum has to be positive, we’ll make the smaller negative: our numbers are -4 and 6. (Check: -4*6 = -24; -4 + 6 = 2.)

The difference has to be 2, because it will be the sum of two factors with different signs. If the product were negative, like the 25 in Brittney’s current problem, we would be looking for the sum. And if the middle coefficient had been negative, we would have taken \(4\) and \(-6\).

Step 2: Factor by grouping

The second step is to replace the middle term (bx) with this sum; namely 2x becomes -4x + 6x:

3x^2 + 2x – 8

3x^2 – 4x + 6x – 8

(I could also have said + 6x – 4x.)

We need to factor this by grouping, which I will assume for the moment you have learned. Factoring the first pair, we find that 3x^2 – 4x = x(3x – 4); and factoring the second pair, we find 6x – 8 = 2(3x – 4). So we now have

x(3x – 4) + 2(3x – 4)

Since the two second factors are both 3x – 4, we can move on; if they were not the same, we know we made a mistake, and go back and fix it.

Now we factor out this common factor:

(x + 2)(3x – 4)

Check that by multiplying, and we’re done!

The check is important, though the final check can be skipped if, as I recommend, you have checked every step, similarly verifying by multiplying that \(x(3x – 4) = 3x^2 – 4x\) and that \(2(3x – 4) = 6x – 8\).

Continuing the original problem

Now, the work you showed is correct as far as you went:

I want to multiply 5*20= 100. Then find two numbers to add to 29.

You want two factors of 100 whose sum is 29. I happened to see the two numbers immediately, because I know numbers pretty well. But if you don’t see it (and I don’t always!), just list factor pairs:

1, 100
2, 50
4, 25
5, 20
10, 10

Which pair works? Then continue.

Once you get the factoring done (or if you don’t finish it and need more help), write back and show you work as far as you get, and we can take you another step forward.

Brittney had stopped there, with the choice of two numbers, and I wanted to see if she had trouble with the next step.

A little struggle …

Brittney answered,

Dr. Peterson,

Thank you so much for your thorough reply. I love how you broke it down so far. I was able to get as far as 5*20=100. But need to figure out how to add to 29.

Also, I got a little lost with the example problem. It’s great, but is the “x” possibly supposed to be a “y”? If so, is there any way you can please resend that entire thing with the “y” instead? I could totally be wrong, and if so I apologize! I’m just a bit confused.

You have NO IDEA how fortunate I am to be able to get help here. I am so grateful. I will need so much more. I am glad my professor is allowing me to turn assignments in late. Otherwise I’d surely fail. I could still yet. But I have never failed, and in fact have an excellent GPA. I am very humbled, and anxious.

Thank you, Brittney

It isn’t clear yet why she couldn’t get the two numbers, or what in my example was confusing; I tried to identify possibilities, and responded:

You said,

I was able to get as far as 5*20=100. But need to figure out how to add to 29.

Here is what I said about that:

You want two factors of 100 whose sum is 29. I happened to see the two numbers immediately, because I know numbers pretty well. But if you don’t see it (and I don’t always!), just list factor pairs:

1, 100
2, 50
4, 25
5, 20
10, 10

Which pair works? Then continue.

I listed all possible pairs of numbers whose product is 100; one of them adds up to 29. Which is it? Then those are the two numbers you have to use.

Apparently you are not quite clear on what has to add to 29, or something like that; it will help if you just do something, even if you’re sure it’s wrong, just so I can see what you are thinking.

This time I put the right pair in bold (hint, hint); some kind of work at this point was needed in order to diagnose the problem.

As for the question about x being a y, I was confused, and guessed that changing between different variables was the issue.

In my problem, the variable is x; in yours, it is y; in another, it might be t. Any variable can be used in an expression.

I deliberately used a different variable than in your problem, just as I used different numbers, to help you get used to seeing such variations. (Incidentally, we so commonly use x that I suspect the author of your problem intentionally used y for exactly the same reason I used x: so it wouldn’t look identical to examples you’ve seen!)

You could, yourself, copy everything I wrote and change every x to a y, if you want (word processors are really good at that); but it’s important to be able to see what is being done without being distracted by extraneous details. So try going through it thinking of “y” as just “the variable”.

Keep working with us, and we’ll get there. You have perseverance; so do we.

Many students do see x so often that they are tripped up when a different variable is used. I don’t think that was really Brittney’s issue, though!

Fixing the factoring

Brittney first answered about getting 29, and got into the factoring part:

Dr. Peterson,

Okay, so the two numbers are 25 and 4. So I write it like:

5y+4y+25y+20.

Then I factor by grouping the first 2 first, then the last 2.

I assume it’s: y(5+4)+5(5y+4)?? Is that right?

Thanks, Brittney

She wrongly copied the first term as \(5y\); that led to a wrong factoring. The work itself is good, and it was right that she stopped where she did, as anything beyond that would have been wrong! So I could honestly be encouraging here:

Good work, almost.

The polynomial is

5y^2+29y+20

You (correctly) replaced the 29y with 4y + 25y and wrote,

5y+4y+25y+20

Presumably you meant

5y^2+4y+25y+20

Then you factored as

y(5+4)+5(5y+4)

When I factor anything, I immediately check by multiplying; when I do that, I get

5y + 4y + 25y + 20

So your factoring is good (and you rightly stopped at that point!), but you were factoring the wrong thing!

You have to be careful at every step to make sure you don’t drop bits like that exponent.

Here is a little saying I often use with students, about how to avoid silly errors:

When you are solving a problem,

    1. Think
    2. Write down what you thought.
    3. Think about what you wrote.
    4. Fix it!

In other words, errors, especially little ones, are normal; what sets apart those who do well is how soon they find and fix them. Every single line I write (when it matters), I check.

Now redo that, and I think you’ll get it right.

As you can see here, when I teach at this level, a major focus of my teaching is on being careful to check constantly. You need a combination of trusting yourself enough to proceed, but distrusting yourself enough to check!

But note that last comment about what I do; it will come back to bite me.

Oops!

While I was writing that, Brittney had added this:

Dr. Peterson,

Thank you! I was confused by the “y” in the problem “3x^2+2y-8″.

I think we’re making progress! Thank you for helping! Forgive me, I have twin six-year-olds and a 1.5 year old who are demanding mama time.

I hastened to reply:

Ah, I see! It was my error, not yours. Sorry!

What I just said about checking every single line I write … I guess I lied.

Yes, I’m human, too!

Brittney responded,

Oh no problem. I just wanted to make sure! I never would have been able to pick that mistake up before, so I guess I am making a little progress. And the way you teach is really helpful! Thank you so much! I see that I forgot to put the squared part on the 5! Is that the only mistake so far?

So it’d be y(5y+4)+5(5y+4). I presume that will turn into (y+5)(5y+4)?

Yes, my expectation that she would get this part was right.

Factoring the second numerator … and finishing!

While I slept, she wrote again, factoring \(5y^2-11y-12\) and completing the problem:

So now I’ve gone to the next numerator. 5*-12=-60. The 2 numbers that work here are -15 and 4. So now it’s written: 5y2-15y+4y-12. Then I took the first two terms and got: 5y(y-3) +4(y-3). I took it down to (5y+4)(y-3).

So (y+5)(y+5)/(y+5)(5y+4) * (5y+4)(y-3)/(y+5).

Then I just assume that whatever is identical in either numerator and either denominator can cancel out since that’d equal one. (Or does this have to only happen on opposite sides, or only on the same side? Meaning the one fraction, not one numerator and the other denominator. Does that make sense?)

So I canceled y+5, y+5, y+5, y+5, 5y+4, and 5y+4. This leaves me with (y-3)/ nothing. So y-3 is the answer?

If I did this right, can you please give me another like it? So I can see if I can do it again? Thank you!

Here is the work, written out:

$$\require{cancel}\frac{y^2+10y+25}{5y^2+29y+20}\cdot\frac{5y^2-11y-12}{y+5} =\\ \frac{(y+5)(y+5)}{(y+5)(5y+4)}\cdot\frac{(5y+4)(y-3)}{y+5} = \\ \frac{\cancel{(y+5)}\xcancel{(y+5)}}{\cancel{(y+5)}\bcancel{(5y+4)}}\cdot\frac{\bcancel{(5y+4)}(y-3)}{\xcancel{y+5}} = y-3$$

In the morning, I answered,

Yes, you’ve got it. I thought you had it in you.

You said,

Then I just assume that whatever is identical in either numerator and either denominator can cancel out since that’d equal one. (Or does this have to only happen on opposite sides, or only on the same side? Meaning the one fraction, not one numerator and the other denominator. Does that make sense?)

Yes, anything on top will cancel with the same factor anywhere on the bottom, because it’s really one big fraction when you multiply:

(y+5)(y+5)(5y+4)(y-3)
---------------------
(y+5)(5y+4)(y+5)

Concluding thoughts

As for providing another example, if you have a textbook, you can probably find more like that with answers in the back, though if this is a review section it might not have a lot.

You can also find sites with problems to try; or try the examples at places like these without looking at the solutions until you’re ready:

https://tutorial.math.lamar.edu/problems/alg/rationalexpressions.aspx

https://www.purplemath.com/modules/rtnlmult.htm

https://mathbitsnotebook.com/Algebra1/RationalEquations/REmultiply.html

https://openstax.org/books/algebra-and-trigonometry/pages/1-6-rational-expressions

https://www.openalgebra.com/2012/11/multiplying-and-dividing-rational.html

I just searched for “multiplying rational expressions” and picked the sites I most respect.

Some of these nicely hide the work for examples, so you can do it without looking, then just click a button and see the solution written out. Here, I was “teaching a student how to fish”, by pointing out sources of practice problems rather than just giving one or two.

In some concluding chat about continuing to use our site, Brittney said,

Can I write to you right here for additional problems? I don’t really want the answers. I just want to know how to do it. I can even give you a problem, and you can change the numbers.

I actually feel like I have made progress in a fraction (no pun intended) of the time it’s taken me elsewhere. The book isn’t as helpful for me as I would hope. My professor is very kind and intelligent, but doesn’t teach in a way that helps me. He assumes others have taken algebra recently, perhaps.

It’s really hard to teach a variety of students; and it’s hard for a book to communicate as clearly as face-to-face. But Brittney has exactly the right idea about getting help. As I replied,

That’s our usual preferred style, if you aren’t able to show any work at all, though we prefer seeing work.

Finally,

Thank you so much for helping me! I completed my section this evening, and couldn’t have done it without your help! Just when I learn one thing, just barely, I have to learn something new!

Hope to talk soon, take care!

Brittney

Hope to see you again soon, Brittney!

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