Algebraic Expression

Algebraic Expression

Any combination of literal numbers or variables and numbers (numerals) connected by +, –, × or ÷ signs is called an algebraic expression.
For example,
5, 6x, a + b × c, 4 × m + n, x – y ÷ z are algebraic expressions. The perimeter P of a triangle whose sides are a, b and c is given by P = a + b + c, area of square is x × x i.e., x2 are algebraic expressions.
A repeated product of a number with itself is written in exponential form.
2 × 2 × 2 × 2 = 24, etc
The same is true for literal numbers also. Thus, if x is a literal, then we have
x × x × x = x3 (Third power of x or x cube)
x × x × x × x = x4 and so on.
Also,           7 × x × x = 7x2
4 × x × x × y × y = 4x2 y2, etc.
x × x × x ……, n times = xn and read as n-th power of x. Here x is called base and n, exponent.

For example :

Algebraic ExpressionNo. of termsTerms
(i)   –32x1–32x
(ii)  2x + 3y22x and 3y
(iii) ax – 5y + cz3ax, –5y and cz
(iv) \(\frac{3}{x}+\frac{y}{7}-\frac{xy}{8}+9\)4 \(\frac{3}{x},\frac{y}{7},-\frac{xy}{8},9\text{ }\And \text{ }so\text{ }on\)

Factors:
When two or more numbers are multiplied together, the numbers themselves are called the factors of the product. The factors of 34 are 2 and 17. The factors of 18x2 are 2, 3 and x, factors of ln are l and n (8 is numerical factor of 8x and x is variable factor of 8x)

Coefficient:
Any factor of a product is the coefficient of the remaining factors. In the product of 8 × 5, the number 8 is coefficient of 5 and 5 is coefficient of 8. In the product 5yz, 5 is the (numerical) coefficient of yz, 5y is the coefficient of z and 5z is coefficient of y and yz is the variable coefficient of 5.
Note : If a letter has no coefficient written before it, the coefficient 1 is understood. Thus y means 1y and yz means 1yz, similarly, –y means –1(y).

Terms

An algebraic expression is a combination of numbers, literals and arithmetical operations. One or more sings (+ and –) separates an expression into several parts. Each part along with its sign is called a term.

Type of Algebraic Expression Definition Examples
1. MonomialA monomial is an expression having one term.6, –5xy, –6x2 etc.
2. BinomialA binomial is an algebraic expression having two terms.2x – 3y, x – y, 3x2 – 6x,
(x – y)2 + 3xy, 3x2 + 5 etc.
3. TrinomialA trinomial is an algebraic expression having three terms.2a – 3b – 5c, 5y2 – 3x + 9, a3 + b3 + c3 etc.
4.Tetranomial  or QuadrinomialA quadrinomial is an algebraic expression having four terms.a + b + c – 3, a3 + b3 + c3 + 3abc, etc.
5. PolynomialBinomial, trinomials and all algebraic expressions having more than three  terms are called polynomials.2a – 3b, x + y – 3yz + 4x2 – 6y2 etc.

Note : 
(i) The words ‘mono’, ‘bi’, ‘tri’ and ‘poly’ mean one, two, three and many.
(ii) A term of an algebraic expression having no literal factor is called a constant term, for example, in x2 + 9x – 8, the constant term is –8.

Example 1: Classify the following algebraic expressions into monomial, binomial, and trinomial:
a + b, 4x + 3y – 7z, y2, 1 + x + x2, z2 + 2
Solution: Monomial: y2
Binomials: a + b, z2 + 2
Trinomials: 4x + 3y – 7z, 1 + x + x2

Example 2: Find the value of a2 + x2, if a = 2 and x = 3.
Solution: a2 + x2 = a × a + x × x
= 2x+ 3x3
= 4 + 9 = 13

Example 3: Write an algebraic expression whose terms are
(a) 7, 4x, 3xy      (b) 4xy2, 3x2y, -9
Solution:
(a) The algebraic expression with terms
7, 4x, and 3xy = 7 + 4x + 3xy
(b) The algebraic expression with terms
4xy2, 3x2y, and – 9 = 4xy2 + 3x2y + (- 9)
= 4xy2 + 3x2y – 9

Like (Similar) and Unlike (Dissimilar ) Terms

These terms are defined as follows:

Like (Similar terms)Unlike (Dissimilar terms)
These are terms whose literal (variable) factors are the same.
For example,
(i) 5x2, – 6x2, + 3x2
(ii) 2(a + b), – 4(a + b), 6(a + b)
(iii) 6xy2, –8xy2, xy2
These are terms whose literal (variable) factors are not same.
For example,
(i) 2x and 5y
(ii) 6xy2 and 8x2y
(iii) (x + y), (x2 + y), 5(x2 + y2)

Finding the Value of an Algebraic Expression

An algebraic expression contains literal (variable) numbers. If we know the numerical values of these variables and substitute them in the given algebraic expression we get a numerical expression which can be simplified by the methods of arithmetic to get a number, called the value of the algebraic expression.

Eg: Evaluate the following expression if a = 2, b = – 1,  c = 1: a2 + b2 + c2 – ab – bc – ca.
Solution:
a2 + b2 + c2 – ab – bc – ca
= (2)2 + (–1)2 + (1)2 – 2 × (–1) – (–1) × 1 – 1 × 2
= 4 + 1 + 1 + 2 + 1 – 2
= 9 – 2 = 7
So, a2 + b2 + c2 – ab – bc – ca at a = 2, b = – 1,  c = 1 has value 7.

Addition and Subtraction of Algebraic Expressions

Like Terms :

  1. The sum of two or more like terms is a like term with a numerical coefficient equal to the sum of the numerical coefficients of all the like terms.
  2. The difference between two like terms is a like term with a numerical coefficient equal to the difference between the numerical coefficients of the two like terms.

Addition and Subtraction of Algebraic Expressions Problems with Solutions

1. Add 2x2, 5x2 and x2.
Solution:
The three like terms are 2x2, 5x2 and x2
Adding these terms, we have 2x2 + 5x2 + x2
=  (2 + 5 + 1) x2      (Distributive property)
=   8x2

2. Add – xy, –5xy, –2xy.
Solution:
Adding the given three like terms, we have
= – xy + (– 5xy) + (– 2xy) = –xy –5xy – 2xy
= (–1 – 5 – 2) xy = – 8xy

3. Add –7x, –5x, 8x, 9x.
Solution:
Positive terms are 8x, 9x
Negative terms are – 7x, – 5x
Sum of positive terms = 8x + 9x = 17x
Sum of negative terms = – 7x + (–5x)
= – 7x – 5x = – 12x
Adding these two terms, we have
17x + (–12)x = 17x – 12x = (17 – 12)x = 5x

4. Add 4x + 3y – 5z, –7z + 5x – 8y and – y –3x + 2z.
Solution:
Column method:
Algebraic Expression 1
Re-write the expressions so that their like terms are in a column as
Horizontal Method :
Sum = (4x + 3y – 5z) + (–7z + 5x – 8y) + (–y –3x + 2z)
= 4x + 3y – 5z –7z + 5x – 8y – y –3x + 2z
= (4x + 5x –3x) + (3y – 8y – y) + ( – 5z – 7z +2z)
= (4 + 5 – 3) x + (3 – 8 – 1) y + (–5 –7 + 2)z
= 6x – 6y –10z

5. Subtract 10x2 from –8x2.
Solution:
Algebraic Expression 2

6. How much is t2 – 5t + 6 greater than t2 + 5t – 6 ?
Solution:
Algebraic Expression 3

Rule for Solving Algebraic Expressions

BODMAS represents the order of Performance of operations namely BODMAS represents the order of Performance of operations namely
B = Brackets;
O = Of;
D = Division;
M = Multiplication;
A = Addition;
S = Subtraction

Examples:

1. Simplify : 2x – [3y – {2x – (y – x)}].
Solution:
We have, 2x – [3y – {2x – (y – x)}]
We first remove the inner most bracket.
2x – [3y – {2x – y + x}]
Next inner most is the curly bracket.
2x – [3y – 2x + y – x]
Now we remove the square bracket.
2x – 3y + 2x – y + x
= (2x + 2x + x) – 3y – y
= (2x + 2x + x) – (3y + y)
= (2 + 2 + 1) x – (3 + 1) y = 5x – 4y

2. Simplify : 5a – [a2 – {2a (1 – a + 4a2) –3a (a2 – 5a – 3)}] – 8a.
Solution:
We first remove the inner most grouping symbol ( ), the { } and then [ ].
Thus we have,
5a – [a2 – {2a (1 – a + 4a2) –3a (a2 – 5a – 3)}] – 8a
= 5a –[a2 –{2a –2a2 + 8a3 – 3a3 + 15a2 + 9a}] – 8a
= 5a –[a2 – 2a + 2a2 – 8a3 + 3a3 – 15a2 – 9a] – 8a
= 5a – a2 + 2a – 2a2 + 8a3 – 3a3 + 15a2 + 9a – 8a
= 5a3 + 12a2 + 8a.

Adding Polynomials

Adding Polynomials

In the previous lesson, we are discussing about polynomials, Types of PolynomialsMonomials, Binomials, and Polynomials . Now lets learn adding polynomials.

Adding Polynomials 1
Add like terms by adding the numerical portion of the terms, following the rules for adding signed numbers.
(The numerical portion of an expression is called the coefficient.)

 

Below are several different ways to attack this example:

  • Using a Horizontal Method to add like terms:

adding-polynomials-1

  • Using a Vertical Method to add like terms:

Adding Polynomials 2

 

 

Polynomials

Polynomials

Algebraic expression containing many terms of the form axn, n being a non-negative integer is called a polynomial. i.e., f(x) = a0 + a1x + a2x2 + a3x3 + …….. + an-1xn-1 + anxn, where x is a variable, a0, a1, a2, …… an are constants and an ≠ 0.
Example: 4x4 + 3x3 – 7x2 + 5x + 3, 3x3 + x2 – 3x + 5.
Polynomials 1
(1) Real polynomial:
f(x) = a0 + a1x + a2x2 + a3x3 + …….. + anxis called real polynomial of real variable x with real coefficients.
Example: 3x3 – 4x2 + 5x – 4, x2 – 2x + 1 etc. are real polynomials.

(2) Complex polynomial:
f(x) = a0 + a1x + a2x2 + a3x3 + …….. + anxn is called complex polynomial of complex variable x with complex coefficients.
Example: 3x2 – (2+4i)x + (5i-4), x3 – 5ix2 + (1+2i)x + 4 etc. are complex polynomials.

(3) Degree of polynomial:
Highest power of variable x in a polynomial is called degree of polynomial.
Example: f(x) = a0 + a1x + a2x2 + a3x3 + …….. + anxn is a n degree polynomial.
f(x) = 4x3 + 3x2 – 7x + 5 is a 3 degree polynomial.
A polynomial of second degree is generally called a quadratic polynomial. Polynomials of degree 3 and 4 are known as cubic and biquadratic polynomials respectively.

(4) Polynomial equation:
If f(x) is a polynomial, real or complex, then f(x) = 0 is called a polynomial equation.

Read More:

Factors And Coefficients Of A Polynomial

Factors And Coefficients Of A Polynomial

Factor: 
When numbers (constants) and variables are multiplied to form a term, then each quantity multiplied is called a factor of the term. A constant factor is called a numerical factor while a variable factor is called a literal factor.

For Example:
(i)   7, x and 7x are factors of 7x, in which
7 is constant (numerical) factor and x is variable (literal) factor.
(ii)  In 5x2y, the numerical factor is –5 and literal factors are : x, y, xy, x2 and x2y.

Coefficient:
Any factor of a term is called the coefficient of the product of the remaining factors.
Factors And Coefficients Of A Polynomial 1
There are two types of coefficients:
1. Numerical coefficient or simply coefficient
2. Literal coefficient

For Example:
(i)   In 7x ; 7 is coefficient of x
(ii) In 7xy, the numerical coefficient of the term 7xy is 7 and the literal coefficient is xy.
In a more general way,
Coefficient of xy = 7
Coefficient of 7x = y
Coefficient of 7y = x
(iii) In (- mn2), the numerical coefficient of the term is (- 1) and the literal coefficient is mn2.
In a more general way,
Coefficient of mn2 = – 1
Coefficient of (-n2) = m
Coefficient of m = (- n2)
(iv)  In –5x2y; 5 is coefficient of ­–x2y; –5 is coefficient of x2y.

Like and unlike terms: Two or more terms having the same algebraic factors are called like terms, and two or more terms having different algebraic factors are called unlike terms.
Factors And Coefficients Of A Polynomial 2
Factors And Coefficients Of A Polynomial 3

Example: In the expression 5x2 + 7xy – 7y – 5xy, look at the terms 7xy and (- 5xy). The factors of 7xy are 7, x, and y and the factors of (- 5xy) are (- 5), x, and y. The algebraic factors (which contain variables) of both terms are x and y. Hence, they are like terms. Other terms 5x2 and (- 7y) have different algebraic factors [5 × x × x and (- 7y)]. Hence, they are unlike terms.

Factors And Coefficients Of A Polynomial With Examples

Example 1:    Write the coefficient of:
(i)   x2 in 3x3 – 5x2 + 7
(ii)  xy in 8xyz
(iii) –y in 2y2 – 6y + 2
(iv) x0 in 3x + 7
Solution:
(i) –5
(ii) 8z
(iii) 6
(iv) Since x0 = 1,
Therefore 3x + 7 = 3x + 7x0
coefficient of x0 is 7.

Example 2: Identify like terms in the following:
2xy, -xy2, x2y, 5y, 8yx, 12yx2, -11xy
Solution:  2xy, 8yx, -11xy are like terms having the same algebraic factors x and y.
x2y and 12yx2 are also like terms having the same algebraic factors x, x and y.

Example 3: State whether the given pairs of terms are like or unlike terms:
(a) 19x, 19y (b) 4m2p, 7pm2
Solution:
(a) 19x and 19y are unlike terms having different algebraic factors, i.e., x and y.
(b) 4m2p, 7pm2 are like terms having the same algebraic factors, i.e., m, m, p.

Algebraic Identities Of Polynomials

Algebraic Identities Of Polynomials

algebraic-identities

You can also read https://cbselibrary.com/ncert-solutions-for-class-10-maths-chapter-2/ for more solved examples.

People also ask

Algebraic Identities Of Polynomials Example Problems With Solutions

Example 1:    Expand each of the following
\(\left( \text{i} \right)\text{ }{{\left( \text{3x — 4y} \right)}^{\text{2}}}\text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{ }\left( \text{ii} \right)\text{ }\!\!~\!\!\text{  }{{\left( \frac{x}{2}+\frac{y}{3} \right)}^{2}}\)
Solution:   (i) We have,
Algebraic Identities Of Polynomials 1

Example 2:    Find the products
(i) (2x + 3y) (2x – 3y)
\(\left( \text{ii} \right)\text{ }\left( x-\frac{1}{x} \right)\left( x+\frac{1}{x} \right)\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)\)
Solution:    (i) We have,
Algebraic Identities Of Polynomials 2

Example 3:    Evaluate each of the following by using identities
(i) 103 × 97         (ii) 103 × 103
(iii) (97)2             (iv) 185 × 185 – 115 × 115
Solution:    (i) We have,
Algebraic Identities Of Polynomials 3

Example 4:    \(\text{If   }x+\frac{1}{x}=6,\text{   find }{{x}^{4}}+\frac{1}{{{x}^{4}}}\)
Solution:    We have,
Algebraic Identities Of Polynomials 4

Example 5:     \(\text{If   }{{x}^{2}}+\frac{1}{{{x}^{2}}}=27,\text{   find the value of the  }x-\frac{1}{x}\)
Solution:    We have,
Algebraic Identities Of Polynomials 5

Example 6:    If x + y = 12 and xy = 32, find the value of x2 + y2
Solution:    We have,
Algebraic Identities Of Polynomials 6

Example 7:    Prove that:
2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = [(a – b)2 + (b – c)2 + (c – a)2]
Solution:    We have,
Algebraic Identities Of Polynomials 7

Example 8:    If a2 + b2 + c2 – ab – bc – ca = 0, prove that a = b = c.
Solution:    We have,
Algebraic Identities Of Polynomials 8

Example 9:    Write the following in expanded form :
(i) (9x + 2y + z)2            (ii) (3x + 2y – z)2
(iii) (x – 2y – 3z)2         (iv) (–x + 2y + z)2
Solution:    Using the identity
Algebraic Identities Of Polynomials 9
Algebraic Identities Of Polynomials 10
Algebraic Identities Of Polynomials 11

Example 10:    If a2 + b2 + c2 = 20 and a + b + c = 0, find ab + bc + ca.
Solution:    
Algebraic Identities Of Polynomials 12

https://www.youtube.com/watch?v=Z5g3kl8JGYc

Example 11:    If a + b + c = 9 and ab + bc + ca = 40, find a2 + b2 + c2.
Solution:    We know that
Algebraic Identities Of Polynomials 13

Example 12:    If a2 + b2 + c2 = 250 and ab + bc + ca = 3, find a + b + c.
Solution:    We know that
Algebraic Identities Of Polynomials 14

Example 13:    Write each of the following in expanded form:
(i) (2x + 3y)3     (ii) (3x ­– 2y)3
Solution:    
Algebraic Identities Of Polynomials 15

Example 14:    If x + y = 12 and xy = 27, find the value of x3 + y3.
Solution:    We know that
Algebraic Identities Of Polynomials 16

Example 15:    If x – y = 4 and xy = 21, find the value of x3 – y3.
Solution:    We know that
Algebraic Identities Of Polynomials 17

Example 16:    \(\text{If   }x+\frac{1}{x}=7,\text{   find the value of }{{x}^{3}}+\frac{1}{{{x}^{3}}}\)
Solution:    We have,
Algebraic Identities Of Polynomials 18

Example 17:    If a + b = 10 and a2 + b2 = 58, find the value of a3 + b3.
Solution:    We know that
Algebraic Identities Of Polynomials 19

Example 18:    \(\text{If   }{{x}^{2}}+\frac{1}{{{x}^{2}}}=7,\text{   find the value of }{{x}^{3}}+\frac{1}{{{x}^{3}}}\)
Solution:    We have,
Algebraic Identities Of Polynomials 20
Algebraic Identities Of Polynomials 21

Example 19:    \(\text{If   }{{x}^{4}}+\frac{1}{{{x}^{4}}}=47,\text{   find the value of }{{x}^{3}}+\frac{1}{{{x}^{3}}}\)
Solution:   We know that
Algebraic Identities Of Polynomials 22
Algebraic Identities Of Polynomials 23

Example 20:    If a + b = 10 and ab = 21, find the value of a3 + b3.
Solution:    We know that
Algebraic Identities Of Polynomials 24

Example 21:     If a – b = 4 and ab = 45, find the value of a3 – b3.
Solution:   We have,
Algebraic Identities Of Polynomials 25

Example 22:    If a + b + c = 0, then prove that a3 + b3 + c3 = 3abc  
Solution:   We know that
Algebraic Identities Of Polynomials 26

Example 23:    Find the following product:
(x + y + 2z) (x2 + y2 + 4z2 – xy – 2yz – 2zx)
Solution:    We have,
Algebraic Identities Of Polynomials 27

Example 24:     If a + b + c = 6 and ab + bc + ca = 11, find the value of a3 + b3 + c3 – 3abc.
Solution:  
We know that
a3 + b3 + c3 – 3abc
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
⇒ a3 + b3 + c3 – 3abc =
(a + b + c) {(a2 + b2 + c2) – (ab + bc + ca)}…(i)
Clearly, we require the values of a + b + c,
a2 + b2 + c2 and ab + bc + ca to obtain the value of a3 + b3 + c3 – 3abc. We are given the values of a + b + c and ab + bc + ca. So, let us first obtain the value of a2 + b2 + c2.
We know that
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca)
⇒ 62 = a2 + b2 + c2 + 2 × 11
[Putting the values of a + b + c and ab + bc + ca]
⇒ 36 = a2 + b2 + c2 + 22
⇒ a2 + b2 + c2 = 36 – 22
⇒ a2 + b2 + c2 = 14
Now, putting a + b + c = 6, ab + bc + ca = 1 and a2 + b2 + c2 = 14 in (i), we get
a3 + b33 + c3 – 3abc = 6 × (14 – 11)
= 6 × 3 = 18.

Example 25:    If x + y + z = 1, xy + yz + zx = –1 and xyz = –1, find the value of x3 + y3 + z3.
Solution:    
We know that :
x3 + y3 + z3 – 3xyz
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
⇒ x3 + y3 + z3 – 3xyz
= (x + y + z) (x2 + y2 + z2 + 2xy + 2yz + 2zx – 3xy – 3yz – 3zx)
[Adding and subtracting 2xy + 2yz + 2zx]
⇒ x3 + y3 + z3 – 3xyz
= (x + y + z) {(x + y + z)2 – 3(xy + yz + zx)}
⇒ x3 + y3 + z3 – 3 × –1 = 1 × {(1)2 – 3 × –1}
[Putting the values of x + y + z, xy + yz + zx and xyz]
⇒ x3 + y3 + z3 + 3 = 4
⇒ x3 + y3 + z3 = 4 – 3
⇒ x3 + y3 + z3 = 1