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Factorization, how can I resolve this ??
- Thread starter danielvnl
- Start date
I'd like to know how can I do these expressions from one to another, like this :
View attachment 2531
How can I develop this ??
Not sure I know what you mean?
Do you know sigma notation?I'd like to know how can I do these expressions from one to another, like this :
View attachment 2531
How can I develop this ??
\(\displaystyle (a + b)^n = \displaystyle \sum_{i=0}^n\binom{n}{i}a^{(n-i)}b^{i} =\left(\sum_{i=0}^n\dfrac{n!}{i! * (n-i)!} * a^{(n-i)}b^i\right).\)
So
\(\displaystyle (a + b)^0 = \displaystyle \left(\sum_{i=0}^0\dfrac{n!}{i! * (n - i)!} * a^{(n-i)}b^i\right) = \dfrac{0!}{0! * (0 - 0)!} * a^{(0 - 0)}b^0 = \dfrac{1}{1 * 1} * 1 * 1 = 1.\)
\(\displaystyle (a + b)^1 = \displaystyle \left(\sum_{i=0}^1\dfrac{n!}{i! * (1-n)!} * a^{(n-i)}b^i\right) = \left(\dfrac{1!}{0! * (1 - 0)!} * a^{(1-0)}b^0\right)+ \left(\dfrac{1!}{1! * (1 - 1)!} * a^{(1-1)}b^1\right) = a + b.\)
\(\displaystyle (a + b)^2 = \displaystyle \left(\sum_{i=0}^2\dfrac{n!}{i! * (n-i)!} * a^{(n-i)}b^i\right) \implies\)
\(\displaystyle (a + b)^2 = \displaystyle \left(\dfrac{n!}{0! * (2 - 0)!} * a^{(2-0)}b^0\right)+ \left(\dfrac{2!}{1! * (2 - 1)!} * a^{(2 - 1)}b^1\right) + \left(\dfrac{2!}{2! * (2 - 2)!} * a^{(2-2)}b^2\right) = a^2 +2ab + b^2.\)
\(\displaystyle (a + b)^3 = \displaystyle \left(\sum_{i=0}^3\dfrac{n!}{i! * (n-i)!} * a^{(n-i)}b^i\right) \implies\)
\(\displaystyle (a + b)^3 = \displaystyle \left(\dfrac{3!}{0! * (3 - 0)!} * a^{(3-0)}b^0\right)+ \left(\dfrac{3!}{1! * (3 - 1)!} * a^{(3 - 1)}b^1\right) + \left(\dfrac{3!}{2! * (3 - 2)!} * a^{(3 - 2)}b^2\right) + \left(\dfrac{3!}{3! * (3 - 3)!} * a^{(3-3)}b^3\right) \implies\)
\(\displaystyle (a + b)^3 = = a^2 + 3a^2b +3ab^2 + b^3.\)
And so on forever. You can use Pascal's triangle to simplify the process.
EDIT Here is a link to Pascal's triangle http://en.wikipedia.org/wiki/Pascal's_triangle
As you can see from it
\(\displaystyle (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.\)
By the way, the term you want for "do from one to another" is "generalize." A lot of math is about generalization and its consequence, abstraction.
SECOND EDIT
\(\displaystyle (a - b)^n = \displaystyle \left(\sum_{i=0}^n\binom{n}{i} * (- 1)^i * a^{(n-i)}b^i\right).\)
Note \(\displaystyle (a - b)^2 = a^2 - 2ab + b^2 \ne a^2 - b^2.\)
There is a different pattern for difference of powers.
\(\displaystyle \displaystyle integer\ n > 1 \implies a^n - b^n = (a - b) * \left(\sum_{i=0}^{n-1}a^{(n-1-i)}b^i\right).\)
So \(\displaystyle a^5 - b^5 = (a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4).\)
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D
Deleted member 4993
Guest
I'd like to know how can I do these expressions from one to another, like this :
View attachment 2531
How can I develop this ??
There are many ways - simplest way is to just multiply the expressions on the right hand side and simplify.
For example:
(a+b)*(a+b) = a*(a+b) + b*(a+b) = a2 + ab + ba + b2 = a2 + 2ab + b2
and so on....
Very good !!! I don't know integers neither zigma notation, but I will try to understand Pascal's triangle ...
Actually it's very simple just multiply the expressions (a+b)*(a+b) to get a2 + 2ab + b2, but the contrary is different. From a2 + 2ab + b2 to get (a+b)*(a+b).
If I had a little more difficult question like transform x2+2xy+y2+5x+5y into a multiplication (x+y)(x+y+5) or transform a2+2ab+b2-c2 into
(a+b+c)(a+b-c), I feel a little difficulty. What should I know to do this ??
Thx you all ...
Actually it's very simple just multiply the expressions (a+b)*(a+b) to get a2 + 2ab + b2, but the contrary is different. From a2 + 2ab + b2 to get (a+b)*(a+b).
If I had a little more difficult question like transform x2+2xy+y2+5x+5y into a multiplication (x+y)(x+y+5) or transform a2+2ab+b2-c2 into
(a+b+c)(a+b-c), I feel a little difficulty. What should I know to do this ??
Thx you all ...
Do you know sigma notation?
\(\displaystyle (a + b)^n = \displaystyle \sum_{i=0}^n\binom{n}{i}a^{(n-i)}b^{i} =\left(\sum_{i=0}^n\dfrac{n!}{i! * (n-i)!} * a^{(n-i)}b^i\right).\)
So
\(\displaystyle (a + b)^0 = \displaystyle \left(\sum_{i=0}^0\dfrac{n!}{i! * (n - i)!} * a^{(n-i)}b^i\right) = \dfrac{0!}{0! * (0 - 0)!} * a^{(0 - 0)}b^0 = \dfrac{1}{1 * 1} * 1 * 1 = 1.\)
\(\displaystyle (a + b)^1 = \displaystyle \left(\sum_{i=0}^1\dfrac{n!}{i! * (1-n)!} * a^{(n-i)}b^i\right) = \left(\dfrac{1!}{0! * (1 - 0)!} * a^{(1-0)}b^0\right)+ \left(\dfrac{1!}{1! * (1 - 1)!} * a^{(1-1)}b^1\right) = a + b.\)
\(\displaystyle (a + b)^2 = \displaystyle \left(\sum_{i=0}^2\dfrac{n!}{i! * (n-i)!} * a^{(n-i)}b^i\right) \implies\)
\(\displaystyle (a + b)^2 = \displaystyle \left(\dfrac{n!}{0! * (2 - 0)!} * a^{(2-0)}b^0\right)+ \left(\dfrac{2!}{1! * (2 - 1)!} * a^{(2 - 1)}b^1\right) + \left(\dfrac{2!}{2! * (2 - 2)!} * a^{(2-2)}b^2\right) = a^2 +2ab + b^2.\)
\(\displaystyle (a + b)^3 = \displaystyle \left(\sum_{i=0}^3\dfrac{n!}{i! * (n-i)!} * a^{(n-i)}b^i\right) \implies\)
\(\displaystyle (a + b)^3 = \displaystyle \left(\dfrac{3!}{0! * (3 - 0)!} * a^{(3-0)}b^0\right)+ \left(\dfrac{3!}{1! * (3 - 1)!} * a^{(3 - 1)}b^1\right) + \left(\dfrac{3!}{2! * (3 - 2)!} * a^{(3 - 2)}b^2\right) + \left(\dfrac{3!}{3! * (3 - 3)!} * a^{(3-3)}b^3\right) \implies\)
\(\displaystyle (a + b)^3 = = a^2 + 3a^2b +3ab^2 + b^3.\)
And so on forever. You can use Pascal's triangle to simplify the process.
EDIT Here is a link to Pascal's triangle http://en.wikipedia.org/wiki/Pascal's_triangle
As you can see from it
\(\displaystyle (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.\)
By the way, the term you want for "do from one to another" is "generalize." A lot of math is about generalization and its consequence, abstraction.
SECOND EDIT
\(\displaystyle (a - b)^n = \displaystyle \left(\sum_{i=0}^n\binom{n}{i} * (- 1)^i * a^{(n-i)}b^i\right).\)
Note \(\displaystyle (a - b)^2 = a^2 - 2ab + b^2 \ne a^2 - b^2.\)
There is a different pattern for difference of powers.
\(\displaystyle \displaystyle integer\ n > 1 \implies a^n - b^n = (a - b) * \left(\sum_{i=0}^{n-1}a^{(n-1-i)}b^i\right).\)
So \(\displaystyle a^5 - b^5 = (a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4).\)
Very good explanation !!!
danielvnl,
All the examples (except the third one) that you provided for factorization are simple expansions of binomial expression, but in the reverse.
JeffM gave excellent detail of the same. If you understand that, you are simply doing the reverse.
In factorization, problems usually are pretty simple once you start with a basic understanding. In fact, the expansions that you provided are the most basic ones, and I encourage you to memorize them. With usage, these expansions will stick to your memory.
In the exam, if you are given a difficult factorization question, then try to use your best guess for a factor and check it. You know you can use simple long division even for expressions to divide expressions. Once you find one factor to the complicated expression, you will notice that the other factor will be a recognizable one, mostly one of the expressions you listed in your examples. If not, just repeat the factorization process as described until you reduce the order of the expression to a recognizable form. When you obtain a recognizable form, you already know what the factors should be.
With practice, it will be easy to make a wise decision as to how you should begin to factor, but the first step is probably to be comfortable with the basic forms that you have provided. This method works well for factoring expanded forms of binomial expressions.
For regular algebraic expressions that needs factorization, again, in my view, the best way is to just guess a factor, use long division or any other method, and see if it works. If it works great, just repeat the same procedure until you can factor it as best as you can.
Finally, always verify your factorization by multiplying your factors to get the original expression.
All the examples (except the third one) that you provided for factorization are simple expansions of binomial expression, but in the reverse.
JeffM gave excellent detail of the same. If you understand that, you are simply doing the reverse.
In factorization, problems usually are pretty simple once you start with a basic understanding. In fact, the expansions that you provided are the most basic ones, and I encourage you to memorize them. With usage, these expansions will stick to your memory.
In the exam, if you are given a difficult factorization question, then try to use your best guess for a factor and check it. You know you can use simple long division even for expressions to divide expressions. Once you find one factor to the complicated expression, you will notice that the other factor will be a recognizable one, mostly one of the expressions you listed in your examples. If not, just repeat the factorization process as described until you reduce the order of the expression to a recognizable form. When you obtain a recognizable form, you already know what the factors should be.
With practice, it will be easy to make a wise decision as to how you should begin to factor, but the first step is probably to be comfortable with the basic forms that you have provided. This method works well for factoring expanded forms of binomial expressions.
For regular algebraic expressions that needs factorization, again, in my view, the best way is to just guess a factor, use long division or any other method, and see if it works. If it works great, just repeat the same procedure until you can factor it as best as you can.
Finally, always verify your factorization by multiplying your factors to get the original expression.
Sigma notation is just a way to write complicated sums quickly. It is also called summation notation. Here is a link http://www.math.montana.edu/frankw/ccp/general/sigma/learn.htmVery good !!! I don't know integers neither zigma notation, but I will try to understand Pascal's triangle ...
Actually it's very simple just multiply the expressions (a+b)*(a+b) to get a2 + 2ab + b2, but the contrary is different. From a2 + 2ab + b2 to get (a+b)*(a+b).
If I had a little more difficult question like transform x2+2xy+y2+5x+5y into a multiplication (x+y)(x+y+5) or transform a2+2ab+b2-c2 into
(a+b+c)(a+b-c), I feel a little difficulty. What should I know to do this ??
Thx you all ...
An integer is one of the numbers ... -3, -2, -1, 0, 1, 2, 3 .... That is, it is either zero or plus or minus a whole number. No fractions allowed in integer land.
Factoring is an art, not a science, and you get better at it with practice. Fortunately, it is seldom if ever necessary to factor an expression though it is frequently very convenient. Moreover, not all expressions can be factored nicely.
The rational root theorem is a technique for factoring a polynomial in one variable with integer coefficients if the polynomial can be factored into rational terms. Here is a link. http://en.wikipedia.org/wiki/Rational_root_theorem
Personally, if I do not see how to factor an expression quickly, I don't bother to do it. Of course, in school, they give you expressions to factor. Kind teachers do not ask you to factor unfactorable expressions so the rational root theorem is a handy thing to know if you are still in school or are learning on your own from a book.
Edit: dsk gave another viewpoint that may be more helpful than mine. In school, you have to factor what you are told to factor and cannot take my casual attitude. It is good to practice factoring because it does simplify many problems if you can do it, but remember that it is not always possible to find a simple factoring.
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