NCERT Solutions for Class 8 गणित Chapter 14 Factorisation

Factorisation Here is the CBSE गणित Chapter 14 for Class 8 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 8 गणित Factorisation Chapter 14 NCERT Solutions for Class 8 गणित Factorisation Chapter 14 The following is a summary in Hindi and English for the academic year 2025-26. You can save these solutions to your computer or use the Class 8 गणित.

Question 1

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Factorise 9x + 18y + 6xy + 27

Solution

Here, we have a common factor 3 in all the terms.

∴    9x + 18y + 6xy + 27 = 3[3x + 6y + 2xy +9]
We find that      3x + 6y = 3(x + 2y) and 2xy + 9 = 1(2xy+9)
i.e. a common factor in both the groups does not exist,
Thus, 3x + 6y + 2xy + 9 cannot be factorised.
On regrouping the terms, we have
   3x + 6y + 2xy + 9 = 3x + 9 + 2xy + 6y
                            = 3(x + 3) + 2y(x + 3)
                            = (x + 3) (3 + 2y)
Now, 3[3x + 6y + 2xy + 9] = 3[(x + 3)(3 + 2y)]
Thus, 9x + 18y + 6xy + 27 = 3(x+3) (2y+3)