OP Malhotra Class 9 Maths Solutions Chapter 4 Factorisation Exercise 4 (C)

Factorise :

Question 1. \(x^2 + 4x + 4\)
Answer:
We need to factorise the expression \(x^2 + 4x + 4\). This expression matches the algebraic identity \(a^2 + 2ab + b^2 = (a+b)^2\).
\(x^2 + 4x + 4 = (x)^2 + 2 \times x \times 2 + (2)^2\)
\( \implies (x + 2)^2 \)
In simple words: We can write this expression as a perfect square. It is the same as \( (x+2) \) multiplied by itself.

🎯 Exam Tip: Identify perfect square trinomials and use the \( (a+b)^2 \) or \( (a-b)^2 \) identity for quick factorisation.

Question 2. \(x^2 + 6x + 9\)
Answer:
To factorise \(x^2 + 6x + 9\), we look for a perfect square pattern. This expression fits the identity \(a^2 + 2ab + b^2 = (a+b)^2\).
\(x^2 + 6x + 9 = (x)^2 + 2 \times x \times 3 + (3)^2\)
\( \implies (x + 3)^2 \)
In simple words: This expression is a perfect square. It can be written as \( (x+3) \) multiplied by itself.

🎯 Exam Tip: Recognise the pattern of \(2ab\) in the middle term to correctly identify a perfect square identity.

Question 3. \(x^2 - 10x + 25\)
Answer:
We need to factorise \(x^2 - 10x + 25\). This expression matches the identity \(a^2 - 2ab + b^2 = (a-b)^2\).
\(x^2 - 10x + 25 = (x)^2 - 2 \times x \times 5 + (5)^2\)
\( \implies (x - 5)^2 \)
In simple words: This expression can be simplified into a perfect square. It is the same as \( (x-5) \) multiplied by itself.

🎯 Exam Tip: Pay close attention to the sign of the middle term; a minus sign indicates the use of the \( (a-b)^2 \) identity.

Question 4. \(4x^2 - 4x + 1\)
Answer:
To factorise \(4x^2 - 4x + 1\), we will use the identity \(a^2 - 2ab + b^2 = (a-b)^2\). First, we identify \(a\) and \(b\). Here, \(a = 2x\) and \(b = 1\).
\(4x^2 - 4x + 1 = (2x)^2 - 2 \times (2x) \times 1 + (1)^2\)
\( \implies (2x - 1)^2 \)
In simple words: This expression is a perfect square of a binomial. It simplifies to \( (2x-1) \) multiplied by itself.

🎯 Exam Tip: Remember to find the square root of the first term (like \(4x^2\)) to correctly identify \(a\) in the identity.

Question 5. \(1 - 8x + 16x^2\)
Answer:
We need to factorise \(1 - 8x + 16x^2\). This is in the form \(a^2 - 2ab + b^2 = (a-b)^2\). Here, \(a = 1\) and \(b = 4x\).
\(1 - 8x + 16x^2 = (1)^2 - 2 \times 1 \times 4x + (4x)^2\)
\( \implies (1 - 4x)^2 \)
In simple words: This expression is a perfect square trinomial. It can be written as \( (1-4x) \) multiplied by itself.

🎯 Exam Tip: Even if the terms are reordered, always look for the square terms and the \(2ab\) term to apply the correct identity.

Question 6. \(49x^4 + 168x^2y^2 + 144y^4\)
Answer:
To factorise \(49x^4 + 168x^2y^2 + 144y^4\), we use the identity \(a^2 + 2ab + b^2 = (a+b)^2\). Here, \(a = 7x^2\) and \(b = 12y^2\).
\(49x^4 + 168x^2y^2 + 144y^4 = (7x^2)^2 + 2 \times 7x^2 \times 12y^2 + (12y^2)^2\)
\( \implies (7x^2 + 12y^2)^2 \)
In simple words: This is a perfect square expression with multiple variables. We can write it as the square of \( (7x^2 + 12y^2) \).

🎯 Exam Tip: When dealing with multiple variables and higher powers, ensure each part of \(a\) and \(b\) is correctly identified, including their exponents.

Question 7. \(x^2 + x + \frac{1}{4}\)
Answer:
We need to factorise \(x^2 + x + \frac{1}{4}\). This expression can be seen as a perfect square using the identity \(a^2 + 2ab + b^2 = (a+b)^2\). Here, \(a = x\) and \(b = \frac{1}{2}\).
\(x^2 + x + \frac{1}{4} = (x)^2 + 2 \times x \times \frac{1}{2} + \left(\frac{1}{2}\right)^2\)
\( \implies \left(x + \frac{1}{2}\right)^2 \)
In simple words: This expression uses a fraction but still follows the perfect square pattern. It simplifies to \( (x + \frac{1}{2}) \) multiplied by itself.

🎯 Exam Tip: Do not be intimidated by fractions; treat them like any other number when applying algebraic identities.

Question 8. \(25p^2 + \frac{5p}{2q} + \frac{1}{16q^2}\)
Answer:
To factorise \(25p^2 + \frac{5p}{2q} + \frac{1}{16q^2}\), we use the identity \(a^2 + 2ab + b^2 = (a+b)^2\). We identify \(a = 5p\) and \(b = \frac{1}{4q}\).
\(25p^2 + \frac{5p}{2q} + \frac{1}{16q^2} = (5p)^2 + 2 \times 5p \times \frac{1}{4q} + \left(\frac{1}{4q}\right)^2\)
\( \implies \left(5p + \frac{1}{4q}\right)^2 \)
In simple words: Even with fractions and multiple variables, this expression follows the perfect square rule. It becomes the square of \( (5p + \frac{1}{4q}) \).

🎯 Exam Tip: When terms involve fractions with variables in the denominator, ensure you correctly identify the square roots for \(a\) and \(b\) and double-check the middle term product.