i want to learn factoring polynomial equation

[monkey3]

where do i start ?

[doogly]

start by multiplying polynomials

[morriswalters]

where do i start ?

Well, what do you know now? Lacking that basic information would cause me to suggest the first grade.

[monkey3]

can i work like this all the time ? is it always possible this way ?

[Carlington]

It might be. Let's find out. What happens using that method on...say, x³ - 3x² + 3x - 1? That seems like a perfectly good polynomial.

[doogly]

It also helps to practice polynomial long division. Like, what is 2x^3 + 3x^2 - x - 2 / x - 4 ?

[Demki]

It also helps to practice polynomial long division. Like, what is 2x^3 + 3x^2 - x - 2 / x - 4 ?

That would be 2x^3 + 3x^2 - x - 2x^(-1) - 4

[Zohar]

I'm not sure if you're being deliberately unhelpful, deliberately obtuse, or you just missed the entire point.

[morriswalters]

can i work like this all the time ? is it always possible this way ?

For a second degree polynomial, assuming that it can be factored normally at all, the answer is yes. For higher degree polynomials using that technique you can get the first term. You can then use polynomial division to find the second term. That is, divide the original polynomial by the first factor.

[doogly]

I'm not sure if you're being deliberately unhelpful, deliberately obtuse, or you just missed the entire point.

It's unlikely we're limiting ourselves to one option here.

[ucim]

The key to solving most problems is to start with the answer.

That's not just a quip; there's something to it. Usually you will know the form of the answer - what it's supposed to "look like"; it's just the specifics that you're trying to find. So, take the "form of the answer" and try to derive the question. Then look at where the parts went - that's where they come from.

Suppose for example you want to factor x² -x + 20. You know the answer (if there is one) will be of the form
(x+a)(x+b) - that's what factoring means. So... given this answer, what if we went backwards? Multiply them together and you get:

x² + (a+b)x + ab.

Well, that looks a lot like the original problem.... where (a+b) is -1, and ab is 20. You've reduced the problem to a different (related) puzzle.

ab=20, so for integers a and b, the possibilites are: (1, 20), (-1, -20), (2, 10) ... (the rest left to the reader)
Of those (small number of possibilities) only one will have the two factors add up to -1.

So... multiplying polynomials (and paying attention to what happens) gives you the keys to factoring them.

The same principle applies when the coefficient of x² is not one (and can't be factored out)

With 2x² -2 + 40 you can factor out a 2 and get the original problem, but with
2x² +3x + 20 you cannot. You'll need to go to the more general form of the answer, which is:

(cx+a)(dx+b)

Do the same trick - multiply this polynomial to see what the "form of the question" would be... and then go from there.

Not all polynomials have integer solutions. However, with some thought you can probably come up with a formula that leads to the general solution by using the insight you've gleaned here multiplying them together. That's the quadratic formula.

For higher order polynomials the idea is the same, just more involved. Look for something you can factor out first, and that leaves you with something you already know. And in some cases they are "simple ones in disguise" (such as
x⁴ - x² + 20
which is just a quadratic but in x², not in x.

Jose

[monkey3]

thanks , it was a process like that , which i don't remember properly right now

is it about the first number and the third number ?
or
is it about the second number and the third number ?

[ucim]

is it about the first number and the third number ?
or
is it about the second number and the third number ?

Try it and see. in the end, it's about all the numbers. The simple cases have a coefficient of 1 for the x² term (or can get there by factoring a constant out of the whole shebang), the less simple ones don't.

In the simple case, (x+a)(x+b) will have a coefficient of 1 in the x² term, because (well, multiply it out. ) So, you are concerned mainly about (what I think you're calling) the second and third numbers... that is, the coefficients of the x term and the constant term. When you've mastered that, then try the less simple ones, where you will also need to be concerned about the coefficients of the x² term. They will of necessity factor into something that is of the form (cx+a)(dx+b), where at least one of c and d are not 1. By that time you'll know what to do.

Spoiler:

Note: in math books you'll probably see (ax+b)(cx+d)... that's just because they are thinking ahead with the alphabet. But of course, being variables, they don't care what you call them, as long as you're consistent. Call them Fred and George if you like.

Apply this technique ("starting from the form of the answer and working backwards") to everything else you see. You'll probably find it's quite (though not universally) insightful.

Jose

[monkey3]

Thanks a lot ucim ,

The whole process of splitting up numbers like this is called factoring or factorization ?

I am a bit confused about the terms .

I have been looking up stuffs from arithmetic itself .

So i am relearning terms like , whole numbers , natural numbers , integers , rational numbers , irrational numbers .

factors ( reducible), prime factors (irreducible), greatest common factor , least common multiple

Monomial , binomial , trinomial , polynomial

simplify , factoring (factorization ) ?

[doogly]

It's really good stuff.

[ucim]

The whole process of splitting up numbers like this is called factoring or factorization ?

Yes. Factors are "things that are multplied together", Terms are "things that are added together. So you are taking a polynomial and breaking it into factors. One of the neat things about this is that if even one of those factors is zero, the whole shebang is zero. So, any value of x that makes one of the (newly found) factors zero, is called a root of the polynomial. It's a place where the graph crosses the x axis - i.e. where y, the value of the polynomial, equals zero.

If you get y=(x+4)(x-3) as your factorization, then -4 and +3 are the roots.

For an ordinary order-2 polynomial (highest term is ax² ) it forms a parabola, which is symmetric. So, halfway between the two roots will be the axis of symmetry. Set x equal to that value and solve for y, that (x,y) coordinate is the vertex. A little more fiddling and you can find the focus and directrix. And with that, you've fully described the parabola and can do tricks with it.

Integers (or whole numbers) are the set { ... -3, -2, -1, 0, 1, 2, 3...}
Rational numbers (from "ratio") are numbers that are equal to an integer divided by another integer. 4/3, -6/13, 3/1 stuff like that. They include the integers. (why?)
Irrational numbers are real numbers that are not rational. Things like pi and the square root of 2 - they cannot be expressed as an integer over an integer.
Transcendental numbers are irrational numbers that are not the roots of any (finite) polynomial.
Real numbers are the numbers you're used to; they include the rational and irrational numbers, and correspond to points on a line. But then some jackass insisted on the answer for "what is the square root of (-1)?", and none of the real numbers would work, so he invented an answer and called it i. This was initially quite unpopular, and these numbers were derisively called "imaginary" numbers, as opposed to what they now had to call "real" numbers (which used to be just called "numbers").

Spoiler:

Well, the negative numbers are just the answer to a different jackass who insisted that
y = 7 - 10
had to have an answer, and called it "negative 3". How can you have -3 apples? And the rational numbers are an ode to the jackass that asked for the answer to "one divided by two"... and called it "a half". How can you even have half a piece of paper? Cut a piece of paper in two and you have two pieces of paper! But that's how mathematicians keep from being bored.

Turns out to not be such a dumb move after all, because it followed all the rules of arithmetic nicely... and the complex numbers are numbers that are the sum of a real number and an imaginary number. They correspond to points on the plane, and there are a lot of tricks you can do with them.

I'll leave the definition of natural numbers to the religious wars section.

{something}nomial is an expression in {some variable, I'll use x} which is a series of terms, each of which is a different (integer) power of x. The highest order term is the one with the biggest exponent, and that defines the {something} in the name. (Cue a religious war as to whether x⁴ should be called a monomial (one term!) or a fourth order polynomial (highest term is of order 4)).

Simplify means to massage the expression until it is more useful to whoever needs it. After work, I need to be simplified. What constitutes "simpler" sometimes depends on the use to which it will be put. For now, it means whatever the teacher wants it to mean.

Hope that helps. Often you can get good ideas about what a word means by simply throwing it into your favorite search engine (I use duckduckgo.com because it doesn't track me), and avoiding the wikipedia entries, which for math stuff can sometimes be incomprehensible if you don't already know most of the answer.

Jose

[monkey3]

Thanks a lot for all that explanation .

So the main thing i should focus on is to learn to how to split into factors ? which is also called factorization ...

Anyway i downloaded two books from the internet .

Algebra 1 for dummies
Practical Algebra: A Self-Teaching Guide

I will try to go through those two books and will come back and post if i have more doubts

[monkey3]

just a bit of an update and some small doubts , please help

i found this really awesome website when i searching for some terms

http://www.mathhands.com/


things i was able to narrow down ,

whole numbers , natural numbers , integers , rational numbers , irrational numbers .

factors ( reducible), prime factors (irreducible), greatest common factor (same as greatest common divisor ?), least common multiple

Monomial , binomial , trinomial , polynomial

simplify , factoring (factorization ) ?


’To factor’ means to break up into multiples.

[ucim]

things i was able to narrow down ,

The list that follows... that is of things you now understand, or things that you still don't get?

’To factor’ means to break up into multiples.

Yes, essentially. In the case of integers, it is to break it into other integers that multiply out to the given number (12 factors into 4x3 or 2x2x3). In the case of polynomials, it is to break it into lower order polynomials that multiply out to the given one (examples upthread).

greatest common factor (same as greatest common divisor ?)

Yes. They are the same. They are primarily useful in saving steps when reducing, although sometimes it's easier to just use the greatest obvious divisor repeatedly. (between 24 and 36, 12 is the greatest common divisor (24 is 2x12 and 36 is 3x12), but if you don't see that right away but recognize that 6 goes into both, you can divide by six to find that 24=4x6 and 36=6x6. From there, you'd recognize that 4 and 6 have a common divisor of 2... so now you have 24=2x2x6 and 36 is 3x2x6; the 2x6, or 12, is your greatest common divisor.

It's related to the least common multiple, which just goes the other way. A common multiple is a target. Given 24 and 36, you can multiply each by (a different) {something} to get the same answer, or "target" number. The obvious case is to multiply each by the other; in both cases you end up with (pulls out calculator) 864. This is useful if you are adding (say)
5/24 + 7/36.
Convert them all to eight-hundred-sixty-fourths, and then you can add them. But this leaves you with unnecessarily big numbers and a lot of extra arithmetic to work out (it's not obvious that 348/864 is the same as {an exercise to the reader}).

But, if you can find a smaller target number (or even better, the smallest) target number, you will save yourself a lot of that extra work.

24x3 is the same as 36x2. Both are 72. That's a much better target, and the least common multiple is the best target. In this case, this is as good as it gets. Now, adding
5/24 + 7/36
is easier, because you have much smaller numbers to work with. Convert them both to seventy-seconds, and you should end up with 29/72, the arithmetic for which you can probably do in your head (3x5, remember it, 2x7, add them up).

The trick to getting the best target number (least common multiple, or LCM) is to re-use factors. Lining them up:

24 = 3 x 2 x 2 x 2
36 = 3 x 3 x 2 x 2

The ideal target (LCM) will have the fewest possible factors that include all the factors needed for each number. So, the bold numbers are already in each one, all we need is to ensure we have all the blue numbers also.

72 = 3 x 3 x 2 x 2 x 2

Mission accomplished.

You don't actually have to split it into prime factors to figure it out; I did so for illustration. If you recognize that
24 = 12 x 2
36 = 12 x 3
you're pretty much there.
72 = 12 x 3 x 2

Note that to get to the target, you multiply by the other missing number:
72 = 24 x 3
72 = 36 x 2

And lookie lookie! The 12 that we are "reusing" is the greatest common divisor!

Jose

[monkey3]

Thanks a lot for that big explanation ...

anyway i have narrowed it down to few things , with the help of this website and tutorials in it http://www.mathhands.com/ when trying to factor polynomials

i thought i would share it here ...


Factoring Polynomials
’To factor’ means to break up into multiples.


Factor by Distributive law method




Factor by grouping




Factor by Splitting



Factor by Very Famous Polynomials







i am happy that i could organize it at least this much .

is that all the methods out there ?

[ucim]

is that all the methods out there ?

Oh, I doubt it, but those are the big ones. Many are special cases; there are as many special cases as there are cats - that is, more than you can skin in a lifetime. But most of those cases aren't all that special and will still succumb to a more general method.

(Note: the key to factoring by grouping is to recognize the useful groups in any given expression. It takes some fiddling sometimes. Given examples are always so neat... real world is less so. And Pascal polynomials come up pretty often; they're especially easy because the coefficients form the pascal triangle, which is easy to construct. {exercise for duckduckgo.com} Sometimes they are disguised however, such as when doing
(x+3y)ⁿ). Another exercise for the reader.

The thing is to understand them... and to do that, multiply them back again, watching where all the pieces end up. You should get an "aha!" moment - "that's why this special case works out so neatly!" And that will be your key for figuring out how to understand all the other stuff that will come up in math, and in lots of other places.

Not all polynomials will factor nicely. For the second order ones that don't, there is the quadratic formula. See if you can derive it, using the following hints:

1: All (second order) polynomials form a parabola; those with real roots cross the x axis (y=0). That's where the roots are.
2: There's a midpoint between the two roots. Call it M. Each root will therefore either be M+k or M-k for some k (which is the distance from the midpoint). So, you already know the answer!

The roots of ax² + bx + c are (M+k) and (M-k). This means that (for some M and k), the polynomial factors into:

(x-(M+k)) (x-(M-k))

In other words,

ax² + bx + c = (x-(M+k)) (x-(M-k))

Now, multiply out the right side... and compare it to the left side. See if you can figure out what M and k would have to be, in terms of a, b, and c. It's a little complex and involves square roots and fractions, but don't worry. It's there. And, it works for all second order polynomials, including the ones that factor nicely. (In those cases, the messy stuff simplifies nicely once you put the numbers in).

(hint - try it first for the special case where a=1. Then generalize it.)

Jose

[monkey3]

Thanks a lot again ucim ,

That doesn't look exactly that easy . Maybe i should stay with the simpler ones and practice on more questions somehow ...

anyway , if i have more doubts i would come back and post more

[Eebster the Great]

The other method for deriving the quadratic formula is by completing the square. One of the Pascal Polynomials you listed is (x+y)² = x²+2xy+y². It's really nice when your polynomial is a square, because then you can just take the square root and get x+y. For instance, if I have (2x+4)² = 4, then I can take the square root of both sides to find that 2x+4 = 2. Solving for x, I find that x = -1.

The problem is that most polynomials are not perfect squares. However, you can always add a number to a polynomial to make it a square. All you have to do is look at the Pascal Polynomial above and make your polynomial fit that pattern. For instance, x²+4x-5 doesn't quite fit the pattern, but x²+4x+4 does. You can check and see that (x+2)² = x²+4x+4.

How can I use this fact to find the roots of x²+4x-5? Well, first I know that a "root" is a value of x that makes the polynomial equal zero, so I write that equation down:

x²+4x-5 = 0

Then I add 9 to both sides so I get my Pascal Polynomial:

x²+4x+4 = 9

Now I can use the rule I learned to factor the left side into a square:

(x+2)² = 9

If I take the square root of both sides, I am almost there:

x+2 = 3

And finally, I subtract 2 from both sides to find that x = 1.

But hold on, shouldn't there be two roots? I only found one! The mistake I made was to assume that because (x+2)² = 9, that means x+2 must equal 3. In fact, there are two numbers whose squares equal 9: 3² = 9, but (-3)² is also equal to 9. So really, I should have said that either x+2 = 3 or x+2 = -3 (we write this as x+2 = ±3). That means that either x = 1 or x = -5.


Of course, in this example, I used a polynomial which can be easily factored anyway into (x-1)(x+5), but this method of completing the squares will work for any quadratic polynomial, even if it doesn't factor. You just usually won't get integer roots (they will usually be irrational).

As a final note, if your polynomial does not start with x² but has some coefficient like 2x² or πx² or whatever, just divide the whole equation by that coefficient first.

If you want to use this to derive the quadratic formula, instead of using numbers as coefficient, use the letters a, b, and c. Therefore, you start with the equation ax²+bx+c = 0 and work from there.

[monkey3]

Thanks a lot Eebster the Great ,

Every point helps me in understanding the factorization methods