These include:

● Factors: Mathematical expressions, variables, and numbers that when divided leave no remainder are called factors. They are the           results of the breakdown of polynomials or binomials.

● Polynomial: The summation of the number of terms present in a mathematical expression is considered a polynomial. Polynomials         break down to form factors.

● Binomial: A mathematical function or a polynomial with two terms is called a binomial.

● Trinomial: A mathematical function or a polynomial with three terms is called a trinomial.

The various methods of factorisation are as follows:

The method of common factors

The method of common factors is frequently used. In this method, the common factor is extracted from a complex term.

For example: 4x + 16y

Factors of 4x = 4 * x

Factors of 16y = 4 * 4 * y

Here, we take 4 as the common factor.

Now, according to the problem,

4x + 16y becomes 4 ( x + 4y)

Factor = 4 ( x + 4y)

The method of reorganising terms

In the absence of a common factor, reorganising the expression to a simpler form eases the factorisation of the equation. After reorganising the expression, the method of common factors can be used.

For example: 3y²x - 9y + 4yx - 12

Factors of 3y²x = 3 * y * y * x

Factors of 9y = 3 * 3 * y

Factors of 4yx = 4 * y * x

Factors of 12 = 3 * 4

Here, when we evaluate the equation, we might not find a common factor to simplify it. Therefore, the expression must be evaluated separately for a common factor.

 

Now, according to the problem,

 3y²x - 9y - (equation 1)

 4yx - 12 - ( equation 2)

3y²x - 9y + 4yx - 12

Taking 3y common in equation 1 and 4 in equation 2

3y (yx - 3) and 4 ( yx - 12)

3y²x - 9y + 4yx - 12 becomes 3y (yx - 3) and 4 ( yx - 3)

Factors = (3y - 4) and (yx - 3)

The method of difference of two squares

This method involves the difference of two squares. In simpler terms, the difference of squares of two terms is equal to the product of the sum of the two terms and their difference.

For example: 25a² - 36b²

Factors of 25a² = 5 * 5 * a * a

Factors of 36b² = 6 * 6 * b * b

Here we see the square of numbers and variables; thus, we square the whole expression for a simpler approach.

Now, according to the problem,

25a² becomes (5a)²

36b² becomes (6b)²

And 25a² - 36b² becomes (5a)² - (6b)²

Applying the formula,

a² - b² = (a + b)( a - b)

(5a)² - (6b)² becomes (5a + 6b)( 5a - 6b)

Factors= (5a + 6b)( 5a - 6b)

These are the methods of factorisation as per Class 8 Maths Chapter 14.

Some of the formulas and their short identities of factorisation are as follows:

 

 

The important questions for factorisation are as follows:

Question 1

Factorise the given expression

x² + 6x - 16

Question 2

Verify the accuracy of the following equations. Rewrite the incorrect ones correctly.

i) (a + 6 )² = a² + 12a + 36

(ii) (2a )² + 5a = 4a + 5a

Question 3

Find the common factor(s) for the following expressions and terms.

(i) 6 xyz, 24 xy², and 12 x²y

(ii) 3x² y³, 10x³ y², and 6x²y² z

Question 4

Factorise the given expression:

(x + y)² – 4xy

Question 5

Factorise the following expressions and mention their factors:

i) 7x – 42

ii) 6p – 12q

iii) 7a² + 14a

iv) -16z + 20z³

v) 5x²y – 15xy²

vi) 10a² – 15b² + 20c²

vii) -4a² + 4ab – 4ca

viii) x²yz + xy²z + xyz²

ix) ax²y + bxy² + cxyz

The solutions to the above-mentioned questions are:

Solution 1

In the question

Factors of x² = x * x

Factors of 6x = 2 * 3 * x

Factors of -16 = -2 * 8

So according to the question,

We know that 6x can be written as +-2x + 8x as the following expression will be equal to 6x.

x² + 6x - 16 = x² - 2x + 8x - 16

x( x – 2 ) + 8( x – 2 )

(x + 8) (x – 2)

Hence the factors of x2 + 6x – 16 = (x + 8) (x – 2)

Solution 2

i) (a + 6) × 2 = a² + 12a + 36

In this equation, the left-hand side (LHS) is

(a + 6)²

Applying the formula of (a + b)²,

(a + 6)² can be written as a² + 12a + 36

We know that the right-hand side (RHS) = a2 + 12a + 36, which is equal to the LHS (a² + 12a + 36)

Hence it is proved that LHS = RHS.

ii) (2a)² + 5a = 4a + 5a

Here, in this equation, we see that

LHS = (2a)² + 5a, which can be further written as 4a² + 5a on opening the brackets of (2a)²

We see that RHS = 4a + 5a

Both sides are not equal. Therefore, LHS ≠ RHS.

Thus the correct equation would be

(2a)² + 5a = 4a2 + 5a

Solution 3

i) The factors for 6 xyz are 2 * 3 * x * y * z

The factors for 24 xy² are 2 * 2 * 2 * 3 * x * y * y

The factors for 12 x²y are 2 * 2 * 3 * x * x * y

In the above simplified factors, we see that the common factors for 6 xyz, 24 xy², and 12 x²y are 2, 3, x, and y:

(2 * 3 * x * y) = 6xy

Common factors = 2, 3, x, y, or 6xy

ii) The factors for 3x² y³ are 3 * x * x * y * y * y

The factors for 10x³ y² are 2 * 5 * x * x * x * y * y

The factors for 6 x² y² z are 3 * 2 * x * x * y * y * z

In the above simplified factors, we see that the common factors for 3x² y³, 10x³ y², and 6x² y² z are x2, y2, and (x2 * y2) = x2 y2

Common factors = x², y², or x² y²

Solution 4

To solve the given expression, it is important to first expand (x + y)²

To solve the expression, use this formula:

(x + y)² = x² + 2xy + y²

(x + y)² – 4xy can be written as

x² + 2xy + y² - 4xy, where we have substituted (x + y)² for x² + 2xy + y².

Thus the following result becomes

x² + y² – 2xy

We also know that the formula for 

(x - y)² is x² + y² - 2xy

Hence we can substitute the result,

x² + y² – 2xy with (x + y)² - 4xy.

Thus, (x + y)² - 4xy becomes (x - y)²

Solution 5

 

i) 7x – 42

Taking 7 common in the equation

Now the factor is 7(x – 6)

 

ii) 6p – 12q

Taking 6 common in the equation

Now the factor is 6(p – 2q)

 

iii) 7a² + 14a

Taking 7a common in the equation

Now the factor is 7a(a + 2)

 

iv) -16z + 20z³

Taking 4z common in the equation

Now the factor is 4z(-4 + 5z²)

 

v) 5x²y – 15xy²

Taking 5xy common in the equation

Now the factor is 5xy(x – 3y)

 

vi) 10a² – 15b² + 20c²

Taking 5 common in the equation

Now the factor is 5(2a² – 3b² + 4c²)

 

vii) -4a2 + 4ab – 4ca

Taking 4a common in the equation

Now the factor is 4a(-a + b – c)

 

viii) x²yz + xy²z + xyz² 

Taking xyz common in the equation

Now the factor is xyz(x + y + z)

 

xi) ax²y + bxy² + cxyz

Taking xy common in the equation

Now the factor is xy(ax + by + cz)

Class 8 Maths Chapter 14 – Factorisation covers topics such as application of formulas, methods of factorisation, and binomials and trinomials. This is all the information one requires to understand factorisation better.