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- Factors of Natural Number
- Factors of algebraic expression
- Factorisation of algebraic expression
- Method of Factorisation
- Division of algebraic expression
- Solved Example on Division
- Common Mistake done in algebraic expression and equation

Consider 16

We can write

$16= 2 \times 8= 4 \times 4=2 \times 2 \times 2 \times 2=1 \times 16$

So 1,2,4,8,16 all are factors of Natural Number 16. Prime Factors are the factors which are prime number.

A number can be expressed as product of prime factors and that form is called prime form

$16=2 \times 2 \times 2 \times 2$

Example

$6xy$

$6xy= 6 \times (xy) = 6 \times(x) \times (y)= 2 \times 3 \times (xy)= 2 \times 3 \times(x) \times (y)$

We can see that 2,3,x,y cannot be further expressed as products of factor. So they are called the prime factors. We called them irreducible factors in terms of algebraic expression

The expression 6x (x - 2). It can be written as a product of factors. 2,3, x and (x - 2)

6x (x - 2). =2×3× x× (x - 2)

The factors 2,3, x and (x +2) are irreducible factors of 6x (x + 2).

2) Then find common factors to factorize the expression

1) $2x+4$

Here we can see 2 is the common factor in the terms

=$2(x+2)$

2) $11x^2y +5x$

First reduce it to irreducible factors form

$=11 \times(x) \times (x) \times y+5 \times (x)$

Now we can see that x is the common factor

$=x(11xy+5)$

2) we look at grouping the terms and check if we find binomial factor from both the groups.

3) Take the common Binomial factor out

1) $2xy + 3x + 2y + 3$

First reduce it to irreducible factors form. There is no common factor across all the terms

$= 2 \times x \times y + 3 \times x + 2 \times y + 3$

Now we think about grouping the terms. We can see that we have common factor between first two terms

$= x \times (2y + 3) + 1 \times (2y + 3)$

$= (2y + 3) (x + 1)$

2) $6xy - 4y + 6 - 9x$

There are no common factors across all the terms, so we think of grouping and rearranging the terms

$=6xy - 4y - 9x+6$

$=2y(3x-2) -3(3x-2)$

$=(2y-3)(3x-2)$

$(a + b)^2 = a^2 + 2ab + b^2$

$(a - b)^2 = a^2 - 2ab + b^2$

$(a + b) (a - b) = a^2 - b^2$

We can write this as

$(2x)^2 -2 \times 2x \times 3 + 3^2$

Which resembles the second identity

Here a=2x and b=3

So

$4x^2 - 12x + 9 =(2x-3)^2$

2) $16x^2 -81$

We can write this as

(4x)

Which resembles the third identity

Here a=4x and b=9

So

$16x^2 -81=(4x-9)(4x+9)$

- $z^{2} - 64 $
- $y^{2} - 49 $
- $x^{2} + 6 x + 9$
- $x^{2} + 14 x + 49$
- $x^{2} - 4 x + 4 $
- $x^{2} + 2 x + 1$

1) we find two factors a and b of q (i.e., the constant term) such that

ab = q and a + b = p

2) Now expression can be written

$x^2+ (a + b) x + ab$

or $x^2 + ax + bx + ab$

or $x(x + a) + b(x + a)$

or $(x + a) (x + b)$ which are the required factors.

1)$x^2 - 7x+ 12$

Comparing to $x^2+ px + q$

We have p=-7 and q=12

Now $12 =(- 3) \times (-4)$ and (-3) + (-4) =- 7

$=x^2 - 3x - 4x + 12$

$= x (x - 3) - 4 (x - 3) = (x - 3) (x - 4)$

2))$y^2+ 11y+ 30$

Comparing to x

We have p=11 and q=30

Now 30 =( 5) × (6) and (5) + (6) =11

$=y^2+ 5y+6y+ 30$

$= y (y +5) +6 (y +5) = (y +5) (y + 6)$

- $x^{2} - 18 x + 80 $
- $y^{2} - 14 y + 40$
- $x^{2} + 4 x - 12$
- $z^{2} + z - 6 $
- $x^{2} - 2 x - 3$
- $y^{2} + y - 2$
- $m^{2} - 36$
- $n^{2} + 2 n - 24$
- $x^{2} - 16 x + 63$
- $a^{2} - 4 a - 5 $

1) Identify the Numerator and denominator

2) Factorise both the Numerator and denominator by the technique of Factorisation using common factor, regrouping, identities and splitting

3) Identify the common factor between numerator and denominator

4) Cancel the common factors and finalize the result

$\frac {48 (x^2yz + xy^2z + xyz^2)}{4xyz}$

$=\frac {48xyz (x + y + z)}{4xyz}$

$= \frac {4 \ times 12 \times xyz (x + y + z)}{4xyz}$

$= 12 (x + y + z)$

Here Dividend=48 (x

Divisor=4xyz

Quotient=12 (x + y + z)

So, we have

Dividend = Divisor × Quotient.

In general, however, the relation is

Dividend = Divisor × Quotient + Remainder

When reminder is not zero

=17×2×a×a×b×c×c / 17×3×a×b×c

Cancelling common factors

= 2ac/3

Here we need to factorize both the numerator and denominator

Let’s take a look at Numerator first

36(x

Here we can factorize this using splitting method

=36(x

=36(x-3)(x-4)

So division will become

36(x

=36(x-3)(x-4)/2(x-4)

=18(x-3)

Coefficient 1 of a term is usually not shown. So student often ignores that. But while adding like terms, we should include it in the sum.

$x+2x+3x$

$=5x$

We ignored the coefficient 1

$x+2x+3x$

$= (1) x+2x+3x$

$=6x$

When you multiply the expression enclosed in a bracket by a constant (or a variable) outside, you usually applied the multiplication to first term only. This is wrong calculation. Each term of the expression has to be multiplied by the constant (or the variable).

$2(2x+3)$

$=4x+3$

We ignored the second term

$2(2x+3)$

$=4x+6$

when you square a monomial, you often ignored the numerical coefficient. This is wrong approach. The numerical coefficient and each factor has to be squared.

$(2x)^2$

$=2x^2$

We ignored the numerical coefficient

$(2x)^2$

$=4x^2$

when you square a binomial, you often do not apply the proper formula. This is wrong approach. The right formula should be used

$(2x+3)^2=4x^2 +9$

We did not use the proper formula

$(2x+3)^2$

$=4x^2+9+12x$

6. In algebraic expressions this is is used in place of prime

7. Terms can be written as product of its

1. A polynomial with three terms

2. A polynomial with one terms

3. it is called the highest power of a polynomial

4. it is an equality, which is true for all values of the variables in the equality

5. A polynomial with two terms.

Check your Answers

**Notes**-
**Worksheets** -
**Ncert Solutions**

Class 8 Maths Class 8 Science

Given below are the links of some of the reference books for class 8 Math.

- Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
- Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
- Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
- Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

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