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Algebraic Identities of Polynomials

  • Last Updated : 21 Feb, 2021

Algebraic Identities are defined for the algebraic expressions.  Algebraic expressions contain variables(a, b, c, x, y, z, etc), numbers(0, 1, 2, 3, 4 …etc) and operators(+, -, *, /….etc). An Algebraic Expression may contain only constants (1, 2, 3, 4, etc), or only variables (x, y, z, etc), or both constant and variable together (5xy, 4p3). Algebraic Identities are basically those Mathematical Equations that make calculations easy in real life.

For example: Consider multiplying two numbers like “989” and “1011”. Now, this is a long calculation, but if you know some identities which suit this kind of problem, It can be solved easily.  Before going into detail about algebraic identities, first, let’s see what is an Identity: 

What is an Identity?

Identity is a relation between two or more than two mathematical expressions, such that they produce the same value for all values of variables. In simple words, it can also be said that the L.H.S of any equation becoming identically equal to the R.H.S, for all values of variables explains an Identity. 

Let’s look at this expression given below,

(x + 2)(x + 4) = x2 + 6x + 8



Evaluate both sides RHS and LHS of this equation for different values of x,

1. x = 5

LHS: (x + 2)(x + 4) = (5 + 2)(5 + 4) = 63

RHS: x2 + 6x + 8 = 52 + 6(5) + 8 = 25 + 30 + 8 = 63

Thus, both sides of this expression are equal for x = 5.

2. x = 10

LHS: (x + 2) (x + 4) = (10 + 2) (10 + 4) = (12)(14) = 168

RHS: x2 + 6x + 8 = 102 + 6(10) + 8 = 100 + 60 + 8 = 168



Thus, both sides of this expression are equal for x = 10.

If we keep on trying this out with different values of x, we will see that L.H.S and R.H.S are equal for every value of x. Such an expression that is true for every value of variables present in it is called Identity.

Note: An equation is only true for some values of variables present in it

For example:

a2 + 3a + 2 = 132

a = 10 satisfies this identity, but a = 5 or – 7 cannot.

Types of Algebraic Expressions

Expressions can be of different types depending on how many terms they contain. There are four different types of expressions that make up identities. 

Monomial Expressions

An expression that contain only one term is called as a Monomial Expression.

For example: 16z2, 8xy, -7m, 11…. etc.

Note: A Monomial Expression can be only a constant, a variable or a combination of both constants and Variables. 



For Example: 4, x3, 15x2

Binomial Expressions

An expression containing only two terms is called a Binomial Expression. 

For example: x + y, 2x + 5z, x2 + 10 .. etc.

Proof of Binomial identities:

Identity 1: (a + b)2 = a2 + 2ab + b2

Proof: L.H.S. = (a + b)2

              L.H.S. = (a + b) (a + b)

          By multiplying each term, we get,

          L.H.S = a2 + ab + ab + b2

         L.H.S. = a2 + 2ab + b2

         L.H.S. = R.H.S.

Identity 2: (a – b)2 = a2 – 2ab + b2

Proof: By taking L.H.S.,

          (a – b)2 = (a – b) (a – b)

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

          (a – b)2 = a2 – 2ab + b2

             L.H.S. = R.H.S.

          Hence, proved.

Identity 3: a2 – b2 = (a + b) (a – b)

Proof: By taking R.H.S and multiplying each term.

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



          (a + b) (a – b) = a2 – b2

            Or

          a2 – b2 = (a + b) (a – b)

         L.H.S. = R.H.S.

         Hence proved. 

Trinomial Expression

An expression containing only three terms is called as Trinomial Expression.

For example: 2a + 3b – 5, a2b – ab2 + b2

Trinomial Identity
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

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

Proof: Taking L.H.S. 

           (a+b+c)2= (a+b+c) × (a+b+c)

           Using Distributive Property:

           (a+b+c)2= a (a+b+c) +b (a+b+c) +c (a+b+c)

                         = a2+ab+ac+ab+b2+bc+ca+cb+c2

               Rearranging the following:

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

          Hence, L.H.S. = R.H.S.

Polynomials

It is a generalization of all three and other types of expression. An expression containing, one or more terms with a non-zero coefficient (with variables having non-negative exponents) is called a polynomial. A polynomial may contain any number of terms, one or more than one.

Ex: x + y, 2a + 3b – 5, 16z2, 2a + 3b – 5 + z.

Now we’re ready for looking into Algebraic Identities. 

Algebraic Identities

It is very important to learn about the basic algebraic identities which are also known as the standard identities. 



Standard Identities
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
a2 – b2 = (a + b)(a – b)

Some other Identities: 

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
(a + b)3 = a3 + b3 + 3ab(a + b)
(a – b)3 = a3 – b3 – 3ab (a – b)
(x + a) (x + b) = x2 + (a + b)x + ab

Let’s look at some examples which use these identities. 

Sample Problems 

Question 1: Find out using identities mentioned above, (4x + 3y)2

Solution:

This can be found out using the identity of (a + b)2 = a2 + b2 + 2ab. 

(4x + 3y)2 = (4x)2 + (3y)2 + 2(4x)(3y) 

                = 16x2 + 9y2 + 24xy

Question 2: Find the values 992

Solution:

Multiplying 99 with 99 will take time and calculation. We can formulate this problem in a form that is easier to calculate. 

We have seen the identity, (a – b)2 = a2 + b2 – 2ab.  

So, 992 = (100 – 1)2 = 1002 + 12 – 2(100)(1) 

                               = 10000 + 1 -200 

                               = 9801

Question 3: Find out 9832 – 172

Solution:

This can take a lot of calculation if we do it the traditional way. We should use the identities to solve this.

We can use a2 – b2 = (a + b)(a -b) 

So, 9832 – 172 = (983 + 17)(983 – 17) 

                      = (1000)(966) 



                      = 966000  

Question 4: Find out (\frac{3}{2}m - \frac{2}{3}n)(\frac{3}{2}m + \frac{2}{3}n)           

Solution:

We can use a2 – b2 = (a + b)(a -b) 

 (\frac{3}{2}m - \frac{2}{3}n)(\frac{3}{2}m + \frac{2}{3}n) \\ \hspace{0.6cm}

= \frac{9}{4}m^{2} - \frac{4}{9}n^{2}

                

Question 5: Find out 10112

Solution:

This problem can be solved using multiple identities. Let’s solve it using the identity with three variables. 

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

10112 = (1000 + 10 + 1)2 = 10002 + 102 + 12 + 2(1000)(10) + 2(10)(1) + 2(1000) 

                                         = 1000000 + 100 + 1 + 20000 + 20 + 2000 

                                         = 1022121

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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