NCERT Solutions for Class 9 Maths Exercise 2.4 Question 4
Understanding the Question 🧐
This question asks us to factorise four different quadratic polynomials. This involves breaking down each polynomial into a product of simpler linear factors. These are essential ncert solutions for mastering algebraic manipulations.
Factorise:
(i) &&12x^2 – 7x + 1&&
(ii) &&2x^2 + 7x + 3&&
(iii) &&6x^2 + 5x – 6&&
(iv) &&3x^2 – x – 4&&
Our Expert’s Approach
To solve this, our experts first identify the core concept, which is the factorisation of quadratic trinomials of the form &&ax^2 + bx + c&&. The most reliable and systematic method for these problems is splitting the middle term. This method is chosen for its clarity and reliability, reflecting years of teaching experience. We will find two numbers that sum to ‘&&b&&’ and multiply to ‘&&ac&&’, then rewrite the polynomial to factor by grouping.
Part (i): Factorise &&12x^2 – 7x + 1&& 📝
Step 1: Identify coefficients
We compare the polynomial &&12x^2 – 7x + 1&& with the general form &&ax^2 + bx + c&&.
Here, &&a = 12&&, &&b = -7&&, and &&c = 1&&.
Step 2: Find the product &&ac&&
We need to find two numbers whose product is &&a \times c = 12 \times 1 = 12&& and whose sum is &&b = -7&&.
Step 3: Split the middle term
Let’s find factors of &&12&& that add up to &&-7&&. The numbers are &&-4&& and &&-3&&.
(Check: &&(-4) \times (-3) = 12&&, and &&(-4) + (-3) = -7&&).
So, we can rewrite &&–7x&& as &&–4x – 3x&&.
Step 4: Factor by grouping
&&12x^2 – 7x + 1 = 12x^2 – 4x – 3x + 1&&
Group the terms: &&(12x^2 – 4x) + (–3x + 1)&&
Factor out the common terms from each group:
&&4x(3x – 1) – 1(3x – 1)&&
Now, factor out the common binomial &&(3x – 1)&&:
&&(3x – 1)(4x – 1)&&
Part (ii): Factorise &&2x^2 + 7x + 3&& 📝
Step 1: Identify coefficients
Comparing with &&ax^2 + bx + c&&, we get &&a = 2&&, &&b = 7&&, and &&c = 3&&.
Step 2: Find the product &&ac&&
The product &&a \times c = 2 \times 3 = 6&&. The sum required is &&b = 7&&.
Step 3: Split the middle term
The two numbers that multiply to &&6&& and add to &&7&& are &&6&& and &&1&&.
So, we rewrite &&7x&& as &&6x + x&&.
Step 4: Factor by grouping
&&2x^2 + 7x + 3 = 2x^2 + 6x + x + 3&&
Group and factor: &&(2x^2 + 6x) + (x + 3)&&
&&2x(x + 3) + 1(x + 3)&&
Factor out the common binomial &&(x + 3)&&:
&&(x + 3)(2x + 1)&&
Part (iii): Factorise &&6x^2 + 5x – 6&& 📝
Step 1: Identify coefficients
Here, &&a = 6&&, &&b = 5&&, and &&c = -6&&.
Step 2: Find the product &&ac&&
The product &&a \times c = 6 \times (-6) = -36&&. The sum required is &&b = 5&&.
Step 3: Split the middle term
We need two numbers that multiply to &&-36&& and have a sum of &&5&&. These numbers are &&9&& and &&-4&&.
(Check: &&9 \times (-4) = -36&&, and &&9 + (-4) = 5&&).
So, we rewrite &&5x&& as &&9x – 4x&&.
Step 4: Factor by grouping
&&6x^2 + 5x – 6 = 6x^2 + 9x – 4x – 6&&
Group and factor: &&(6x^2 + 9x) + (–4x – 6)&&
&&3x(2x + 3) – 2(2x + 3)&&
Factor out the common binomial &&(2x + 3)&&:
&&(2x + 3)(3x – 2)&&
Part (iv): Factorise &&3x^2 – x – 4&& 📝
Step 1: Identify coefficients
Here, &&a = 3&&, &&b = -1&&, and &&c = -4&&.
Step 2: Find the product &&ac&&
The product &&a \times c = 3 \times (-4) = -12&&. The sum required is &&b = -1&&.
Step 3: Split the middle term
We need two numbers that multiply to &&-12&& and add to &&-1&&. These numbers are &&-4&& and &&3&&.
(Check: &&(-4) \times 3 = -12&&, and &&-4 + 3 = -1&&).
So, we rewrite &&-x&& as &&–4x + 3x&&.
Step 4: Factor by grouping
&&3x^2 – x – 4 = 3x^2 – 4x + 3x – 4&&
Group and factor: &&(3x^2 + 3x) + (–4x – 4)&&
&&3x(x + 1) – 4(x + 1)&&
Factor out the common binomial &&(x + 1)&&:
&&(x + 1)(3x – 4)&&
Conclusion and Key Points ✅
We have successfully factorised all four quadratic polynomials by using the ‘splitting the middle term’ method. This technique is fundamental in algebra for solving quadratic equations and simplifying expressions.
- Part (i): The factors of &&12x^2 – 7x + 1&& are &&(3x – 1)(4x – 1)&&.
- Part (ii): The factors of &&2x^2 + 7x + 3&& are &&(x + 3)(2x + 1)&&.
- Part (iii): The factors of &&6x^2 + 5x – 6&& are &&(2x + 3)(3x – 2)&&.
- Part (iv): The factors of &&3x^2 – x – 4&& are &&(x + 1)(3x – 4)&&.
- Always check your factorisation by multiplying the factors back. The result should be the original polynomial.
- The ‘splitting the middle term’ method works for factorising trinomials of the form &&ax^2 + bx + c&&.
- Be careful with signs when grouping terms, especially when factoring out a negative number.
FAQ
What is the ‘splitting the middle term’ method for factorisation?
It is a technique used to factorise a quadratic trinomial of the form &&ax^2 + bx + c&&. The goal is to find two numbers that add up to ‘&&b&&’ (the middle term’s coefficient) and multiply to ‘&&ac&&’ (the product of the first and last term’s coefficients). The middle term is then rewritten as the sum of these two numbers, allowing the polynomial to be factored by grouping.
How do you find the two numbers for splitting the middle term?
First, calculate the product of the coefficient of &&x^2&& (which is ‘&&a&&’) and the constant term (‘&&c&&’). This gives you ‘&&ac&&’. Then, list the factor pairs of ‘&&ac&&’. Finally, identify the pair that adds or subtracts to give you the coefficient of the middle term, ‘&&b&&’.
What is the general form of a quadratic polynomial?
The general form of a quadratic polynomial is &&ax^2 + bx + c&&, where ‘&&a&&’, ‘&&b&&’, and ‘&&c&&’ are real numbers and ‘&&a&&’ is not equal to zero.
Why is it important for the coefficient ‘a’ not to be zero in a quadratic polynomial?
If ‘&&a&&’ were zero, the term &&ax^2&& would become zero, and the polynomial would become &&bx + c&&, which is a linear polynomial, not a quadratic one. The &&x^2&& term is what defines a polynomial as quadratic.
Can all quadratic polynomials be factorised by splitting the middle term?
No, not all quadratic polynomials can be easily factorised using integer pairs for splitting the middle term. Some may have irrational or complex roots, requiring other methods like completing the square or the quadratic formula.
Further Reading
For more practice and a deeper understanding of polynomials, you can refer to the official NCERT textbook available at the NCERT website.
