GSEB Class 8 Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Question 1. Use a suitable identity to get each of the following products?
(x + 3)(x + 3)
(2y + 5)(2y + 5)
3. (2a - 7)(2a – 7)
4. (3a-\( \frac { 1 }{ 2 } \)) (3a - \( \frac { 1 }{ 2 } \))
5. (1.1m-0.4)(1.1m + 0.4)
6. (a² + b²)(-a² + b²)
7. (6x - 7)(6x + 7)
8. (-a + c)(-a + c)
9. (\( \frac { x }{ 2 } \) + \( \frac { 3y }{ 4 } \)) (\( \frac { x }{ 2 } \) + \( \frac { 3y }{ 4 } \))
10. (7a – 9b) (7a – 9b)
Answer:
1. Using the identity, \( (x + a)(x + b) = x^2 + (a + b)x + ab \), we can find the product.
\( (x + 3)(x + 3) = x^2 + (3 + 3)x + (3 \times 3) \)
\( = x^2 + 6x + 9 \)
2. Using the identity, \( (x + a)(x + b) = x^2 + (a + b)x + ab \), we have
\( (2y + 5)(2y + 5) \)
\( = (2y)^2 + (5 + 5)2y + (5 \times 5) \)
\( = 4y^2 + (10)2y + 25 \)
\( = 4y^2 + 20y + 25 \)
3. Using the identity, \( (a – b)^2 = a^2 – 2ab + b^2 \)
\( (2a – 7)(2a – 7) = (2a – 7)^2 \)
\( = (2a)^2 – 2(2a)(7) + (7)^2 \)
\( = 4a^2 - 28a + 49 \)
4. Using the identity, \( (a – b)^2 = a^2 – 2ab + b^2 \), we have
\( (3a - \frac{1}{2}) (3a - \frac{1}{2}) = (3a - \frac{1}{2})^2 \)
\( = (3a)^2 - 2(3a) (\frac{1}{2}) + (\frac{1}{2})^2 = 9a^2 – 3a + \frac{1}{4} \)
5. Using the identity, \( (a + b)(a - b) = a^2 – b^2 \), we have
\( (1.1m – 0.4)(1.1m + 0.4) \)
\( = (1.1m)^2 – (0.4)^2 = 1.21m^2 – 0.16 \)
6. Given: \( (a^2 + b^2)(-a^2 + b^2) = (b^2 + a^2)(b^2 – a^2) \)
Using the identity, \( (a + b)(a – b) = a^2 – b^2 \), we have
\( (b^2 + a^2)(b^2 – a^2) = (b^2)^2 – (a^2)^2 = b^4 – a^4 \)
7. Given: \( (6x – 7)(6x + 7) = (6x + 7)(6x – 7) \)
Using the identity, \( (a + b)(a – b) = a^2 – b^2 \), we have
\( (6x + 7)(6x – 7) = (6x)^2 – (7)^2 \)
\( = 36x^2 - 49 \)
8. Using the identity, \( (a + b)^2 = a^2 + 2ab + b^2 \), we have
\( (-a + c)(-a + c) = (-a + c)^2 \)
\( = (-a)^2 + 2(-a)(c) + (c)^2 \)
\( = a^2 + 2(-ac) + c^2 \)
\( = a^2 – 2ac + c^2 \)
9. Using the identity, \( (a + b)^2 = a^2 + b^2 + 2ab \), we have
\( (\frac{x}{2} + \frac{3y}{4}) (\frac{x}{2} + \frac{3y}{4}) = (\frac{x}{2} + \frac{3y}{4})^2 \)
\( = (\frac{x}{2})^2 + 2(\frac{x}{2})(\frac{3y}{4}) + (\frac{3y}{4})^2 \)
\( = \frac{x^2}{4} + \frac{3xy}{4} + \frac{9y^2}{16} \)
10. Using the identity \( (a – b)^2 = a^2 – 2ab + b^2 \), we have
\( (7a - 9b)(7a – 9b) = (7a – 9b)^2 \)
\( = (7a)^2 – 2(7a)(9b) + (9b)^2 \)
\( = 49a^2 - 126ab + 81b^2 \)
In simple words: To find these products, we use special algebraic formulas called identities. Each identity helps us multiply the terms quickly. We just need to put the numbers and letters into the correct places in the formula.

Exam Tip: Understand which identity applies to which product form. For repeated factors like `(x+3)(x+3)`, use `(a+b)^2`; for different signs like `(1.1m-0.4)(1.1m+0.4)`, use `(a-b)(a+b)`. Always be careful with negative signs when applying the identities.

Question 2. Use the identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \) to find the following products.
1. (x + 3)(x + 7)
2. (4x + 5)(4x + 1)
3. (4x - 5)(4x – 1)
4. (4x + 5)(4x – 1)
5. (2x + 5y)(2x + 3y)
6. (2a² + 9)(2a² + 5)
7. (xyz – 4)(xyz – 2)
Answer:
1. \( (x + 3)(x + 7) = x^2 + (3 + 7)x + (3 \times 7) \)
\( = x^2 + 10x + 21 \)
2. \( (4x + 5)(4x + 1) \)
\( = (4x)^2 + (5 + 1)4x + (5 \times 1) \)
\( = 16x^2 + 24x + 5 \)
3. \( (4x - 5)(4x – 1) \)
\( = (4x)^2 + (-5 – 1)4x + [(-5) \times (-1)] \)
\( = 16x^2 + (-6)4x + (5) \)
\( = 16x^2 - 24x + 5 \)
4. \( (4x + 5)(4x – 1) \)
\( = (4x)^2 + (5 – 1)4x + [5 \times (-1)] \)
\( = 16x^2 + (4)4x + (-5) \)
\( = 16x^2 + 16x – 5 \)
5. \( (2x + 5y)(2x + 3y) \)
\( = (2x)^2 + (5y + 3y)2x + (5y \times 3y) \)
\( = 4x^2 + (5 + 3)y \times 2x + 15y^2 \)
\( = 4x^2 + [8y \times 2x] + 15y^2 \)
\( = 4x^2 + 16xy + 15y^2 \)
6. \( (2a^2 + 9)(2a^2 + 5) \)
\( = [2a^2]^2 + (9 + 5)2a^2 + 9 \times 5 \)
\( = 4a^4 + 14 \times 2a^2 + 45 \)
\( = 4a^4 + 28a^2 + 45 \)
7. \( (xyz – 4)(xyz – 2) \)
\( = (xyz)^2 + [(-4) + (-2)]xyz + (-4) \times (-2) \)
\( = (xyz)^2 + (-6)xyz + 8 \)
\( = x^2y^2z^2 – 6xyz + 8 \)
In simple words: We use a specific algebraic rule here: when you multiply two terms like `(x + a)` and `(x + b)`, the answer is `x` squared, plus the sum of `a` and `b` times `x`, plus the product of `a` and `b`. This rule helps us quickly find the products for these problems.

Exam Tip: Remember to apply the identity \( (x + a)(x + b) \) carefully, identifying the 'x', 'a', and 'b' terms correctly in each expression, especially when 'x' is itself an expression like `4x` or `xyz`.

Question 3. Find the following squares by using the identities?
1. (b – 7)2
2. (xy + 3z)2
3. (6x2 – 5y)²
4. (\( \frac{2}{3}m + \frac{3}{2}n \))²
5. (0.4p – 0.5q)2
6. (2xy + 5y)²
Answer:
1. \( (b – 7)^2 \)
Using \( (a – b)^2 = a^2 – 2ab + b^2 \), we have
\( (b – 7)^2 = (b)^2 – 2(b)(7) + (7)^2 \)
\( = b^2 - 14b + 49 \)
2. \( (xy + 3z)^2 \)
Using \( (a + b)^2 = a^2 + 2ab + b^2 \), we have
\( (xy + 3z)^2 = (xy)^2 + 2(xy)(3z) + (3z)^2 \)
\( = x^2y^2 + 6xyz + 9z^2 \)
3. \( (6x^2 – 5y)^2 \)
Using the identity \( (a – b)^2 = a^2 – 2ab + b^2 \), we have
\( (6x^2 – 5y)^2 = (6x^2)^2 – 2(6x^2)(5y) + (5y)^2 \)
\( = 36x^4 – 60x^2y + 25y^2 \)
4. \( (\frac{2}{3}m + \frac{3}{2}n)^2 \)
Using \( (a + b)^2 = a^2 + 2ab + b^2 \), we have
\( (\frac{2}{3}m + \frac{3}{2}n)^2 \)
\( = (\frac{2}{3}m)^2 + 2(\frac{2}{3}m)(\frac{3}{2}n) + (\frac{3}{2}n)^2 \)
\( = \frac{4}{9}m^2 + 2mn + \frac{9}{4}n^2 \)
5. \( (0.4p – 0.5q)^2 \)
Using \( (a – b)^2 = a^2 – 2ab + b^2 \), we have
\( (0.4p – 0.5q)^2 \)
\( = (0.4p)^2 – 2(0.4p)(0.5q) + (0.5q)^2 \)
\( = 0.16p^2 – 0.4pq + 0.25q^2 \)
6. \( (2xy + 5y)^2 \)
Using \( (a + b)^2 = a^2 + 2ab + b^2 \), we have
\( (2xy + 5y)^2 = (2xy)^2 + 2(2xy)(5y) + (5y)^2 \)
\( = 4x^2y^2 + 20xy^2 + 25y^2 \)
In simple words: To find the square of an expression, we apply one of two key identities: `(a + b)` squared or `(a - b)` squared. These rules help us expand the expressions into three terms: the square of the first term, plus or minus two times the product of both terms, plus the square of the second term.

Exam Tip: Pay close attention to the sign between the terms (plus or minus) to pick the correct identity `(a+b)^2` or `(a-b)^2`. Ensure you square the entire term, including any coefficients and variables.

Question 4. Simplify:
1. (a²-b²)2
2. (2x + 5)² - (2x - 5)²
3. (7m – 8n)² + (7m + 8n)²
4. (4m + 5n)² + (5m + 4n)2
5. (2.5p – 1.5q)² – (1.5p – 2.5q)²
6. (ab + bc)² – 2ab²c
7. (m² – n²m)² + 2m³n²
Answer:
1. \( (a^2-b^2)^2 = (a^2)^2 – 2a^2 \times b^2 + (b^2)^2 \)
[Using \( (a – b)^2 = a^2 – 2ab + b^2 \)]
\( = a^4 – 2a^2b^2 + b^4 \)
2. \( (2x + 5)^2 – (2x – 5)^2 \)
\( = [(2x)^2 + 2(2x)(5) + (5)^2] – [(2x)^2 – 2(2x)(5) + (5)^2] \)
\( = [4x^2 + 20x + 25] – [4x^2 – 20x + 25] \)
\( = 4x^2 + 20x + 25 – 4x^2 + 20x – 25 \)
\( = (4 – 4)x^2 + (20 + 20)x + (25 – 25) \)
\( = (0)x^2 + 40x + 0 = 40x \)
3. \( (7m – 8n)^2 + (7m + 8n)^2 \)
\( = [(7m)^2 – 2(7m)(8n) + (8n)^2] + [(7m)^2 + 2(7m)(8n) + (8n)^2] \)
\( = [49m^2 – 112mn + 64n^2] + [49m^2 + 112mn + 64n^2] \)
\( = 49m^2 – 112mn + 64n^2 + 49m^2 + 112mn + 64n^2 \)
\( = (49 + 49)m^2 + (-112 + 112)mn + (64 + 64)n^2 \)
\( = 98m^2 + (0)mn + 128n^2 = 98m^2 + 128n^2 \)
4. \( (4m + 5n)^2 + (5m + 4n)^2 \)
\( = [(4m)^2 + 2(4m)(5n) + (5n)^2] + [(5m)^2 + 2(5m)(4n) + (4n)^2] \)
\( = 16m^2 + 40mn + 25n^2 + 25m^2 + 40mn + 16n^2 \)
\( = (16 + 25)m^2 + (40 + 40)mn + (25 + 16)n^2 \)
\( = 41m^2 + 80mn + 41n^2 \)
5. \( (2.5p – 1.5q)^2 – (1.5p – 2.5q)^2 \)
\( = [(2.5p)^2 – 2(2.5p)(1.5q) + (1.5q)^2] – [(1.5p)^2 – 2(1.5p) (2.5q) + (2.5q)^2] \)
\( = 6.25p^2 – 7.5pq + 2.25q^2 – [2.25p^2 – 7.5pq + 6.25q^2] \)
\( = 6.25p^2 – 7.5pq + 2.25q^2 – 2.25p^2 + 7.5pq – 6.25q^2 \)
\( = (6.25 - 2.25)p^2 + (-7.5 + 7.5)pq + (2.25 - 6.25)q^2 \)
\( = 4p^2 + 0pq - 4q^2 \)
\( = 4p^2 – 4q^2 \)
6. \( (ab + bc)^2 – 2ab^2c \)
\( = (ab)^2 + 2(ab)(bc) + (bc)^2 – 2ab^2c \)
\( = a^2b^2 + 2ab^2c + b^2c^2 – 2ab^2c \)
\( = a^2b^2 + (2 – 2)ab^2c + b^2c^2 \)
\( = a^2b^2 + (0)ab^2c + b^2c^2 \)
\( = a^2b^2 + 0 + b^2c^2 = a^2b^2 + b^2c^2 \)
7. \( (m^2 – n^2m)^2 + 2m^3n^2 \)
\( = (m^2)^2 - 2(m^2)(n^2m) + (n^2m)^2 + 2m^3n^2 \)
\( = m^4 – 2m^3n^2 + n^4m^2 + 2m^3n^2 \)
\( = m^4 + (-2 + 2)m^3n^2 + n^4m^2 \)
\( = m^4 + (0)m^3n^2 + n^4m^2 = m^4 + m^2n^4 \)
In simple words: To make these expressions simpler, we first use the correct algebraic identity to expand any squared terms. Then, we gather similar terms together and add or subtract their coefficients. This process helps us reduce complex expressions into their simplest forms.

Exam Tip: When simplifying, always expand squared terms first using the appropriate identity. Be extremely careful with signs, especially when subtracting an entire expanded expression as in sub-part 2 and 5.

Question 5. Show that
1. \( (3x + 7)^2 – 84x = (3x – 7)^2 \)
2. \( (9p – 5q)^2 + 180pq = (9p + 5q)^2 \)
3. \( (\frac{4}{3}m - \frac{3}{4}n)^2 + 2mn = \frac{16}{9}m^2 + \frac{9}{16}n^2 \)
4. \( (4pq + 3q)^2 – (4pq – 3q)^2 = 48pq^2 \)
5. \( (a – b)(a + b) + (b − c)(b + c) + (c − a)(c + a) = 0 \)
Answer:
1. LHS \( = (3x + 7)^2 – 84x \)
\( = (3x)^2 + 2(3x)(7) + (7)^2 – 84x \)
\( = 9x^2 + 42x + 49 – 84x \)
\( = 9x^2 + (42 – 84)x + 49 \)
\( = 9x^2 – 42x + 49 \)
RHS \( = (3x - 7)^2 \)
\( = (3x)^2 - 2(3x)(7) + (7)^2 \)
\( = 9x^2 – 42x + 49 \)
Since, LHS = RHS
Therefore, \( (3x + 7)^2 – 84x = (3x – 7)^2 \)
2. LHS \( = (9p – 5q)^2 + 180pq \)
\( = (9p)^2 - 2(9p)(5q) + (5q)^2 + 180pq \)
\( = 81p^2 – 90pq + 25q^2 + 180pq \)
\( = 81p^2 + (-90 + 180)pq + 25q^2 \)
\( = 81p^2 + 90pq + 25q^2 \)
RHS \( = (9p + 5q)^2 \)
\( = (9p)^2 + 2(9p)(5q) + (5q)^2 \)
\( = 81p^2 + 90pq + 25q^2 \)
Since, LHS = RHS
Therefore, \( (9p – 5q)^2 + 180pq = (9p + 5q)^2 \)
3. LHS \( = [\frac{4}{3}m – \frac{3}{4}n]^2 + 2mn \)
\( = (\frac{4}{3}m)^2 - 2(\frac{4}{3}m)(\frac{3}{4}n) + (\frac{3}{4}n)^2 + 2mn \)
\( = \frac{16}{9}m^2 - 2mn + \frac{9}{16}n^2 + 2mn \)
\( = \frac{16}{9}m^2 + (-2 + 2)mn + \frac{9}{16}n^2 \)
\( = \frac{16}{9}m^2 + (0)mn + \frac{9}{16}n^2 \)
\( = \frac{16}{9}m^2 + \frac{9}{16}n^2 \)
RHS \( = \frac{16}{9}m^2 + \frac{9}{16}n^2 \)
Since, LHS = RHS
Therefore, \( (\frac{4}{3}m – \frac{3}{4}n)^2 + 2mn = \frac{16}{9}m^2 + \frac{9}{16}n^2 \)
4. LHS \( = (4pq + 3q)^2 – (4pq – 3q)^2 \)
\( = [(4pq)^2 + 2(4pq)(3q) + (3q)^2] – [(4pq)^2 – 2(4pq)(3q) + (3q)^2] \)
\( = [16p^2q^2 + 24pq^2 + 9q^2] – [16p^2q^2 – 24pq^2 + 9q^2] \)
\( = 16p^2q^2 + 24pq^2 + 9q^2 – 16p^2q^2 + 24pq^2 – 9q^2 \)
\( = (16 - 16)p^2q^2 + (24 + 24)pq^2 + (9 – 9)q^2 \)
\( = (0)p^2q^2 + 48pq^2 + (0)q^2 = 48pq^2 \)
RHS \( = 48pq^2 \)
Since, LHS = RHS
Therefore, \( (4pq + 3q)^2 – (4pq – 3q)^2 = 48pq^2 \)
5. LHS \( = (a – b)(a + b) + (b − c)(b + c) + (c − a)(c + a) \)
\( = (a^2 – b^2) + (b^2 – c^2) + (c^2 – a^2) \)
\( = a^2 - b^2 + b^2 - c^2 + c^2 – a^2 \)
\( = a^2 – a^2 + b^2 – b^2 + c^2 – c^2 = 0 \)
RHS \( = 0 \)
Since, LHS = RHS
Therefore, \( (a – b)(a + b) + (b − c)(b + c) + (c − a)(c + a)= 0 \)
In simple words: To show that these equations are true, we work with both sides of the equation. We expand the expressions on the Left Hand Side (LHS) and the Right Hand Side (RHS) separately using algebraic identities. If both sides simplify to the same expression, then the original equation is proven to be correct.

Exam Tip: When showing that two expressions are equal, always work with the LHS and RHS separately. Do not mix terms across the equals sign until you have simplified each side. Clearly state which identity is being used in each step.

Question 6. Using identities, evaluate:
1. \( 71^2 \)
2. \( 99^2 \)
3. \( 102^2 \)
4. \( 998^2 \)
5. \( 5.2^2 \)
6. \( 297 \times 303 \)
7. \( 78 \times 82 \)
8. \( 8.9^2 \)
9. \( 1.05 \times 9.5 \)
Answer:
1. We can write \( 71 = 70 + 1 \)
So, \( (71)^2 = (70 + 1)^2 \)
\( = (70)^2 + 2(70)(1) + (1)^2 \) [Using \( (a + b)^2 = a^2 + 2ab + b^2 \)]
\( = 4900 + 140 + 1 = 5041 \)
2. We have \( 99 = 100 – 1 \)
So, \( (99)^2 = (100 – 1)^2 \)
\( = (100)^2 – 2(100)(1) + (1)^2 \) [Using \( (a – b)^2 = a^2 – 2ab + b^2 \)]
\( = 10000 - 200 + 1 = 9801 \)
3. We have \( 102 = 100 + 2 \)
So, \( (102)^2 = (100 + 2)^2 \)
\( = (100)^2 + 2(100)(2) + (2)^2 \) [Using \( (a + b)^2 = a^2 + 2ab + b^2 \)]
\( = 10000 + 400 + 4 = 10404 \)
4. We have \( 998 = 1000 - 2 \)
So, \( (998)^2 = (1000 – 2)^2 \)
\( = (1000)^2 – 2(1000)(2) + (2)^2 \) [Using \( (a – b)^2 = a^2 – 2ab + b^2 \)]
\( = 1000000 - 4000 + 4 = 996004 \)
5. We have \( 5.2 = 5 + 0.2 \)
So, \( (5.2)^2 = (5 + 0.2)^2 \)
\( = (5)^2 + 2(5)(0.2) + (0.2)^2 \) [Using \( (a + b)^2 = a^2 + 2ab + b^2 \)]
\( = 25 + 2 + 0.04 = 27.04 \)
6. We write \( 297 = 300 - 3 \) and \( 303 = 300 + 3 \)
So, \( 297 \times 303 = (300 – 3)(300 + 3) \)
\( = (300)^2 – (3)^2 \) [Using \( (a + b)(a – b) = a^2 – b^2 \)]
\( = 90000 - 9 = 89991 \)
7. We write \( 78 = 80 - 2 \) and \( 82 = 80 + 2 \)
So, \( 78 \times 82 = (80 – 2)(80 + 2) \)
\( = 80^2 – (2)^2 \) [Using \( (a + b)(a – b) = a^2 – b^2 \)]
\( = 6400 - 4 = 6396 \)
8. We have \( 8.9 = 9 – 0.1 \)
So, \( (8.9)^2 = (9 – 0.1)^2 \)
\( = (9)^2 – 2(9)(0.1) + (0.1)^2 \) [Using \( (a – b)^2 = a^2 – 2ab + b^2 \)]
\( = 81-1.8 + 0.01 \)
\( = 81.01 - 1.80 = 79.21 \)
9. We have \( 1.05 = 1 + 0.05 \)
And \( 9.5 \)
So, \( (1.05) \times 9.5 = (1 + 0.05) \times 9.5 \)
\( = (1 \times 9.5) + (9.5 \times 0.05) \)
\( = 9.500 + 0.475 = 9.975 \)
In simple words: To evaluate these expressions using identities, we first rewrite the numbers in a form that fits one of our algebraic identities, like `(a + b)` or `(a - b)`. Then, we apply the identity to simplify the calculation, making it easier to find the final answer without direct multiplication.

Exam Tip: For numerical evaluations, always try to express the given numbers as a sum or difference from a convenient base (e.g., 10, 100, 1000) or as a decimal sum, to effectively use the `(a+b)^2`, `(a-b)^2`, or `(a+b)(a-b)` identities.

Question 7. Using \( a^2 – b^2 = (a + b)(a – b) \), find:
1. \( 51^2-49^2 \)
2. \( (1.02)^2 – (0.98)^2 \)
3. \( 153^2-147^2 \)
4. \( 12.1^2-7.9^2 \)
Answer:
1. \( 51^2 – 49^2 = (51 + 49)(51 – 49) \)
\( = (100) \times (2) = 200 \)
2. \( (1.02)^2 – (0.98)^2 \)
\( = (1.02 + 0.98)(1.02 – 0.98) \)
\( = (2.0) \times (0.04) = 0.08 \)
3. \( 153^2-147^2 \)
\( = (153 + 147)(153 – 147) \)
\( = (300) \times (6) = 1800 \)
4. \( (12.1)^2 – (7.9)^2 \)
\( = (12.1 + 7.9)(12.1 – 7.9) \)
\( = (20.0) \times (4.2) \)
\( = 84.0 \)
In simple words: To quickly calculate these differences of squares, we use the identity `a` squared minus `b` squared equals `(a + b)` multiplied by `(a - b)`. This means we just add the two numbers together, subtract the second from the first, and then multiply those two new results.

Exam Tip: The identity \( a^2 – b^2 = (a + b)(a – b) \) is very efficient for mental math and quick calculations. Always identify 'a' and 'b' correctly, then perform the addition and subtraction before multiplying.

Question 8. Using \( (x + a)(x + b) = x^2 + (a + b)x + ab \), find:
1. \( 103 \times 104 \)
2. \( 5.1 \times 5.2 \)
3. \( 103 \times 98 \)
4. \( 9.7 \times 9.8 \)
Answer:
1. \( 103 \times 104 = (100 + 3) \times (100 + 4) \)
\( = (100)^2 + (3 + 4) \times 100 + (3 \times 4) \)
\( = 10000 + 700 + 12 \)
\( = 10712 \)
2. \( 5.1 \times 5.2 = (5 + 0.1) \times (5 + 0.2) \)
\( = (5)^2 + (0.1 + 0.2) \times 5 + (0.1 \times 0.2) \)
\( = 25 + (0.3) \times 5 + 0.02 \)
\( = 25 + 1.5 + 0.02 \)
\( = 26.52 \)
3. \( 103 \times 98 = (100 + 3) \times (100 – 2) \)
\( = (100)^2 + (3 – 2)100 + [3 \times (- 2)] \)
\( = 10000 + 100-6 \)
\( = 10094 \)
4. \( 9.7 \times 9.8 = (10 – 0.3) \times (10 – 0.2) \)
\( = (10)^2 + [(-0.3) + (-0.2)] 10 + [(-0.3) \times (-0.2)] \)
\( = 100 + [- 0.5] \times 10 + 0.06 \)
\( = 100-5 + 0.06 \)
\( = 95.06 \)
In simple words: To find these products using the given identity, we first rewrite each number as a sum or difference from a common base (like 10 or 100). Then, we substitute these into the identity `x` squared plus `(a + b)` times `x`, plus `a` times `b`, and simplify to get the answer.

Exam Tip: When using the \( (x + a)(x + b) \) identity for numerical products, choose a convenient 'x' value (e.g., 10, 100, 10 for decimals) that makes 'a' and 'b' small and easy to work with. Be extra careful with negative values for 'a' or 'b' as they affect sums and products.