### i want to learn factoring polynomial equation

Posted:

**Thu Apr 20, 2017 4:58 pm UTC**where do i start ?

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Posted: **Thu Apr 20, 2017 4:58 pm UTC**

where do i start ?

Posted: **Thu Apr 20, 2017 7:56 pm UTC**

start by multiplying polynomials

Posted: **Fri Apr 21, 2017 12:42 am UTC**

Well, what do you know now? Lacking that basic information would cause me to suggest the first grade.monkey3 wrote:where do i start ?

Posted: **Fri Apr 21, 2017 7:20 am UTC**

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

Posted: **Fri Apr 21, 2017 8:08 am UTC**

It might be. Let's find out. What happens using that method on...say, x^{3} - 3x^{2} + 3x - 1? That seems like a perfectly good polynomial.

Posted: **Fri Apr 21, 2017 12:17 pm UTC**

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

Posted: **Fri Apr 21, 2017 1:54 pm UTC**

doogly wrote: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

Posted: **Fri Apr 21, 2017 2:04 pm UTC**

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

Posted: **Fri Apr 21, 2017 2:28 pm UTC**

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.monkey3 wrote:can i work like this all the time ? is it always possible this way ?

Posted: **Fri Apr 21, 2017 2:30 pm UTC**

Zohar wrote: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.

Posted: **Fri Apr 21, 2017 2:38 pm UTC**

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 x^{2} -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:

x^{2} + (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 x^{2} is not one (and can't be factored out)

With 2x^{2} -2 + 40 you can factor out a 2 and get the original problem, but with

2x^{2} +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

x^{4} - x^{2} + 20

which is just a quadratic but in x^{2}, not in x.

Jose

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 x

(x+a)(x+b) - that's what factoring means. So... given this answer, what if we went backwards? Multiply them together and you get:

x

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 x

With 2x

2x

(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

x

which is just a quadratic but in x

Jose

Posted: **Fri Apr 21, 2017 5:53 pm UTC**

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 ?

is it about the first number and the third number ?

or

is it about the second number and the third number ?

Posted: **Sat Apr 22, 2017 3:34 am UTC**

Try it and see. in the end, it's about all the numbers. The simple cases have a coefficient of 1 for the xmonkey3 wrote:is it about the first number and the third number ?

or

is it about the second number and the third number ?

In the simple case, (x+a)(x+b) will have a coefficient of 1 in the x

Jose

Posted: **Sun Apr 23, 2017 2:45 am UTC**

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 ) ?

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 ) ?

Posted: **Sun Apr 23, 2017 3:11 am UTC**

It's really good stuff.

Posted: **Sun Apr 23, 2017 3:27 am UTC**

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.monkey3 wrote:The whole process of splitting up numbers like this is called factoring or factorization ?

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 ax

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").

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 x

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

Posted: **Sun Apr 23, 2017 4:29 am UTC**

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

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

Posted: **Sun Apr 23, 2017 10:58 am UTC**

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.

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.

Posted: **Sun Apr 23, 2017 2:31 pm UTC**

The list that follows... that is of things you now understand, or things that you still don't get?monkey3 wrote:things i was able to narrow down ,

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).monkey3 wrote:’To factor’ means to break up into multiples.

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.monkey3 wrote:greatest common factor (same as 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

Posted: **Sun Apr 23, 2017 4:34 pm UTC**

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 ?

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 ?

Posted: **Sun Apr 23, 2017 5:07 pm UTC**

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.monkey3 wrote:is that all the methods out there ?

(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)

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 ax

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

In other words,

ax

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

Posted: **Sun Apr 23, 2017 5:48 pm UTC**

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

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

Posted: **Mon Apr 24, 2017 1:31 am UTC**

The other method for deriving the quadratic formula is by completing the square. One of the Pascal Polynomials you listed is (x+y)^{2} = x^{2}+2xy+y^{2}. 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, x^{2}+4x-5 doesn't quite fit the pattern, but x^{2}+4x+4 does. You can check and see that (x+2)^{2} = x^{2}+4x+4.

How can I use this fact to find the roots of x^{2}+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:

x^{2}+4x-5 = 0

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

x^{2}+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: 3^{2} = 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 x^{2} but has some coefficient like 2x^{2} or πx^{2} 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 ax^{2}+bx+c = 0 and work from there.

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, x

How can I use this fact to find the roots of x

x

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

x

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

(x+2)

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)

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 x

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 ax

Posted: **Mon Apr 24, 2017 8:13 am UTC**

Thanks a lot Eebster the Great ,

Every point helps me in understanding the factorization methods

Every point helps me in understanding the factorization methods