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

S Chand Class 9 ICSE Maths Solutions Chapter 4 Factorisation Ex 4(D)

Factorise the following :

Question 1. \(x^2-4\)
Answer: We need to factorise the expression \(x^2 - 4\). This can be written as the difference of two squares.
\(x^2 - 4 = (x)^2 - (2)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=x\) and \(b=2\):
\( = (x + 2)(x - 2) \)
This identity is a fundamental tool for factoring expressions that are a perfect square minus another perfect square.
In simple words: To factor this, we see it's like "something squared minus something else squared." We find what numbers were squared (which are \(x\) and 2), then write them as two brackets: one with a plus sign and one with a minus sign in the middle.

🎯 Exam Tip: Always recognize expressions in the form \(a^2 - b^2\) as they can be easily factored into \((a-b)(a+b)\). This is a common pattern in algebra.

Question 2. \(y^2 – 25\)
Answer: We need to factorise the expression \(y^2 - 25\). This is also in the form of the difference of two squares.
\(y^2 - 25 = (y)^2 - (5)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=y\) and \(b=5\):
\( = (y + 5)(y - 5) \)
Recognizing perfect squares like 25 (\(5^2\)) is crucial for applying this factorization method efficiently.
In simple words: This problem also uses the rule of "difference of two squares." You find the square root of each part (which are \(y\) and 5), then write them in two brackets, one with a plus and one with a minus.

🎯 Exam Tip: Familiarize yourself with common perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144) to quickly identify \(b^2\) in difference of squares problems.

Question 3. \(a^2-1\)
Answer: We need to factorise the expression \(a^2 - 1\). This can be seen as the difference of two squares.
\(a^2 - 1 = (a)^2 - (1)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=a\) and \(b=1\):
\( = (a + 1)(a - 1) \)
The number 1 is unique because it is its own square (\(1^2 = 1\)), which often simplifies factorization problems. This makes it easy to apply the formula.
In simple words: Here, we have 'a' squared minus 1. Since 1 is also 1 squared, we can use the same pattern as before. Write 'a' and '1' in two brackets, one with a plus and one with a minus.

🎯 Exam Tip: Always remember that 1 can be written as \(1^2\), which is very useful for factoring expressions like \(x^2-1\).

Question 4. \(9z^2 – 64\)
Answer: We need to factorise the expression \(9z^2 - 64\). This is a difference of two squares.
\(9z^2 - 64 = (3z)^2 - (8)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=3z\) and \(b=8\):
\( = (3z + 8)(3z - 8) \)
For terms like \(9z^2\), it is important to take the square root of both the coefficient (9) and the variable (\(z^2\)) to find the correct 'a' term.
In simple words: First, figure out what was squared to get \(9z^2\) (that's \(3z\)). Then, figure out what was squared to get 64 (that's 8). Once you have these, put them into two brackets: one with a plus and one with a minus sign.

🎯 Exam Tip: When dealing with terms like \(9z^2\), ensure you correctly identify the square root as \(3z\), not just \(3\) or \(z\).

Question 5. \(9x^2-b^2\)
Answer: We need to factorise the expression \(9x^2 - b^2\). This is a classic example of the difference of two squares.
\(9x^2 - b^2 = (3x)^2 - (b)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=3x\) and \(b=b\):
\( = (3x + b)(3x - b) \)
This problem shows that the variables in the terms can be different, but the factorization rule remains exactly the same.
In simple words: Similar to the last one, find what was squared to make \(9x^2\) (it's \(3x\)) and what was squared to make \(b^2\) (it's \(b\)). Then, put these two parts into two brackets, one with a plus and one with a minus.

🎯 Exam Tip: The difference of squares formula works with any terms, whether they are numbers, variables, or combinations, as long as both terms are perfect squares.

Question 6. \(25 – x^2y^2\)
Answer: We need to factorise the expression \(25 - x^2y^2\). This fits the pattern of the difference of two squares.
\(25 - x^2y^2 = (5)^2 - (xy)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=5\) and \(b=xy\):
\( = (5 + xy)(5 - xy) \)
You can write \(x^2y^2\) as \((xy)^2\), which makes it very clear how to apply the difference of squares formula.
In simple words: We have 25, which is \(5 \times 5\). And \(x^2y^2\) is \((xy) \times (xy)\). So, we can write it as \( (5)^2 - (xy)^2 \). This lets us use the same rule: one bracket with plus, one with minus.

🎯 Exam Tip: Remember that \((xy)^2\) is the same as \(x^2y^2\). This property helps identify perfect squares involving multiple variables.

Question 7. \(81a^2x^2 - 49b^2y^2\)
Answer: We need to factorise the expression \(81a^2x^2 - 49b^2y^2\). This is a complex but still applicable case of the difference of two squares.
\(81a^2x^2 - 49b^2y^2 = (9ax)^2 - (7by)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=9ax\) and \(b=7by\):
\( = (9ax + 7by)(9ax - 7by) \)
This example demonstrates that the difference of squares rule applies effectively even to terms with multiple variables and coefficients, just find the square root of each combined term.
In simple words: This looks a bit big, but it's the same idea! Find the square root of \(81a^2x^2\) (that's \(9ax\)). Find the square root of \(49b^2y^2\) (that's \(7by\)). Then, put them into two brackets, one with a plus and one with a minus sign.

🎯 Exam Tip: Break down complex terms like \(81a^2x^2\) into their individual square roots: \(\sqrt{81}=9\), \(\sqrt{a^2}=a\), \(\sqrt{x^2}=x\), to correctly form \(a\).

Question 8. \(x^2 - \frac{1}{4}\)
Answer: We need to factorise the expression \(x^2 - \frac{1}{4}\). This is a difference of two squares involving a fraction.
\(x^2 - \frac{1}{4} = (x)^2 - \left(\frac{1}{2}\right)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=x\) and \(b=\frac{1}{2}\):
\( = \left(x + \frac{1}{2}\right)\left(x - \frac{1}{2}\right) \)
This shows the difference of squares formula works perfectly even when one or both terms are fractions, as long as they are perfect squares.
In simple words: Here we have 'x' squared, and then a fraction. The square root of \(1/4\) is \(1/2\). So we write it as \( (x)^2 - (1/2)^2 \), and then use the same plus and minus bracket rule.

🎯 Exam Tip: Don't be intimidated by fractions; find the square root of the numerator and the denominator separately to find the square root of the fraction.

Question 9. \(-25 + \frac{1}{64}b^2\)
Answer: We need to factorise the expression \(-25 + \frac{1}{64}b^2\). First, we rewrite it to match the standard form \(a^2 - b^2\):
\( -25 + \frac{1}{64}b^2 = \frac{1}{64}b^2 - 25 \)
Now, express each term as a square:
\( = \left(\frac{1}{8}b\right)^2 - (5)^2 \)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=\frac{1}{8}b\) and \(b=5\):
\( = \left(\frac{1}{8}b + 5\right)\left(\frac{1}{8}b - 5\right) \)
Rearranging terms to fit a known formula like the difference of squares is a common and effective strategy in factorization problems.
In simple words: We first flip the terms around to make it easier to see the pattern. Then, we find what was squared to get \( \frac{1}{64}b^2 \) (which is \( \frac{1}{8}b \)) and what was squared to get 25 (which is 5). Finally, we write them in two brackets, one with a plus and one with a minus.

🎯 Exam Tip: If an expression isn't immediately in the \(a^2 - b^2\) format, try rearranging terms to see if it fits the pattern, especially with negative signs.

Question 10. \(\frac{a^2}{9}-\frac{b^2}{16}\)
Answer: We need to factorise the expression \(\frac{a^2}{9} - \frac{b^2}{16}\). This is a difference of two squares involving fractional terms.
\( \frac{a^2}{9} - \frac{b^2}{16} = \left(\frac{a}{3}\right)^2 - \left(\frac{b}{4}\right)^2 \)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=\frac{a}{3}\) and \(b=\frac{b}{4}\):
\( = \left(\frac{a}{3} + \frac{b}{4}\right)\left(\frac{a}{3} - \frac{b}{4}\right) \)
This demonstrates that even when both terms are fractions, the fundamental principle of squaring both numerator and denominator applies correctly.
In simple words: For this, we find the square root of each part. For \( \frac{a^2}{9} \), it's \( \frac{a}{3} \). For \( \frac{b^2}{16} \), it's \( \frac{b}{4} \). Once we have these, we use the same two-bracket rule, one with a plus and one with a minus.

🎯 Exam Tip: When dealing with algebraic fractions that are perfect squares, take the square root of each component (\(a^2\), 9, \(b^2\), 16) separately to correctly identify \(a\) and \(b\).

Question 11. \(2.25a^2-b^2\)
Answer: We need to factorise the expression \(2.25a^2 - b^2\). This is a difference of two squares where one term has a decimal coefficient.
\(2.25a^2 - b^2 = (1.5a)^2 - (b)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=1.5a\) and \(b=b\):
\( = (1.5a + b)(1.5a - b) \)
Remember that \(1.5^2 = 2.25\), so decimals can also be part of perfect squares that fit this factorization pattern.
In simple words: Here, \(2.25a^2\) is the same as \( (1.5a)^2 \) because \(1.5 \times 1.5 = 2.25\). And \(b^2\) is just \( (b)^2 \). So, we use our difference of squares rule: \( (1.5a + b)(1.5a - b) \).

🎯 Exam Tip: Be comfortable working with squares of decimals; for example, \(1.5^2 = 2.25\) and \(0.5^2 = 0.25\).

Question 12. \(36a^8 - 121\)
Answer: We need to factorise the expression \(36a^8 - 121\). This is another instance of the difference of two squares.
\(36a^8 - 121 = (6a^4)^2 - (11)^2\)
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), where \(a=6a^4\) and \(b=11\):
\( = (6a^4 + 11)(6a^4 - 11) \)
When finding the square root of a variable with an exponent, you always halve the exponent; for example, \(\sqrt{a^8} = a^4\).
In simple words: For \(36a^8\), the square root is \(6a^4\). For 121, the square root is 11. Now, we put them into two brackets, one with a plus sign and one with a minus sign in between.

🎯 Exam Tip: For exponents, the square root of \(x^{2n}\) is always \(x^n\). This is a common point of confusion, so be careful.

Evaluate the following :

Question 13. \(54^2-36^2\)
Answer: We need to evaluate \(54^2 - 36^2\). We can use the difference of squares formula, \( a^2 - b^2 = (a+b)(a-b) \).
Here, \( a = 54 \) and \( b = 36 \).
So, \( (54)^2 - (36)^2 = (54 + 36)(54 - 36) \)
First, calculate the sums and differences inside the brackets:
\( 54 + 36 = 90 \)
\( 54 - 36 = 18 \)
Now, multiply these results:
\( = 90 \times 18 = 1620 \)
This method simplifies calculation greatly, avoiding the need to calculate large squares individually and then subtract them.
In simple words: To find the value of 54 squared minus 36 squared, we use a special math trick. We add the two numbers together (\(54+36\)) and then subtract the second number from the first (\(54-36\)). Then, we multiply those two new numbers. This gives us the answer much faster than squaring each number separately.

🎯 Exam Tip: Applying the difference of squares formula for numerical evaluations can save a lot of time and reduce the chances of calculation errors compared to squaring large numbers directly.

Question 14. \((3.2)^2 – (1.8)^2\)
Answer: We need to evaluate \((3.2)^2 - (1.8)^2\). We will apply the difference of squares formula, \( a^2 - b^2 = (a+b)(a-b) \).
Here, \( a = 3.2 \) and \( b = 1.8 \).
So, \( (3.2)^2 - (1.8)^2 = (3.2 + 1.8)(3.2 - 1.8) \)
First, calculate the sums and differences inside the brackets:
\( 3.2 + 1.8 = 5.0 \)
\( 3.2 - 1.8 = 1.4 \)
Now, multiply these results:
\( = 5.0 \times 1.4 = 7.0 = 7 \)
This factorization technique is very useful for mental math or simplifying calculations involving decimals.
In simple words: To solve \( (3.2)^2 - (1.8)^2 \), we use the same trick as before. We add 3.2 and 1.8 to get 5.0. Then we subtract 1.8 from 3.2 to get 1.4. Finally, we multiply 5.0 by 1.4 to get the answer, which is 7.

🎯 Exam Tip: When evaluating numerical expressions like this, always look for opportunities to use algebraic identities such as the difference of squares, as they simplify calculations significantly, especially with decimals.