This document provides an introduction to factorizing algebraic expressions for an 8th grade math presentation. It defines factorizing as writing an expression as a product of factors, which can be numbers, variables, or expressions. It outlines three methods for factorizing: using common factors, regrouping terms, and identities. It then discusses how to factorize quadratic expressions of the form x^2 + (a + b)x + ab by using the identity (x + a)(x + b) = x^2 + (a + b)x + ab. An example is worked through to demonstrate this process. Finally, it previews that division of algebraic expressions will be covered next.

2. MATHSPOWER POINT PRESENTATION
DONE BY: VARUN SANTHOSH- VIII E

3. INTRODUCTION TO FACTORISATION
When we factorize an algebraic expression, we write it as a product of factors.
These factors may be numbers, algebraic variables or algebraic expressions.
Expressions like 3xy, 5 , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor
form. Their factors can be just read off from them, as we already know. On the
other hand consider expressions like 2x + 4, 3x + 3y. It is not obvious what
their factors are. We need to develop systematic methods to factorize these
expressions, i.e., to find their factors. This is what we shall do now.

4. METHODS OF FACTORISATION
• METHOD OF COMMON FACTORS
• FACTORISATION BY REGROUPING TERMS
• FACTORISATION USING IDENTITIES

5. FACTORS OF A FORM
• Let us now discuss how we can factorize expressions in one variable, like x 2 + 5x + 6, y 2 – 7y + 12, z 2 –
4z – 12, 3m2 + 9m + 6, etc. Observe that these expressions are not of the type (a + b) 2 or (a – b) 2 , i.e.,
they are not perfect squares. For example, in x 2 + 5x + 6, the term 6 is not a perfect square. These
expressions obviously also do not fit the type (a 2 – b2 ) either. They, however, seem to be of the type x
2 + (a + b) x + a b. We may therefore, try to use Identity IV studied in the last chapter to factorize these
expressions: (x + a) (x + b) = x 2 + (a + b) x + ab (IV) For that we have to look at the coefficients of x and
the constant term. Let us see how it is done in the following example. Example 9: Factories x 2 + 5x + 6
Solution: If we compare the R.H.S. of Identity (IV) with x 2 + 5x + 6, we find ab = 6, and a + b = 5. From
this, we must obtain a and b. The factors then will be (x + a) and (x + b). If a b = 6, it means that a and b
are factors of 6. Let us try a = 6, b = 1. For these values a + b = 7, and not 5, So this choice is not right.
Let us try a = 2, b = 3. For this a + b = 5 exactly as required. The factorized form of this given expression
is then (x +2) (x + 3).

6. DIVISION OF ALGEBRAIC EXPRESSION
We have learnt how to add and subtract algebraic expressions. We also know how to
multiply two expressions. We have not however, looked at division of one algebraic
expression by another. This is what we wish to do in this section. We recall that
division is the inverse operation of multiplication. Thus, 7 × 8 = 56 gives 56 ÷ 8 = 7 or
56 ÷ 7 = 8.

7. CAN YOU FIND THE ERROR

8. VIDEO TIME
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