Factoring: Some special cases.
Here are some algebraic formulas, which can be used for factoring and simplifications of expressions. All of them can be proved disclosing of brackets and collecting like terms of items.
The binomials formulas
(a + b)^{2} = a^{2} + 2ab + b^{2} | – square of the sum | |
(a – b)^{2} = a^{2} – 2ab + b^{2} | – square of the difference | |
a^{2} – b^{2} = (a – b)(a + b) | – difference of squares | |
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2ac + 2bc |
The trinomials formulas
(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3} | – cubes of the sum |
(a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3} | – cubes of the difference |
a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2}) | – sum of cubes |
a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2}) | – difference of cubes |
The formulas for the fourth degree
(a + b)^{4} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4} |
(a – b)^{4} = a^{4} – 4a^{3}b + 6a^{2}b^{2} – 4ab^{3} + b^{4} |
a^{4} – b^{4} = (a – b)(a + b)(a^{2} + b^{2}) |
Binomial theorem
(a + b)^{n} = a^{n} + na^{n – 1}b + |
(a - b)^{n} = a^{n} - na^{n – 1}b + |
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