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Algebraic Factorization Formula: nth Power Formula, Notes

Algebraic formulae

Introduction to Algebra
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Algebraic Factorization Formula:

To solve an algebraic equation it is necessary to break the given equation into its constituent parts which will make it easy to solve. Suppose the equation formed is in a higher degree we can say more than 4 than it becomes difficult to deal with the same equation containing a lower degree. If an expression is factorized in terms of algebraic variables of lower degree from what it is by use of simple formula than it becomes easy to express and deal with the same.

So, here algebra needed factorization. In earlier classes, we studied simple factorization and related problems which were in generally of 2-degree equation. Now we will study the equation of 2, 3 and higher degree equation.

The most important thing about these formulae is that you don’t have to remember all but you should understand and remember the pattern included in it so that you can solve any questions related to it confidently.

Our experts made it easy for you to understand it as they have arranged a list of necessary formulas in a great order to make it easy for you to understand and learn it in given order.

Here is the list of algebraic factorization formula for nth power:

an + bn = (a – b)(b0an-1 – b1an-2 + …… – bn-2a1 + bn-1a0)

an – bn = (a – b)(b0an-1 + b1an-2 + …… + bn-2a1 + bn-1a0)

Factors of definite powers:

a– b2 = (a + b)(a – b)

a2 + b2 = 1/2[(a + b)2 + (a – b)2]

a3 – b= (a – b)(a2 + ab + b2)

a3 + b= (a + b)(a2 – ab + b2)

a4 – b4 = (a – b)(a + b)(a2 – ab + b2)

a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)

Simple factorization formula:

(a + b)= a2 + 2ab + b2

(a − b)= a2 − 2ab + b2

(a + b)3=a3 + b3 + 3ab(a + b)

(a – b)= a3 – b3 – 3ab(a – b)

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a − b)4 = a4 − 4a3b + 6a2b2 − 4ab3 + b4

(a + b + c)2 = a2 + b2 +c2 + 2ab + 2ac + 2bc

(a + b + c +…)2 = a2 + b2 + c2 + … + 2(ab+ac+bc+…)

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