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Contact your math's tutor because your math's is PATHETIC,sir..
try findin the answer in r.s aggarwal
Posted on Mar 11, 2008
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Here, We deal with Some Special Products in Polynomials.
Certain products of Polynomials occur more often
in Algebra. They are to be considered specially.
These are to be remembered as Formulas in Algebra.
Remembering these formulas in Algebra is as important
as remembering multiplication tables in Arithmetic.
We give a list of these Formulas and Apply
them to solve a Number of problems.
We give Links to other Formulas in Algebra.
Here is the list of Formulas in
Polynomials which are very useful in Algebra.
Formulas in Polynomials :
Algebra Formula 1 in Polynomials:
Square of Sum of Two Terms:
(a + b)2 = a2 + 2ab + b2
Algebra Formula 2 in Polynomials:
Square of Difference of Two Terms:
(a - b)2 = a2 - 2ab + b2
Algebra Formula 3 in Polynomials:
Product of Sum and Difference of Two Terms:
(a + b)(a - b) = a2 - b2
Algebra Formula 4 in Polynomials:
Product giving Sum of Two Cubes:
(a + b)(a2 - ab + b2) = a3 + b3
Algebra Formula 5 in Polynomials:
Cube of Difference of Two Terms:
(a - b)3 = a3 - 3a2b + 3ab2 - b3 = a3 - 3ab(a - b) - b3
Algebra Formula 8 in Polynomials:
Each of the letters in fact represent a TERM.
e.g. The above Formula 1 can be stated as
(First term + Second term)2
= (First term)2 + 2(First term)(Second term) + (Second term)2
Jul 02, 2011 | Computers & Internet
Jul 01, 2011 | Computers & Internet
Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:
The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.
The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.
This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.
This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y
This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2
This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3
This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4
This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5
This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6
Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.
binomial coefficient (n over k) for (x+y)^6
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