Maharashtra Board Class 8 Maths part 1 Chapter 6 Factorisation of Algebraic expressions PDF Download

Factorisation Of Algebraic Expressions

Let's Recall

In the previous standard we have learnt to factorise the expressions of the form \(ax + ay\) and \(a^2 - b^2\)

For example:

1. \(4xy + 8xy^2 = 4xy(1 + 2y)\)

2. \(p^2 - 9q^2 = (p)^2 - (3q)^2 = (p + 3q)(p - 3q)\)

Let's Learn

Factors Of A Quadratic Trinomial

An expression of the form \(ax^2 + bx + c\) is called a quadratic trinomial.

We know that \((x + a)(x + b) = x^2 + (a + b)x + ab\)

Therefore, the factors of \(x^2 + (a + b)x + ab\) are \((x + a)\) and \((x + b)\).

To find the factors of \(x^2 + 5x + 6\), by comparing it with \(x^2 + (a + b)x + ab\) we get, \(a + b = 5\) and \(ab = 6\). So, let us find the factors of 6 whose sum is 5.

Then writing the trinomial in the form \(x^2 + (a + b)x + ab\), find its factors.

\(x^2 + 5x + 6 = x^2 + (3 + 2)x + 3 \times 2\) ........ \(x^2 + (a + b)x + ab\)

\(= x^2 + 3x + 2x + 6\) ........ multiply (3 + 2) by x, make two groups of the four terms obtained.

\(= x(x + 3) + 2(x + 3) = (x + 3)(x + 2)\)

Study the following examples to know how a given trinomial is factorised.

Example 1

Factorise: \(2x^2 - 9x + 9\)

Solution: First we find the product of the coefficient of the square term and the constant term. Here the product is \(2 \times 9 = 18\).

Now, find factors of 18 whose sum is -9, that is equal to the coefficient of the middle term.

\(18 = (-6) \times (-3); (-6) + (-3) = -9\)

Write the term \(-9x\) as \(-6x - 3x\)

\(2x^2 - 9x + 9 = 2x^2 - 6x - 3x + 9\)

\(= 2x(x - 3) - 3(x - 3)\)

\(= (x - 3)(2x - 3)\)

Teacher's Note

Factorisation helps us break down big algebraic expressions into smaller parts. It is like breaking a big number into its prime factors, just like 12 = 2 × 2 × 3.

Exam Trick

Remember the steps: First find the product of first and last coefficient. Then find two numbers whose product equals this and whose sum equals the middle coefficient. This is the key to solving all trinomial factorisation problems.

Points To Remember

\(ax^2 + bx + c\) is a quadratic trinomial.
Find the product of coefficient of \(x^2\) and constant term.
Find factors of this product whose sum equals the middle term coefficient.
Split the middle term using these factors.
Group and take out common factors to get the final answer.

Example 2

Factorise: \(2x^2 + 5x - 18\)

Solution: \(2x^2 + 5x - 18\)

\(= 2x^2 + 9x - 4x - 18\)

\(= x(2x + 9) - 2(2x + 9)\)

\(= (2x + 9)(x - 2)\)

Example 3

Factorise: \(x^2 - 10x + 21\)

Solution: \(x^2 - 10x + 21\)

\(= x^2 - 7x - 3x + 21\)

\(= x(x - 7) - 3(x - 7)\)

\(= (x - 7)(x - 3)\)

Example 4

Find the factors of \(2y^2 - 4y - 30\)

Solution: \(2y^2 - 4y - 30\)

\(= 2(y^2 - 2y - 15\)) ........ taking out the common factor 2

\(= 2(y^2 - 5y + 3y - 15\))

\(= 2[y(y - 5) + 3(y - 5)]\)

\(= 2(y - 5)(y + 3)\)

Teacher's Note

Always look for common factors first before factorising. In example 4, we took out 2 first, which made the problem easier to solve.

Exam Trick

Before solving any problem, check if all terms have a common factor. If yes, take it out first. This reduces your calculation work and reduces errors.

Points To Remember

Check for common factors in all terms first.
Take out the common factor before factorising the trinomial.
Follow the steps of splitting the middle term carefully.
Always verify your answer by multiplying back the factors.
Practice different types of problems to master the technique.

Let's Learn

Factors Of \(a^3 + b^3\)

We know that \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\), which we can write as \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)

Now, \(a^3 + b^3 + 3ab(a + b) = (a + b)^3\) ........ interchanging the sides.

Therefore, \(a^3 + b^3 = (a + b)^3 - 3ab(a + b) = [(a + b)(a + b)^2] - 3ab(a + b)\)

\(= (a + b)[(a + b)^2 - 3ab] = (a + b)(a^2 + 2ab + b^2 - 3ab)\)

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

Therefore, \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

Example 1

\(x^3 + 27y^3 = x^3 + (3y)^3\)

\(= (x + 3y)[x^2 - x(3y) + (3y)^2]\)

\(= (x + 3y)[x^2 - 3xy + 9y^2]\)

Example 2

\(8p^3 + 125q^3 = (2p)^3 + (5q)^3 = (2p + 5q)[(2p)^2 - 2p \times 5q + (5q)^2]\)

\(= (2p + 5q)(4p^2 - 10pq + 25q^2)\)

Teacher's Note

The formula \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) is very important. You can use it whenever you see the sum of two cubes, just like in a real market when you bundle two cubic boxes together.

Exam Trick

Remember: The middle term in the answer always has a negative sign. So \(a^3 + b^3\) gives you \((a + b)(a^2 - ab + b^2)\) with a minus sign in the bracket.

Points To Remember

\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
First write the expression as sum of two cubes.
Find the cube roots of both terms.
Use the formula to get the factorised form.
Check that the middle term has a negative sign.

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