NCERT Solutions for Class 9 Maths Exercise 2.4 Question 4
Understanding the Question 🧐
In this question, we are asked to factorise four different quadratic polynomials. A quadratic polynomial is an expression of the form &&ax^2 + bx + c&&. We will use the method of splitting the middle term.
Here’s how the method works:
- Identify the coefficients &&a&&, &&b&&, and &&c&&.
- Calculate the product &&a \times c&&.
- Find two numbers, let’s call them &&p&& and &&q&&, such that their sum is &&p + q = b&& and their product is &&p \times q = a \times c&&.
- Rewrite the middle term &&bx&& as &&px + qx&&.
- Factor the resulting four-term expression by grouping.
Factorise the following:
Part (i): &&12x^2 – 7x + 1&& 📝
Step 1: Identify coefficients and find the product &&ac&&.
Here, &&a = 12&&, &&b = -7&&, and &&c = 1&&. The product is &&a \times c = 12 \times 1 = 12&&.
Step 2: Find two numbers that sum to &&b&& and multiply to &&ac&&.
We need two numbers that sum to &&-7&& and multiply to &&12&&. The numbers are &&-4&& and &&-3&&.
(Check: &&-4 + (-3) = -7&& and &&(-4) \times (-3) = 12&&).
Step 3: Split the middle term.
We rewrite &&-7x&& as &&-4x – 3x&&.
&&12x^2 – 4x – 3x + 1&&
Step 4: Factor by grouping.
Group the first two terms and the last two terms.
&& (12x^2 – 4x) + (-3x + 1) &&
Factor out the greatest common factor from each group.
&&4x(3x – 1) – 1(3x – 1)&&
Now, factor out the common bracket &&(3x – 1)&&.
Final Answer: &&(3x – 1)(4x – 1)&&
Part (ii): &&2x^2 + 7x + 3&& 📝
Step 1: Identify coefficients and find the product &&ac&&.
Here, &&a = 2&&, &&b = 7&&, and &&c = 3&&. The product is &&a \times c = 2 \times 3 = 6&&.
Step 2: Find two numbers.
We need two numbers that sum to &&7&& and multiply to &&6&&. The numbers are &&6&& and &&1&&.
(Check: &&6 + 1 = 7&& and &&6 \times 1 = 6&&).
Step 3: Split the middle term.
&&2x^2 + 6x + 1x + 3&&
Step 4: Factor by grouping.
&& (2x^2 + 6x) + (x + 3) &&
&&2x(x + 3) + 1(x + 3)&&
Final Answer: &&(x + 3)(2x + 1)&&
Part (iii): &&6x^2 + 5x – 6&& 📝
Step 1: Identify coefficients and find the product &&ac&&.
Here, &&a = 6&&, &&b = 5&&, and &&c = -6&&. The product is &&a \times c = 6 \times (-6) = -36&&.
Step 2: Find two numbers.
We need two numbers that sum to &&5&& and multiply to &&-36&&. The numbers are &&9&& and &&-4&&.
(Check: &&9 + (-4) = 5&& and &&9 \times (-4) = -36&&).
Step 3: Split the middle term.
&&6x^2 + 9x – 4x – 6&&
Step 4: Factor by grouping.
&& (6x^2 + 9x) + (-4x – 6) &&
&&3x(2x + 3) – 2(2x + 3)&&
Final Answer: &&(2x + 3)(3x – 2)&&
Part (iv): &&3x^2 – x – 4&& 📝
Step 1: Identify coefficients and find the product &&ac&&.
Here, &&a = 3&&, &&b = -1&&, and &&c = -4&&. The product is &&a \times c = 3 \times (-4) = -12&&.
Step 2: Find two numbers.
We need two numbers that sum to &&-1&& and multiply to &&-12&&. The numbers are &&-4&& and &&3&&.
(Check: &&-4 + 3 = -1&& and &&(-4) \times 3 = -12&&).
Step 3: Split the middle term.
&&3x^2 – 4x + 3x – 4&&
Step 4: Factor by grouping.
&& (3x^2 – 4x) + (3x – 4) &&
&&x(3x – 4) + 1(3x – 4)&&
Final Answer: &&(3x – 4)(x + 1)&&
Conclusion and Key Points ✅
Factorising quadratic polynomials by splitting the middle term is a systematic process. By finding two numbers that satisfy the sum and product conditions, you can break down the polynomial into a four-term expression that is easily factorised by grouping.
- If the product &&ac&& is positive, both numbers you are looking for will have the same sign as the middle term &&b&&.
- If the product &&ac&& is negative, the two numbers will have different signs. The larger number will take the same sign as the middle term &&b&&.
- Write the polynomial in the standard form &&ax^2 + bx + c&&.
- Calculate the product of the first and last coefficients, &&ac&&.
- Find two numbers, &&p&& and &&q&&, where &&p+q=b&& and &&pq=ac&&.
- Rewrite the polynomial as &&ax^2 + px + qx + c&&.
- Factor by taking the common factor from the first two terms and the last two terms.
- Factor out the common binomial term.
FAQ
Q: What is the ‘splitting the middle term’ method?
A: It is a technique used to factorise a quadratic polynomial of the form &&ax^2 + bx + c&&. The method involves rewriting the middle term ‘&&bx&&’ as a sum of two terms, &&px + qx&&, where the sum of &&p&& and &&q&& is ‘&&b&&’ and their product is ‘&&ac&&’. This allows for factorisation by grouping.
Q: How do you factorise &&12x^2 – 7x + 1&&?
A: You need two numbers that add to &&-7&& and multiply to &&12&&. These numbers are &&-4&& and &&-3&&. Split the middle term to get &&12x^2 – 4x – 3x + 1&&. Factor by grouping to get the final answer: &&(3x – 1)(4x – 1)&&.
Q: How do you decide the signs of the two numbers when splitting the middle term?
A: If the product ‘&&ac&&’ is positive, both numbers have the same sign as the middle term ‘&&b&&’. If ‘&&ac&&’ is negative, the numbers have opposite signs, with the larger number taking the same sign as ‘&&b&&’.
Q: Does the order of the split terms matter?
A: No, the order does not matter. For example, writing &&12x^2 – 4x – 3x + 1&& or &&12x^2 – 3x – 4x + 1&& will both lead to the same final factors after grouping.
Q: What if you can’t find two integers that fit the sum and product criteria?
A: If you cannot find two integers that satisfy the conditions, it means the quadratic polynomial cannot be factorised using this method with integers. Other methods, like the quadratic formula (which you will learn in higher classes), might be needed.
Further Reading
To practice more problems on factorisation, refer to your Class 9 Maths textbook. The official NCERT textbooks are an excellent resource and are available on their website: https://ncert.nic.in/.
