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ikerous
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What tricks do you use for factoring binomials where the coefficient for the x2 != 1?

For example, 20x2 -7x - 6. Normally I'd just make a guess for the two coefficients for the x's (4x ) and (5x ) and then just play around with the numbers to get the rest of the terms. However this is a pretty crappy method when you're trying to teach an algebra student how to factor. Obviously I could use the rational roots theorem or use the quadratic formula and use the roots to solve for the factors. But again, that's just way too tough for basic algebra. And I'm certain theres an easier way.

I tutor and I know some students are taught some method using the factors of a*c and b or something. I haven't seen it enough to catch on. So if you know that method or preferably something really simple and fool-proof I'd really appreciate it. Hopefully you guys have some clever tricks for this. I don't have a clue how I was taught this stuff and I've just used the guess and check method since then. I think I prefer that to anything else but it's definitely not useful when you're trying to teach someone.

quintopia
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### Re: Factoring Binomials

If you want a rigorous methodical way, just divide all coeffs by a, then do it the usual way when a=1, then multiply by 20 again at the end.

Mathmagic
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### Re: Factoring Binomials

ikerous wrote:use the quadratic formula and use the roots to solve for the factors. But again, that's just way too tough for basic algebra.

Way too tough for basic algebra? The quadratic formula is basic algebra...
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Vhailor
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### Re: Factoring Binomials

I'll show you how I was thought...
first, find two numbers (k and l) so that
k*l=a*c
and
k+l=b
in ax2+bx+c

then, replace the "b" by those two numbers like so:
ax2+ kx +lx + c

then, all that's left is simple factoring, I'll use your equation as an example...

20x2-7x-6
find two numbers so that
k*l=-120
k+l=-7
those numbers are -15 and 8
so...
20x2-15x+8x-6
5x(4x-3)-2(4x-3)
(5x-2)(4x-3)

ikerous
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### Re: Factoring Binomials

mathmagic wrote:
ikerous wrote:use the quadratic formula and use the roots to solve for the factors. But again, that's just way too tough for basic algebra.

Way too tough for basic algebra? The quadratic formula is basic algebra...

If I try and get them to use the quadratic formula for basic factoring I'll get blank stares... it's just a bad idea. Especially when they know theres an easier way.

Vhailor wrote:I'll show you how I was thought...
first, find two numbers (k and l) so that
k*l=a*c
and
k+l=b
in ax2+bx+c

then, replace the "b" by those two numbers like so:
ax2+ kx +lx + c

then, all that's left is simple factoring, I'll use your equation as an example...

20x2-7x-6
find two numbers so that
k*l=-120
k+l=-7
those numbers are -15 and 8
so...
20x2-15x+8x-6
5x(4x-3)-2(4x-3)
(5x-2)(4x-3)

Wow that's exactly what they've been doing. Thanks a ton for the explanation.

That's pretty much what I was looking for. Does anyone know any other tricks or anything any easier? Not sure if it can get any easier than this method

Vhailor
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### Re: Factoring Binomials

when I learnt it the only method I found to make it simpler was to write a brute force program on my TI calculator to find the two numbers of the sum and product... for very big numbers it can get tricky

Mathmagic
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### Re: Factoring Binomials

What grade level is this? Are you just tutoring?
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ikerous
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### Re: Factoring Binomials

mathmagic wrote:What grade level is this? Are you just tutoring?

It's college level but the classes range from basic addition and subtraction to differentials. A lot of the people taking the algebra level classes are just fulfilling GE requirements and care more about doing the problems the easiest way possible and a lot less about understanding what they're doing. Theres also a few levels of algebra classes all of which have different math abilities to draw from. But this is the method the teachers are using so it's probably what I should be tutoring with anyways. Unless theres some way to make it even easier on them which is always appreciated

Token
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### Re: Factoring Binomials

Just to be annoyingly pedantic, a binomial is, technically, a polynomial with two terms. A degree-two polynomial is called a quadratic. So, really, you're asking about factoring non-monic quadratics, not binomials.
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Nimz
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I've seen another simple method of factoring trinomial quadratics with leading coefficient other than one, but it involves doing three mathematically invalid operations that end up cancelling each other out. It's somewhat akin to reducing 16/64 to 1/4 by cancelling the sixes, but more dangerous because for students at the level of factoring, it's not transparent why the "mistakes" cancel each other out. If I remember right, I saw the method in the Fallacies, Flaws and Flimflam (FFF) section of The College Mathematics Journal (CMJ).

ax2+bx+c = x2+bx+ac (first "mistake")
x2+bx+ac = (x+k)(x+l) (same k and l from Vhailor)
divide both sides of (x+k)(x+l) by a (second "mistake") to get (x+k/a)(x+l/a)
(x+k/a)(x+l/a) = (x+p/q)(x+r/s) (reducing k/a and l/a to lowest terms)
(x+p/q)(x+r/s) = (qx+p)(sx+r) (third "mistake": you put the denominators in front, but now you have the correct factorisation)

An example might be more convincing, and to that end, 20x2-7x-6 is as good an example as any.
20x2-7x-6 = x2-7x-120 = (x-15)(x+8) = (x-15/20)(x+8/20) = (x-3/4)(x+2/5) = (4x-3)(5x+2).
[EDIT] If the point of the factoring is to find zeros, (x-3/4)(x+2/5) is a better stopping place. From there it's easy to see that x = 3/4 or x = -2/5. [/EDIT]

I tutor a fair amount at this level, too, but I've never show my students this. What I do show them is how to get the k and l for factoring by grouping, the method Vhailor showed. I have my students make a table with the product ac at the top and all the possible factorisations under it. So for -120 and -7 the table would look something like:
-120
1*-120
2*-60
3*-40
4*-30
5*-24
6*-20
7* (7 doesn't go into -120 evenly)
8*-15
9*
10*-12
11*
12 is already on the list (as -12), so the rest will be the same, but with a reversal of signs. Only one pair has a sum of -7 (8 and -15), so that's what k and l are. Often students see that that pair works before they try to divide -120 by 9.
LOWA

ikerous
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### Re: Factoring Binomials

Wow that is amazing. I'll have to play around with it some, see why it even works. Thanks a lot. That's very cool.

Govalant
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### Re: Factoring Binomials

f(x)ax^2 + bx + c can be expresed in cannonical form (I think. I never remember names actually) which is:

f(x) = a(x-z1)(x-z2)

where z1 and z2 are the function's zeros.

If you replace z1 and z2 with the quadratic equation (one adding, one substracting) and simplify, you'll get right back to ax^2+bx+c.

Maybe it helps to imagine the graph. It's clear that at x = z1 and x = z2 the function is zero. And changing a expands or contracts the parabola.

EDIT:

For 20x^2 -7x - 6 it's:

20(x-.75)(x+0.4)
20 (x^2 - .75x + 0.4x - 0.3)
20 (x^2 - 0.35x - 0.3)
20x^2 - 7x + 6

EDIT2:

Well, I'm sorry i havent read everything. I think you should anyway use the quadratic equation. There's no need to prove it, so it could be used easily.

And a tiny help. If c is 0. You can factor x. (You probably knew that anyway)
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Nimz
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### Re: Factoring Binomials

ikerous wrote:Wow that is amazing. I'll have to play around with it some, see why it even works. Thanks a lot. That's very cool.
To see why it works, it helps to see where it does NOT work. Specifically, it doesn't work when there is a common factor that can be factored out first. And of course, as with any factoring method (besides quadratic formula and the like), it is difficult to apply if the factors have irrational or complex coefficients. I don't necessarily see that as a bad thing, though.

An example where the three-mistakes method fails:
20x2-2x-6 = x2-2x-120 = (x-12)(x+10) = (x-12/20)(x+10/20) = (x-3/5)(x+1/2) = (5x-3)(2x+1). As you can see, it's off by a factor of 2 - the common factor. Just as before, though, if all you are doing is looking for zeros, this can get you there since the constant factor doesn't come into play.

Something I always tell my students when they are learning factoring is to always look for common factors before doing any other factoring. If d=gcd(a,b,c), instead of listing all the factors of ac, you only need to list all the factors of ac/d2. Factoring whole numbers is alot easier for most humans if the numbers are smaller. Unfortunately, I don't have that data available for raptors.

Govalant wrote:I think you should anyway use the quadratic equation. There's no need to prove it, so it could be used easily.

There are a couple of problems I can see with using the quadratic formula to teach factoring. Primarily, the quadratic formula is usually proven using (a very special case of) factoring. Using the quadratic formula without proof seems to me to support the kind of rote memorisation that tends toward a lack of imagination in maths. The second problem is that all quadratic expressions can be factored using the quadratic formula. That's a problem because that can interfere with the idea that a quadratic can be prime. Certainly no polynomial with degree>1 is prime in C[x], but in Q[x] any quadratic with irrational or complex zeros is prime. Using abstract algebra to explain that x2+4 and x2-5 are prime is a bit overkill. The fact that there is no pair of (rational) factors that have a product of 4 (or -5) and sum of 0 is a much more transparent way to make that point.

I apologise if that came across as confrontational. Using the quadratic formula as a tool for factoring is perfectly valid. I don't think it should be the first method learned for factoring, though.
LOWA

The Sleeping Tyrant
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### Re: Factoring Binomials

One way I was taught to factor ax2 + bx + c was to find the two numbers k and l such that kl = ac and k + l = b, and then set up a pair of ratios (a:k and a:l).

When you reduce the ratios to lowest form, they give (or so I've been taught) the coefficients of the factors. So, for example:

Your two numbers, k and l, are 15 and 56.

a:k = 24:15 = 8:5, so your first factor is (8x + 5).

a:l = 24:56 = 3:7, so the second factor is (3x + 7).

So, 24x2 + 71x + 35 factors to (8x+5)(3x+7).

quintopia
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### Re: Factoring Binomials

Tyrant: look up there ^

The Sleeping Tyrant
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### Re: Factoring Binomials

quintopia wrote:Tyrant: look up there ^

I saw it, and in fact the method described up there is how I do it personally (or at least was before I learned the factor theorem).

However, for some people who aren't necessarily so strong at factoring in the first place, setting it up as a ratio where all you have to worry about is finding the numbers and dividing right can be helpful.

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### Re: Factoring Binomials

Vhailor wrote::lol:
when I learnt it the only method I found to make it simpler was to write a brute force program on my TI calculator to find the two numbers of the sum and product... for very big numbers it can get tricky
I wrote one of those! Unfortunately, I had completely mastered the art by the time I finished algebra I, where we had to show our work, and thus have never had a use for it.

I don't remember the method from Algebra I (all those years ago), but it involved a table and drawing lines on the paper.

Today, for 20x2 - 7x - 6 = (5x+2)(4x-3), I factor the coefficient on the squared term and fill in (5x- )(4x- ), then look for the correct factors of 6 to fill in the other #'s, (5x-2)(4x-3), then turn the correct - into a + (easy on paper). If 5 & 4 don't work, I would then erase and write (2x- )(10x- ), then (1x- )(20x- ), etc.
I generally pick the the correct factors on the 1st or 2nd try though, unless we haven't used quadratics in a very long time, such as right now in calc AB, where you want to expand polynomials, not factor them .
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mikegoo
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### Re: Factoring Binomials

Another variation* on the "it works (provided there is no common factor) just don't worry about the details" that some of my coworkers teach that makes me want to cry and bang my head against the wall.
I'll just work through the same example everyone else has been doing:

20x^2-7x-6

Just like some of the other methods find 2 numbers that multiply to ac (-120) and add to b (-7) getting -15 and 8. Now we go ahead and make our 2 groupings both with 20x and seperate the numbers into em like so:

(20x - 15)(20x + 8)

Anything common in each of the groupings just gets factored out and thrown away so we factor out and throw away a 5 from the first group and a 4 from the second leaving:

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

ta da!

Again, this method doesn't work if there is a common factor in the original problem.

I'm a big fan of guess and check myself (emphasising the idea of working a multiplication problem in reverse vs. remembering a trick that works if I remember it right since I"m not sure exactly why it works but the teacher showed it to me and it worked so I'll use it).

* I realize this isn't actually a different method than 2 of the others presented before...it just takes the last 2 steps a different path that some people find easier to do (not understand).

scarletmanuka
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### Re: Factoring Binomials

So, in summary, if you're setting problems in factoring quadratics, you should always include several with common factors.

Hefty One
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### Re: Factoring Binomials

We were taught this way, and it even has a cool rhyming name. By the way, its the same thing you just mentioned but with a cool* name

It's called the "slide and divide method"

Basically, you take the expression:
ax^2+bx+c
20x^2-7x-6

"slide the 'a' term to the right and multiply it by the 'c' term

x^2-7x-120

find the factors that you can multiply to get c (-120) and add to get b (-7)

(x-15)(x+8)

then you divide by what you slid

(x-15/20)(x+8/20)

which leaves
(x-3/4)(x+2/5)

then fully factor it by sliding the denominators to the coefficient position of x
(4x-3)(5x+2)

doublez
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### Re: Factoring Binomials

i always found the quadratic formula the easiest (and fastest) way to factor quadratics... even some really obvious ones like x^2+3x+2. maybe i'm just odd.

SimonM
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### Re: Factoring Binomials

doublez wrote:i always found the quadratic formula the easiest (and fastest) way to factor quadratics... even some really obvious ones like x^2+3x+2. maybe i'm just odd.

Errm, how does the quadratic formula hope you factor? Are you just finding the roots and sticking them back in?
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doublez
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### Re: Factoring Binomials

SimonM wrote:
doublez wrote:i always found the quadratic formula the easiest (and fastest) way to factor quadratics... even some really obvious ones like x^2+3x+2. maybe i'm just odd.

Errm, how does the quadratic formula hope you factor? Are you just finding the roots and sticking them back in?

er... yes. it's faster for me lol.

antonfire
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### Re: Factoring Binomials

I do that too. It's because the quadratics I encounter usually don't factor over the rationals anyway, so I don't waste time checking whether the do.
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### Re: Factoring Binomials

The Cubic and Quartic Formula's are awsome.

Try using them sometime, to find roots. Its Hillariously fun, I felt mathematically handicaped the first time I tried.
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Nimz
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### Re: Factoring Binomials

doublez wrote:i always found the quadratic formula the easiest (and fastest) way to factor quadratics... even some really obvious ones like x^2+3x+2. maybe i'm just odd.

With a name like doublez, I'm not sure how you could be odd, unless z = n + 1/2, for some n in the set of integers. Furthermore, quadratics are never odd, though they are sometimes even. You could be odd if z = 1, because 2 is the oddest prime.

On topic, if the goal is factor a quadratic rather than find the zeros of the quadratic (e.g. reducing rational expressions), the quadratic formula seems like more work to me, than using one of the product/sum factoring techniques. If the quadratic does not factor over the rationals, it's probably better in most cases to write it unfactored, anyway. If the goal is to find the zeros, go ahead and use the formula. Taking the trouble to find the zeros and then write the factored form doesn't make sense to me, in that case.

I do have a bias toward product/sum factoring. I can usually see quickly whether a given quadratic can be factored over the rationals by shuffling the factors of the product around in my head a bit. I also have a slight bias against the quadratic formula right now. That's just the rebellious side of me coming out, though. I recently heard someone say that the only way to find the zeros of a particular quadratic with irrational zeros was by using the quadratic formula. I solved it by completing the square even though I usually prefer the quadratic formula.
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imagxz
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### Re: Factoring Quadratics into Binomials

[spamminess redacted]
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antonfire
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### Re: Factoring Binomials

Nimz wrote:I solved it by completing the square even though I usually prefer the quadratic formula.
But.. the quadratic formula is completing the square.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Mathmagic
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### Re: Factoring Binomials

antonfire wrote:
Nimz wrote:I solved it by completing the square even though I usually prefer the quadratic formula.
But.. the quadratic formula is completing the square.

Funny story:

One day, I decided to find a general formula for the roots of a second-degree polynomial that wasn't the quadratic formula. I started by completing the square, and solving for x. I ended up with the quadratic formula. I head-desked. Then I laughed.
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Yesila
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### Re: Factoring Quadratics into Binomials

imagxz wrote:I'm surprised you guys haven't played with cross factoring. If you can figure it out, its 100 times easier than traditional method. See if you can grasp it . http://lifejelly.org/archives/23 . This is the easiest trick there is

It seems to me like aside from some cumbersome notation it's "100 times" exactly the same as the traditional method.

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### Re: Factoring Quadratics into Binomials

Yesila wrote:
imagxz wrote:I'm surprised you guys haven't played with cross factoring. If you can figure it out, its 100 times easier than traditional method. See if you can grasp it . http://lifejelly.org/archives/23 . This is the easiest trick there is

It seems to me like aside from some cumbersome notation it's "100 times" exactly the same as the traditional method.

The only difference between the traditional method and the "cross-factoring" is you're writing down what most people (or at least *I* do) do in their heads (i.e. "What two numbers multiply to the x0 term and add to the x coefficient").
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Nimz
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### Re: Factoring Binomials

mathmagic wrote:
antonfire wrote:
Nimz wrote:I solved it by completing the square even though I usually prefer the quadratic formula.
But.. the quadratic formula is completing the square.

Funny story:

One day, I decided to find a general formula for the roots of a second-degree polynomial that wasn't the quadratic formula. I started by completing the square, and solving for x. I ended up with the quadratic formula. I head-desked. Then I laughed.

The quadratic formula is what you end up with when you complete the square on an arbitrary quadratic equation, as demonstrated in that funny story, but it is a formula. Completing the square is a process. Perhaps an analogy: solving a Rubik's cube by sequence A of moves is not the same as solving it by sequence B of moves. You get to the same place by doing essentially the same thing, but those two sequences are distinct.

Also, little quibbles about the "cross-factoring" thing - those are quadratic expressions, not quadratic equations. I will third the notion that that method isn't any easier or indeed that different than stuff that's already been said here (i.e. guess and check, which the OP mentioned).
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Mathmagic
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### Re: Factoring Quadratics into Binomials

Nimz wrote:
mathmagic wrote:
antonfire wrote:
Nimz wrote:I solved it by completing the square even though I usually prefer the quadratic formula.
But.. the quadratic formula is completing the square.

Funny story:

One day, I decided to find a general formula for the roots of a second-degree polynomial that wasn't the quadratic formula. I started by completing the square, and solving for x. I ended up with the quadratic formula. I head-desked. Then I laughed.

The quadratic formula is what you end up with when you complete the square on an arbitrary quadratic equation, as demonstrated in that funny story, but it is a formula. Completing the square is a process. Perhaps an analogy: solving a Rubik's cube by sequence A of moves is not the same as solving it by sequence B of moves. You get to the same place by doing essentially the same thing, but those two sequences are distinct.

I think a better analogy would be:

You want a pizza. You can either order it and get it delivered, or make it yourself. You end up with the same thing in the end, but the first option has all the work cut out of it.
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Fafnir43
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### Re: Factoring Quadratics into Binomials

But you have to remember the pizza delivery place's number.
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Mathmagic
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### Re: Factoring Quadratics into Binomials

Fafnir43 wrote:But you have to remember the pizza delivery place's number.

Just like how you have to remember the quadratic formula?
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Fafnir43
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### Re: Factoring Quadratics into Binomials

Exactly!
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prbeacon
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### Re: Factoring .. [correction]

Vhailor wrote:
...
20x2-7x-6
find two numbers so that
k*l=-120
k+l=-7
those numbers are -15 and 8
so...
20x2-15x+8x-6
5x(4x-3)-2(4x-3)
(5x-2)(4x-3)

there is a small mistake in each of the final two lines, a negative which should be positive.