 Hello and welcome to the session. In this session we will discuss division of algebraic expressions. We know that division is inverse operation of multiplication like if we say that 2 into 3 is 6, then we can also write this as 6 divided by 3 is equal to 2 or 6 divided by 2 is equal to 3. Now for algebraic expressions also division is inverse operation of multiplication like if you have the 2x multiplied by 5x is equal to 10x square then this could be written as 10x square divided by 2x gives us 5x or 10x square divided by 5x gives us 2x. Now let's consider division of a monomial by another monomial. Consider the monomials 10x cube and 5x square we need to divide 10x cube by 5x square for this first what we do is we write both these monomials in irreducible factor forms. So we have 10x cube is written as 2 into 5 into x into x into x and then 5x square is written as 5 into x into x. Now we have 10x cube divided by 5x square is equal to 10x cube upon 5x square that is 2 into 5 into x into x into x upon 5 into x into x. Now we cancel the common terms in the numerator and denominator that is this cancels with this and we are left with 2 into x. So we have 10x cube divided by 5x square is equal to 2x. So this is how when we are given 2 monomials we can divide them easily. Next we shall consider division of a polynomial by a monomial this can be done in 2 ways either by common factor method or by dividing each term of the polynomial by the monomial. First let's see how we do by the common factor method. Consider the polynomial 5x square minus 6x we divide this by the monomial 3x. Now let's see which is the first way of doing this that is by the common factor method in this we express each term of the polynomial in the factor form consider the polynomial 5x square minus 6x. Now this could be written as 5 into x into x minus 6 into x. Now we see that x is the common factor in both these terms. So we separate the x from both these terms. So this can be written as x into 5 into x minus x into 6. Now this is further written as x into 5x minus 6. That is the given polynomial 5x square minus 6x we have written as x into 5x minus 6. Now then we need to divide 5x square minus 6x by 3x that is we have x into 5x minus 6 upon 3x. This x and x gets cancelled and we are left with 5x minus 6 upon 3. So this is how we divide a polynomial by a monomial by common factor method. Now let's see what is the second way of dividing a polynomial by a monomial. In this case we divide each term of the polynomial by the monomial. Now the polynomial that we had considered was 5x square minus 6x and this needs to be divided by the monomial 3x. We write this as 5x square minus 6x upon 3x. Now this is equal to 5x square upon 3x minus 6x upon 3x. Now this x and 1x in the numerator gets cancelled and this x gets cancelled with this x and 3 2 times is 6. So we are left with 5x upon 3 minus 2 which is same as 5x minus 6 upon 3. So this is how we can divide a polynomial by a monomial by 2 methods. Next we see division of polynomial by polynomial. Now let's see how we do this. In this case we cannot divide each term in the dividend polynomial by the divisor polynomial. Instead we will factorize both the polynomials and we will cancel the common factors. Like let's consider the polynomial y square plus 7y plus 10 and we need to divide this by the polynomial y plus 5. Now let's factorize the polynomial y square plus 7y plus 10. This could be written as y square plus 2y plus 5y plus 10. Now this is equal to y into y plus 2 plus 5 into y plus 2. So we say that y square plus 7y plus 10 is equal to y plus 5 into y plus 2. Now the other polynomial that is y plus 5 cannot be factorized further since it is already in the factor form. So now we have y square plus 7y plus 10 upon y plus 5 is equal to y plus 5 into y plus 2 upon y plus 5. Now y plus 5 and y plus 5 gets cancelled and we are left with y plus 2. So this is how we can divide a polynomial by another polynomial. Now let's discuss finding errors while solving the exercises of algebra. These two links make many common errors and these errors need to be avoided. Like consider the mathematical statement 4 into x minus 5 equal to 4x minus 5. Now this statement is not correct. We know that while multiplying an expression enclosed in a bracket by a constant or you can say a variable also, outside that bracket each term of the expression should be multiplied by the constant or the variable outside the bracket. But in this case as you can see we have multiplied only 4 that is the constant term outside the bracket with the variable x but we have not multiplied 4 with this 5. So this is not a correct statement. The correct statement is given by 4 into x minus 5 is equal to 4x minus 4 into 5 that is 20. This is the correct mathematical statement. In this way we can come across many mathematical statements or many algebraic statements but we should avoid making errors in such statements. This completes the session. Hope you have understood the division of algebraic expressions.