Components of Algebraic Expression
Monomials, Binomials and Polynomials
Steps to add or Subtract Algebraic Expression
Example 1: Add 15p2 – 4p + 5 and 9p – 11
Solution:
Write down the expressions in separate rows with like terms in the same column and add.
Example 2: Subtract 5a2 – 4b2 + 6b – 3 from 7a2 – 4ab + 8b2 + 5a – 3b.
Solution:
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.
Example3: The product of 4x and 3x will be 12x2.
Example 4: The product of 5x, 3x and 4x will be 60x3.
Example5: The product of 2p and 3q will be 6pq
Example 6: The product of 2x2y, 3x and 9 will be 54x3y
1. Multiplying Two Monomials
While multiplying two polynomials the resultant variable will come by
Example 7:
Example 8:
While multiplying three or more monomial the criterion will remain the same.
Example 9:
Example 10: Find the volume of each rectangular box with given length, breadth, and height.
To multiply a monomial with a binomial we have to multiply the monomial with each term of the 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.
Example 14: Simplify (3a + 4b) × (2a + 3b)
Solution:
(3a + 4b) × (2a + 3b)
= 3a × (2a + 3b) + 4b × (2a + 3b) [distributive law]
= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)
= 6 a2 + 9ab + 8ba + 12b2
= 6 a2 + 17ab + 12b2 [Since ba = ab]
In this also we have to multiply each term of the binomial with every term of trinomial.
Example 14: Simplify (p + q) (2p – 3q + r) – (2p – 3q) r.
Solution:
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) [distributive law]
= 2p2 – 3pq + pr + 2pq – 3q2 + qr
= 2p2 – pq – 3q2 + qr + pr (–3pq and 2pq are like terms)
(2p – 3q) r = 2pr – 3qr
Therefore,
(p + q) (2p – 3q + r) – (2p – 3q) r
= 2p2 – pq – 3q2 + qr + pr – (2pr – 3qr)
= 2p2 – pq – 3q2 + qr + pr – 2pr + 3qr
= 2p2 – pq – 3q2 + (qr + 3qr) + (pr – 2pr)
= 2p2 – 3q2 – 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, x2 = 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 = a2 + 2ab + b2
(a - b)2 = a2 – 2ab + b2
a2 – b2 = (a + b) (a - b)
(x + a) (x + b) = x2 + (a + b)x + ab
(a + b + c)2 = a2 + b2 + c2 + 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.
What Have We Discussed?
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1. What are algebraic expressions and how are they formed? |
2. How do you add and subtract algebraic expressions? |
3. What is the process for multiplying a monomial by a monomial? |
4. How do you multiply a polynomial by a polynomial? |
5. What are standard identities in algebra? |
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