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Any mathematical expression which consists of numbers, variables and operations are called Algebraic Expression.
1. Terms
Every expression is separated by an operation which is called Terms. Like 7n and 2 are the two terms in the above figure.
2. Factors
Every term is formed by the product of the factors.7n is the product of 7 and n which are the factors of 7n.
3. Coefficient
The number placed before the variable or the numerical factor of the term is called Coefficient of that variable.7 is the numerical factor of 7n so 7 is coefficient here.
4. Variable
Any letter like x, y etc. are called Variables. The variable in the above figure is n.
5. Operations
Addition, subtraction etc. are the operations which separate each term.
6. Constant
The number without any variable is constant. 2 is constant here.
An expression can be represented on the number line.
Example
How to represent x + 5 and x – 5 on the number line?
Solution:
First, mark the distance x and then x + 5 will be 5 unit to the right of x.
In the case of x – 5 we will start from the right and move towards the negative side. x – 5 will be 5 units to the left of x.
The algebraic expression having only one term.
5x^{2}
The algebraic expression having two terms.
5x^{2 }+ 2y
The algebraic expression having three terms.
5x^{2 }+ 2y + 9xy
The algebraic expression having one or more terms with the variable having non-negative integers as an exponent.
5x^{2 }+ 2y + 9xy + 4 and all the above expressions are also polynomial.
Terms having the same variable are called Like Terms.
2x and -9x
24xy and 5yx
6x^{2} and 12x^{2}
The terms having different variable are called, Unlike Terms.
2x and - 9y
24xy and 5pq
6x^{2} and 12y^{2}
Steps to add or Subtract Algebraic Expression
First of all, we have to write the algebraic expressions in different rows in such a way that the like terms come in the same column.
Add them as we add other numbers.
If any term of the same variable is not there in another expression then write is as it is in the solution.
Add 15p^{2} – 4p + 5 and 9p – 11
Write down the expressions in separate rows with like terms in the same column and add.
Subtract 5a^{2} – 4b^{2 }+ 6b – 3 from 7a^{2} – 4ab + 8b^{2} + 5a – 3b.
For subtraction also write the expressions in different rows. But to subtract we have to change their signs from negative to positive and vice versa.
While multiplying we need to take care of some points about the multiplication of like and unlike terms.
The coefficients will get multiplied.
The power will not get multiplied but the resultant variable will be the addition of the individual powers.
The product of 4x and 3x will be 12x^{2}.
The product of 5x, 3x and 4x will be 60x^{3}.
The power will remain the same if the variable is different.
If some of the variables are the same then their powers will be added.
The product of 2p and 3q will be 6pq
The product of 2x^{2}y, 3x and 9 will be 54x^{3}y
1. Multiplying Two Monomials
While multiplying two polynomials the resultant variable will come by
The coefficient of product = Coefficient of the first monomial × Coefficient of the second monomial
The algebraic factor of product = Algebraic factor of the first monomial × Algebraic factor of the second monomial.
25y × 3xy = 125xy^{2}
2. Multiplying Three or More Monomials
While multiplying three or more monomial the criterion will remain the same.
4xy × 5x^{2}y^{2} × 6x^{3}y^{3} = (4xy × 5x^{2}y^{2}) × 6x^{3}y^{3}
= 20x^{3}y^{3} × 6x^{3}y^{3}
= 120x^{3}y^{3} × x^{3}y^{3}
= 120 (x^{3} × x^{3}) × (y^{3} × y^{3})
= 120x^{6} × y^{6}
= 120x^{6}y^{6}
We can do it in other way also
4xy × 5x^{2}y^{2} × 6x^{3 }y^{3}
= (4 × 5 × 6) × (x × x^{2} × x^{3}) × (y × y^{2} × y^{3})
= 120 x^{6}y^{6}
1. Multiplying a Monomial by a Binomial
To multiply a monomial with a binomial we have to multiply the monomial with each term of the binomial.
Multiplication of 8 and (x + y) will be (8x + 8y).
Multiplication of 3x and (4y + 7) will be (12xy + 21x).
Multiplication of 7x^{3} and (2x^{4 }+ y^{4}) will be (14x^{7}+ 7x^{3}y^{4}).
2. Multiplication of Monomial by a trinomial
This is also the same as above.
Multiplication of 8 and (x + y + z) will be (8x + 8y + 8z).
Multiplication of 4x and (2x + y + z) will be (8x^{2} + 4xy + 4xz).
Multiplication of 7x^{3} and (2x^{4}+ y^{4}+ 2) will be (14x^{7 }+ 7x^{3}y^{4 }+ 14x^{3}).
1. Multiplying a Binomial by a Binomial
We use the distributive law of multiplication in this case. Multiply each term of a binomial with every term of another binomial. After multiplying the polynomials we have to look for the like terms and combine them.
Simplify (3a + 4b) × (2a + 3b)
(3a + 4b) × (2a + 3b)
= 3a × (2a + 3b) + 4b × (2a + 3b) [distributive law]
= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)
= 6 a^{2 }+ 9ab + 8ba + 12b^{2}
= 6 a^{2} + 17ab + 12b^{2} [Since ba = ab]
2. Multiplying a Binomial by a Trinomial
In this also we have to multiply each term of the binomial with every term of trinomial.
Simplify (p + q) (2p – 3q + r) – (2p – 3q) r.
We have a binomial (p + q) and one trinomial (2p – 3q + r)
(p + q) (2p – 3q + r)
= p(2p – 3q + r) + q (2p – 3q + r)
= 2p^{2} – 3pq + pr + 2pq – 3q^{2} + qr
= 2p^{2} – pq – 3q^{2} + qr + pr (–3pq and 2pq are like terms)
(2p – 3q) r = 2pr – 3qr
Therefore,
(p + q) (2p – 3q + r) – (2p – 3q) r
= 2p^{2} – pq – 3q^{2} + qr + pr – (2pr – 3qr)
= 2p^{2} – pq – 3q^{2} + qr + pr – 2pr + 3qr
= 2p^{2} – pq – 3q^{2} + (qr + 3qr) + (pr – 2pr)
= 2p^{2} – 3q^{2} – pq + 4qr – pr
An identity is an equality which is true for every value of the variable but an equation is true for only some of the values of the variables.
So an equation is not an identity.
Like, x^{2} = 1, is valid if x is 1 but is not true if x is 2.so it is an equation but not an identity.
(a + b)^{2 }= a^{2} + 2ab + b^{2}
(a - b)^{2} = a^{2} – 2ab + b^{2}
a^{2} – b^{2} = (a + b) (a - b)
(x + a) (x + b) = x^{2} + (a + b)x + ab
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca
These identities are useful in carrying out squares and products of algebraic expressions. They give alternative methods to calculate products of numbers and so on.
(4x – 3y)^{2}
= (4x)^{2} – 2(4x) (3y) + (3y)^{2 }
= 16x^{2} – 24xy + 9y^{2}
Use the Identity (x + a) (x + b) = x^{2} + (a + b) x + ab to find the value of 501 × 502
501 × 502
= (500 + 1) × (500 + 2)
= 500^{2} + (1 + 2) × 500 + 1 × 2
= 250000 + 1500 + 2
= 251502
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