Answer sheets of meritorious students of class 12th’ 2012 M.P Board – All Subjects

February 14, 201339 Insightful Publications

**Factorisation Formulas**: Factorisation, also known as factoring, is a process of breaking down a large number into several small numbers. When these small numbers are multiplied, we will get the actual or original number. Usually, students are introduced to the concepts of Factorisation in Grade 6.

Factorisation is one of the important methods that is used to break down an Algebraic or Quadratic Equation into a simple form. Thus, to break down the complex equation, one should be aware of Factorisation Formulas. In this article, we will provide you with all the necessary information related to different Factorisation Formulas for Polynomials, Trigonometry, Algebra, and Quadratic Equations. Students are also provided with the Factorisation Formulas PDF, which they can download from this article.

When an Algebraic Equation or Quadratic Equation is reduced into a simpler equation with the help of Factorisation Method, the simpler equation is treated as **Product of Factors**. The Product of Factors of an equation can be an Integer, Variable or Expression itself.

The main approach of Factorisation Method is that we won’t be expanding the brackets further.

Also, check:

Numbers can be factorised into different combinations and applying factorisation methods to numbers is easy, whereas finding the factors of an equation is a little challenging.

The numbers 1, 3, 5, and 15 are **Factors of 15 **as it can be divided the number 15 itself.

1 X 15 = 15 3 X 5 = 15 5 X 3 = 15 15 X 1 = 15 |

The same Factoring Method is applied for Polynomials, Algebraic and Quadratic equations as well. The important Polynomials, Algebra, Quadratic Equation Factorisation Formulas are given below.

(a + b)^{2 } = a^{2 }+ 2ab + b^{2}

(a − b)^{2 } = a^{2 }− 2ab + b^{2}

(a + b)^{3} =a^{3 }+ b^{3 }+ 3ab(a + b)

(a – b)^{3 } = a^{3 }– b^{3 }– 3ab(a – b)

(a + b)^{4} = a^{4 }+ 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}

(a − b)^{4} = a^{4} − 4a^{3}b + 6a^{2}b^{2} − 4ab^{3} + b^{4}

(a + b + c)^{2} = a^{2} + b^{2} +c^{2} + 2ab + 2ac + 2bc

(a + b + c +…)^{2} = a^{2} + b^{2} + c^{2 }+ … + 2(ab+ac+bc+…)

a^{2 }– b^{2} = (a + b)(a – b)

a^{2 }+ b^{2} = 1/2[(a + b)^{2 }+ (a – b)^{2}]

a^{3 }– b^{3 }= (a – b)(a^{2 }+ ab + b^{2})

a^{3 }+ b^{3 }= (a + b)(a^{2 }– ab + b^{2})

a^{4 }– b^{4} = (a – b)(a + b)(a^{2 }– ab + b^{2})

a^{5} – b^{5} = (a – b)(a^{4} + a^{3}b + a^{2}b^{2} + ab^{3 }+ b^{4})

Except the first 2 formulas from the above list, all other comes under Factoring Cubic Polynomials Formulas as well.

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a^{n } b^{n} = (a – b)(b^{0}a^{n-1} + b^{1}a^{n-2} + …… + b^{n-2}a^{1} + b^{n-1}a^{0})

a^{n }+ b^{n} = (a – b)(b^{0}a^{n-1} – b^{1}a^{n-2} + …… b^{n-2}a^{1} + b^{n-1}a^{0})

Factorisation or Factor Formula Trigonometry is given below

- sin
*A*+sin*B*= 2sin \(\frac{A+B\ }{2}\)cos\(\frac{A-B\ }{2}\)

- sin
*A*–sin*B*= 2cos \(\frac{A+B\ }{2}\)sin\(\frac{A-B\ }{2}\)

- cos
*A*+cos*B*= 2cos \(\frac{A+B\ }{2}\)cos\(\frac{A-B\ }{2}\)

- cos
*A*-cos*B*= –2sin \(\frac{A+B\ }{2}\)sin\(\frac{A-B\ }{2}\)

Factorisation is the process of finding factors of the given number, whether it is a Prime or Composite number. Whereas, **Prime Factorisation** is the process of finding Prime Factors of a given composite number. That is, the Prime Factorisation Method can be applied only for the Composite number. There are 2 methods to find Prime Factors of the number. To know everything about how to find Prime Factors of a given number click the link below.

**CLICK HERE TO KNOW ABOUT PRIME FACTORISATION**

Few solved examples using Factorisation are given below:

1. Express the following as in the form of (a+b)(a-b)(i) a^{2} – 64(ii) 20a ^{2} – 45b^{2}Answer: For representing the expressions in (a+b)(a-b) form, we will have to use the following formula: a^{2} – b^{2} = (a+b)(a-b) (i) a^{2} – 64 = a^{2} – 8^{2 }= (a + 8)(a – 8) (ii) 20a^{2} – 45b^{2} = 5(4a^{2} – 9b^{2}) = 5(2a + 3b)(2a – 3b) |

2 How to Find the Factor of an equation give below(i) 54x^{3}y + 81x^{4}y^{2}Answer: We can Factorise the expression 54x^{3}y + 81x^{4}y^{2} in the following way:= 2 × 3 × 3 × 3 × x × x × x × y + 3 × 3 × 3 × 3 × x × x × x × x × y × y = 3 × 3 × 3 × x × x × x × y × (2 + 3 xy) = 27x ^{3}y (2 + 3 xy) |

3. Solve the Quadratic Equation by Factorisation Method (x + y) ^{2} – 4xyAnswer: To solve this expression, expand (x + y)^{2}Use the formula: (x + y)^{2} = x^{2} + 2xy + y^{2}(x + y) ^{2} – 4xy = x^{2} + 2xy + y^{2} – 4xy= x ^{2} + y^{2} – 2xyWe know, (x – y) ^{2} = x^{2} + y^{2} – 2xySo, factorisation of (x + y) ^{2} – 4xy = (x – y)^{2} |

With the help of above-solved examples, you will get an idea on how to Factorise Quadratic Equations.

The frequently asked questions regarding Factorisation Formulas are given below:

**Q. What is the Factorisation method in Math?**

A. Factorisation is the reverse of multiplying out. Complex Algebraic, Polynomials, or Quadratic Equations are broken down to simpler Equations. The simpler equation when multiplied back gives the actual equation. This process is known as Factorisation.

**Q. Define Factorisation.**

A. Factorisation can be defined as the resolution of an entity into factors such that when multiplied together they give the original entity.

**Q. What is the first method to solve a Quadratic Equation?**

A. The first method to solve a Quadratic Equation is Factoring. Followed by Factoring, we will have to apply Quadratic Formula and Complete the Square.

**Q. How do you find the HCF?**

A. The **highest common factor** (HCF) is found by multiplying all the factors of that particular number.

Now, you are provided with all the necessary information regarding Factorisation Formulas. Students can make use of **NCERT Solutions** provided by Embibe for your exam preparation.

**DOWNLOAD FACTORISATION FORMULAS PDF FROM HERE**

We hope this detailed article on Factorisation Formulas helps you.

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