NCERT Solutions for Class 9 Maths Exercise 2.4 Question 8
Understanding the Question 🧐
This question asks us to factorise several polynomials. The key to solving these problems is to recognize that they are the expanded forms of the cubic identities &&(x+y)^3&& or &&(x-y)^3&&. These ncert solutions will show you how to identify the pattern and find the factors.
Factorise each of the following:
We need to match the given polynomials with one of these two identities:
- &&(x+y)^3 = x^3 + y^3 + 3x^2y + 3xy^2&&
- &&(x-y)^3 = x^3 – y^3 – 3x^2y + 3xy^2&&
Part (i): Factorise &&8a^3 + b^3 + 12a^2b + 6ab^2&& 📝
Step 1: Identify the pattern.
All the terms are positive, so this expression should match the identity &&(x+y)^3&&.
Step 2: Identify &&x&& and &&y&& from the cubic terms.
We can write the cubic terms as:
&&8a^3 = (2a)^3&&
&&b^3 = (b)^3&&
This suggests that &&x = 2a&& and &&y = b&&.
Step 3: Verify the other terms.
Let’s check if the remaining terms match &&3x^2y&& and &&3xy^2&& using &&x = 2a&& and &&y = b&&.
&&3x^2y = 3(2a)^2(b) = 3(4a^2)(b) = 12a^2b&&. (Matches!)
&&3xy^2 = 3(2a)(b)^2 = 6ab^2&&. (Matches!)
Step 4: Write the final factored form.
Since the expression matches the identity &&x^3 + y^3 + 3x^2y + 3xy^2&&, we can write it in its factored form &&(x+y)^3&&.
Answer: &&8a^3 + b^3 + 12a^2b + 6ab^2 = (2a + b)^3 = (2a+b)(2a+b)(2a+b)&&
Part (ii): Factorise &&8a^3 – b^3 – 12a^2b + 6ab^2&& 📝
Step 1: Identify the pattern.
The presence of negative terms (&&-b^3&& and &&-12a^2b&&) suggests this expression matches the identity &&(x-y)^3&&.
Step 2: Identify &&x&& and &&y&&.
The cubic terms are &&8a^3 = (2a)^3&& and &&b^3 = (b)^3&&.
This gives us &&x = 2a&& and &&y = b&&.
Step 3: Verify the other terms.
The identity &&(x-y)^3&& has middle terms &&-3x^2y&& and &&+3xy^2&&.
&&-3x^2y = -3(2a)^2(b) = -3(4a^2)(b) = -12a^2b&&. (Matches!)
&&+3xy^2 = 3(2a)(b)^2 = 6ab^2&&. (Matches!)
Step 4: Write the final factored form.
The polynomial perfectly matches the expansion of &&(x-y)^3&&.
Answer: &&8a^3 – b^3 – 12a^2b + 6ab^2 = (2a – b)^3 = (2a-b)(2a-b)(2a-b)&&
Part (iii): Factorise &&27 – 125a^3 – 135a + 225a^2&& 📝
Step 1: Identify the pattern.
With mixed signs, this should match &&(x-y)^3&&.
Step 2: Identify &&x&& and &&y&&.
The cubic terms are &&27 = (3)^3&& and &&125a^3 = (5a)^3&&.
So, let’s take &&x = 3&& and &&y = 5a&&.
Step 3: Verify the other terms.
Let’s check the terms &&-3x^2y&& and &&+3xy^2&&.
&&-3x^2y = -3(3)^2(5a) = -3(9)(5a) = -135a&&. (Matches!)
&&+3xy^2 = 3(3)(5a)^2 = 9(25a^2) = 225a^2&&. (Matches!)
Step 4: Write the final factored form.
The expression &&27 – 125a^3 – 135a + 225a^2&& is the expansion of &&(3-5a)^3&&.
Answer: &&27 – 125a^3 – 135a + 225a^2 = (3 – 5a)^3 = (3-5a)(3-5a)(3-5a)&&
Part (iv): Factorise &&64a^3 – 27b^3 – 144a^2b + 108ab^2&& 📝
Step 1: Identify the pattern.
The negative signs point to the identity &&(x-y)^3&&.
Step 2: Identify &&x&& and &&y&&.
The perfect cubes are &&64a^3 = (4a)^3&& and &&27b^3 = (3b)^3&&.
This suggests &&x = 4a&& and &&y = 3b&&.
Step 3: Verify the other terms.
Let’s check the middle terms.
&&-3x^2y = -3(4a)^2(3b) = -3(16a^2)(3b) = -144a^2b&&. (Matches!)
&&+3xy^2 = 3(4a)(3b)^2 = 12a(9b^2) = 108ab^2&&. (Matches!)
Step 4: Write the final factored form.
The expression matches the expansion perfectly.
Answer: &&64a^3 – 27b^3 – 144a^2b + 108ab^2 = (4a – 3b)^3 = (4a-3b)(4a-3b)(4a-3b)&&
Part (v): Factorise &&27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p&& 📝
Step 1: Identify the pattern.
The negative signs indicate the &&(x-y)^3&& identity.
Step 2: Identify &&x&& and &&y&&.
The cubic terms, including the fraction, are:
&&27p^3 = (3p)^3&&
&&\frac{1}{216} = (\frac{1}{6})^3&&
This means &&x = 3p&& and &&y = \frac{1}{6}&&.
Step 3: Verify the other terms.
Now, verify the middle terms involving fractions. Be careful with cancellation!
&&-3x^2y = -3(3p)^2(\frac{1}{6}) = -3(9p^2)(\frac{1}{6}) = -\frac{27p^2}{6} = -\frac{9}{2}p^2&&. (Matches!)
&&+3xy^2 = 3(3p)(\frac{1}{6})^2 = 9p(\frac{1}{36}) = \frac{9p}{36} = \frac{1}{4}p&&. (Matches!)
Step 4: Write the final factored form.
The polynomial is the expansion of &&(3p – \frac{1}{6})^3&&.
Answer: &&27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p = (3p – \frac{1}{6})^3 = (3p – \frac{1}{6})(3p – \frac{1}{6})(3p – \frac{1}{6})&&
- First, rewrite the given polynomial in the standard form if it’s jumbled.
- Identify the two perfect cube terms to find your potential &&x&& and &&y&&.
- Always verify the middle two terms (&&3x^2y&& and &&3xy^2&&) before concluding the answer.
- Don’t be afraid of fractions! The same rules apply.
FAQ
Q: What is the main technique used to factorise these polynomials?
A: The main technique is to recognize that the given polynomials are the expanded forms of the cubic algebraic identities &&(x+y)^3&& or &&(x-y)^3&&. By identifying the pattern, we can easily write the factored form.
Q: How can you tell whether to use the &&(x+y)^3&& or &&(x-y)^3&& identity?
A: The signs of the terms are the key. If all terms in the polynomial are positive, it matches the expansion of &&(x+y)^3&&. If the terms with odd powers of the second variable (like &&y^3&& and the &&x^2y&& term) are negative, it matches the expansion of &&(x-y)^3&&.
Q: What are the first terms to identify in the polynomial?
A: You should first identify the two terms that are perfect cubes. This helps you determine the values of ‘x’ and ‘y’ for the identity. For example, in &&8a^3 + b^3 + …&&, the perfect cubes are &&8a^3&& and &&b^3&&.
Q: How do you handle fractions in the factorisation, like in the last subpart?
A: Fractions are handled in the exact same way. You find the cube root of the fractional term. For example, the cube root of &&\frac{1}{216}&& is &&\frac{1}{6}&&. This value then becomes your ‘y’ in the identity.
Q: What is the full expansion of the identity &&(x-y)^3&&?
A: The full expansion is &&(x-y)^3 = x^3 – y^3 – 3x^2y + 3xy^2&&.
Q: What does it mean to ‘factorise’ a polynomial?
A: To factorise a polynomial means to express it as a product of its factors, which are typically simpler polynomials. In this case, we express a four-term polynomial as the cube of a binomial, like &&(2a+b)^3&&.
Further Reading
For more practice on factorisation and other concepts of polynomials, you can refer to the official NCERT textbook available on their website: https://ncert.nic.in/
