NCERT Solutions for Class 9 Maths Exercise 2.3 Question 4
Understanding the Question 🧐
In this problem, we are asked to factorise four different quadratic polynomials. A quadratic polynomial is an expression of the form &&ax^2 + bx + c&&, where &&a \neq 0&&. Factorisation is the process of breaking down the polynomial into a product of its factors (simpler polynomials, usually linear).
The most common method for this type of factorisation is splitting the middle term. This technique involves rewriting the middle term, &&bx&&, as a sum of two terms in such a way that we can then factor the expression by grouping. These detailed ncert solutions will guide you through every step.
Factorise:
(i) &&12x^2 – 7x + 1&&
(ii) &&2x^2 + 7x + 3&&
(iii) &&6x^2 + 5x – 6&&
(iv) &&3x^2 – x – 4&&
Part (i): Factorise &&12x^2 – 7x + 1&& 📝
Step 1: Identify the coefficients and find the product of &&a&& and &&c&&.
The given polynomial is &&12x^2 – 7x + 1&&. Comparing it with &&ax^2 + bx + c&&, we have:
- &&a = 12&&
- &&b = -7&&
- &&c = 1&&
Now, calculate the product of &&a&& and &&c&&:
&&a \times c = 12 \times 1 = 12&&
Step 2: Find two numbers that multiply to &&ac&& (12) and add up to &&b&& (-7).
We need to find two numbers whose product is &&12&& and whose sum is &&-7&&. Let’s think about the factors of 12:
- &&1 \times 12 = 12&&, but &&1+12 = 13&&.
- &&2 \times 6 = 12&&, but &&2+6 = 8&&.
- &&3 \times 4 = 12&&, and &&3+4 = 7&&. This is close!
Since the sum needs to be &&-7&& and the product is positive, both numbers must be negative.
Let’s try &&-3&& and &&-4&&:
Sum: &&(-3) + (-4) = -7&&
Product: &&(-3) \times (-4) = 12&&
These are our numbers: &&-3&& and &&-4&&.
Step 3: Split the middle term and factor by grouping.
We rewrite &&-7x&& as &&-3x – 4x&& (or &&-4x – 3x&&, the order doesn’t matter).
&&12x^2 – 7x + 1 = 12x^2 – 4x – 3x + 1&&
Now, group the terms:
&&= (12x^2 – 4x) + (-3x + 1)&&
Factor out the greatest common factor from each group:
&&= 4x(3x – 1) – 1(3x – 1)&&
Note: When factoring out from &&(-3x+1)&&, we take out &&-1&& to make the term inside the bracket match the first one.
Now, factor out the common binomial term &&(3x – 1)&&:
&&= (3x – 1)(4x – 1)&&
Final Answer: The factors are &&(3x – 1)&& and &&(4x – 1)&&.
Part (ii): Factorise &&2x^2 + 7x + 3&& 📝
Step 1: Identify coefficients and find the product &&ac&&.
Here, &&a = 2&&, &&b = 7&&, &&c = 3&&.
Product &&a \times c = 2 \times 3 = 6&&.
Step 2: Find two numbers that multiply to 6 and add to 7.
We need a product of &&6&& and a sum of &&7&&. The numbers are clearly &&6&& and &&1&&.
Sum: &&6 + 1 = 7&&
Product: &&6 \times 1 = 6&&
The numbers are &&6&& and &&1&&.
Step 3: Split the middle term and factor.
Rewrite &&7x&& as &&6x + x&&:
&&2x^2 + 7x + 3 = 2x^2 + 6x + x + 3&&
Group the terms:
&&= (2x^2 + 6x) + (x + 3)&&
Factor out the common factors:
&&= 2x(x + 3) + 1(x + 3)&&
Factor out the common binomial &&(x + 3)&&:
&&= (x + 3)(2x + 1)&&
Final Answer: The factors are &&(x + 3)&& and &&(2x + 1)&&.
Part (iii): Factorise &&6x^2 + 5x – 6&& 📝
Step 1: Identify coefficients and find the product &&ac&&.
Here, &&a = 6&&, &&b = 5&&, &&c = -6&&.
Product &&a \times c = 6 \times (-6) = -36&&.
Step 2: Find two numbers that multiply to -36 and add to 5.
Since the product is negative (&&-36&&), the two numbers must have opposite signs. Since the sum is positive (&&+5&&), the larger number must be positive.
Let’s check factors of 36: (9 and 4), (12 and 3), (6 and 6)…
Let’s try 9 and -4:
Sum: &&9 + (-4) = 5&&
Product: &&9 \times (-4) = -36&&
The numbers are &&9&& and &&-4&&.
Step 3: Split the middle term and factor.
Rewrite &&5x&& as &&9x – 4x&&:
&&6x^2 + 5x – 6 = 6x^2 + 9x – 4x – 6&&
Group the terms:
&&= (6x^2 + 9x) + (-4x – 6)&&
Factor out the common factors:
&&= 3x(2x + 3) – 2(2x + 3)&&
Factor out the common binomial &&(2x + 3)&&:
&&= (2x + 3)(3x – 2)&&
Final Answer: The factors are &&(2x + 3)&& and &&(3x – 2)&&.
Part (iv): Factorise &&3x^2 – x – 4&& 📝
Step 1: Identify coefficients and find the product &&ac&&.
Here, &&a = 3&&, &&b = -1&&, &&c = -4&&.
Product &&a \times c = 3 \times (-4) = -12&&.
Step 2: Find two numbers that multiply to -12 and add to -1.
The product is negative, so the numbers have opposite signs. The sum is negative, so the larger number must be negative.
Let’s try factors of 12: (4 and 3).
Let’s try &&-4&& and &&3&&:
Sum: &&(-4) + 3 = -1&&
Product: &&(-4) \times 3 = -12&&
The numbers are &&-4&& and &&3&&.
Step 3: Split the middle term and factor.
Rewrite &&-x&& as &&-4x + 3x&&:
&&3x^2 – x – 4 = 3x^2 – 4x + 3x – 4&&
Group the terms:
&&= (3x^2 + 3x) + (-4x – 4)&& (Rearranging for easier factoring)
&&= 3x(x + 1) – 4(x + 1)&&
Factor out the common binomial &&(x + 1)&&:
&&= (x + 1)(3x – 4)&&
Final Answer: The factors are &&(x + 1)&& and &&(3x – 4)&&.
Conclusion and Key Points ✅
We have successfully factorised all four quadratic polynomials using the splitting the middle term method. This powerful technique works for any factorisable quadratic polynomial of the form &&ax^2 + bx + c&&. The key is to find two numbers that multiply to give &&a \times c&& and add to give &&b&&. Once found, you can rewrite the polynomial, group terms, and find the common factors.
- If &&ac&& is positive, both numbers you are looking for will have the same sign as &&b&&.
- If &&ac&& is negative, the two numbers will have opposite signs, and the larger one will have the same sign as &&b&&.
- A quadratic polynomial is of the form &&ax^2 + bx + c&&, where &&a \neq 0&&.
- Factorisation means expressing the polynomial as a product of its factors.
- The ‘splitting the middle term’ method is based on finding two numbers, say &&p&& and &&q&&, such that &&p+q=b&& and &&pq=ac&&.
- Always check your final answer by multiplying the factors back together to see if you get the original polynomial.
FAQ
Q: What is the method to factorise &&12x^2 – 7x + 1&&?
A: The method is called splitting the middle term. We find two numbers that multiply to &&12 \times 1 = 12&& and add to &&-7&&. These numbers are &&-4&& and &&-3&&. We rewrite the expression as &&12x^2 – 4x – 3x + 1&& and factor by grouping to get the final answer: &&(3x – 1)(4x – 1)&&.
Q: How do you split the middle term in &&2x^2 + 7x + 3&&?
A: First, calculate the product &&a \times c = 2 \times 3 = 6&&. Then, find two numbers that multiply to &&6&& and add to &&7&& (the coefficient of the middle term). The numbers are &&6&& and &&1&&. So, you split &&7x&& into &&6x + x&&. The polynomial becomes &&2x^2 + 6x + x + 3&&, which factors into &&(x + 3)(2x + 1)&&.
Q: What are the factors of &&6x^2 + 5x – 6&&?
A: The factors are &&(2x + 3)&& and &&(3x – 2)&&. This is found by finding two numbers that multiply to &&6 \times (-6) = -36&& and add to &&5&&. The numbers are &&9&& and &&-4&&. So, we split &&5x&& into &&9x – 4x&& and then factor by grouping.
Q: Can you explain the factorisation of &&3x^2 – x – 4&&?
A: To factorise &&3x^2 – x – 4&&, we look for two numbers that have a product of &&3 \times (-4) = -12&& and a sum of &&-1&&. These numbers are &&-4&& and &&3&&. We rewrite the expression as &&3x^2 – 4x + 3x – 4&&. After grouping and factoring, we get the result &&(x + 1)(3x – 4)&&.
Q: What is the general form of a quadratic polynomial?
A: The general form is &&ax^2 + bx + c&&, where &&a, b,&& and &&c&& are real numbers, and most importantly, &&a \neq 0&&. If &&a=0&&, the term &&x^2&& disappears, and it’s no longer a quadratic polynomial.
Further Reading
For a deeper understanding of polynomials and to access the original textbook questions, you can visit the official NCERT website. The concepts of factorisation are fundamental to algebra and are explained thoroughly in the Class 9 Maths NCERT textbook.
