Algebraic Identities of Polynomials
Last Updated : 07 Aug, 2024
Algebraic identities are equations that hold true for all values of the variables involved. In the context of polynomials, these identities are particularly useful for simplifying expressions and solving equations.
Algebraic Identities of PolynomialsWhat are Algebraic Identities?
Algebraic Identities are defined for the algebraic expressions that 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.
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 |
Key concepts in Algebraic Identities of Polynomials
Function Definition: A real function is a rule that assigns each element ? in a subset of the real numbers R to exactly one element ?(?) in ?.
Operations on Functions:
- Addition: (?+?)(?)=?(?)+?(?)
- Subtraction: (?−?)(?)=?(?)−?(?)
- Multiplication: (?⋅?)(?)=?(?)⋅?(?)
- Division (where ?(?)≠0): (f(g/x)) = f(x)/g(x)
- Composition: (?∘?)(?)=?(?(?))
Sample Problems on Algebraic Identities of Polynomials
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
Worksheet
Problem 1: Solve for x: 3x − 7 = 2x + 5
Problem 2: Solve for y: 4y + 3 = 9
Problem 3: Solve for x: 5x/2 = 15 − 2x
Problem 4: Solve for x: x2 − 4x − 12 = 0
Problem 5: Solve for x: 2x2 + 5x − 3 = 0
Problem 6: Solve for x: x2 + 6x + 9 = 0
Conclusion
Algebraic equations are mathematical statements that use algebraic expressions set equal to one another. Solving algebraic equations is a fundamental skill in mathematics. The process generally involves isolating the variable, simplifying the equation, and verifying solutions. Understanding the properties of different types of equations and the methods for solving them is crucial for progressing in mathematics and its applications.
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