Ganita Manjari Class 9 Maths Chapter 4: Exploring Algebraic Identities — NCERT Solutions

Explanation: x² − 16 = x² − 4², which is a difference of two squares. Using a² − b² = (a + b)(a − b), we get (x + 4)(x − 4). Checking: (x + 4)(x − 4) = x² − 4x + 4x − 16 = x² − 16.

Explanation: Compare with (a + b)² = a² + 2ab + b². Here x² + 6x + 9 = x² + 2 × x × 3 + 3², so it equals (x + 3)². The option (x + 9)(x − 3) gives the right middle term 6x but the wrong constant −27, and (x + 1)(x + 9) gives the right constant but the wrong middle term 10x.

Explanation: Write each term as a square: 4x² = (2x)² and 25 = 5². By the identity a² − b² = (a + b)(a − b), 4x² − 25 = (2x + 5)(2x − 5). Checking: (2x + 5)(2x − 5) = 4x² − 10x + 10x − 25 = 4x² − 25.

Explanation: Compare with (a − b)² = a² − 2ab + b². Here x² − 10x + 25 = x² − 2 × x × 5 + 5², so it equals (x − 5)². Since the middle term is negative, (x + 5)² is wrong, and (x + 5)(x − 5) would give x² − 25 with no middle term at all.

Explanation: Using x² + (a + b)x + ab, we need two numbers whose product is −24 and whose sum is −5. These are −8 and +3, since (−8) × 3 = −24 and −8 + 3 = −5. So x² − 5x − 24 = (x − 8)(x + 3). The option (x + 8)(x − 3) is the common sign slip; it expands to x² + 5x − 24.

Explanation: 8x³ + 27 = (2x)³ + 3³, a sum of cubes. Using a³ + b³ = (a + b)(a² − ab + b²) with a = 2x and b = 3, we get (2x + 3)(4x² − 6x + 9). The middle term of the second bracket is −ab = −6x, not +6x, and the option (2x − 3)(4x² + 6x + 9) is actually the factorisation of 8x³ − 27.

Explanation: Write 998 = 1000 − 2 and use (a − b)² = a² − 2ab + b². Then 998² = 1000² − 2 × 1000 × 2 + 2² = 1000000 − 4000 + 4 = 996004. Forgetting the factor 2 in 2ab gives 998004, and a sign slip on b² gives 995996.

Explanation: Check the perfect-square pattern a² − 2ab + b² with a = 3x and b = 4y. Then a² = 9x², b² = 16y², and 2ab = 2 × 3x × 4y = 24xy, which matches the middle term with a minus sign. So 9x² − 24xy + 16y² = (3x − 4y)². The option (3x + 4y)(3x − 4y) would give 9x² − 16y² with no middle term.

Explanation: Multiply the leading coefficient and the constant: 2 × 3 = 6, and we need two numbers with product 6 and sum 7, which are 6 and 1. Split the middle term: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). The tempting option (2x + 3)(x + 1) expands to 2x² + 5x + 3, which has the wrong middle term.

Explanation: Use the identity a³ − b³ = (a − b)³ + 3ab(a − b). Substituting, a³ − b³ = 3³ + 3 × 10 × 3 = 27 + 90 = 117. As a check, a = 5 and b = 2 satisfy a − b = 3 and ab = 10, and 5³ − 2³ = 125 − 8 = 117. Answering 27 means only (a − b)³ was computed and the 3ab(a − b) term was ignored.

Explanation: First apply a² − b² with a = x² and b = 4: x⁴ − 16 = (x² + 4)(x² − 4). The bracket x² − 4 is again a difference of squares and factors further as (x + 2)(x − 2), while x² + 4 cannot be factorised further. So the complete factorisation is (x² + 4)(x + 2)(x − 2); stopping at (x² + 4)(x² − 4) is incomplete.

Explanation: Group the last three terms: x² − y² − 6y − 9 = x² − (y² + 6y + 9) = x² − (y + 3)². This is now a difference of squares with a = x and b = y + 3, so it factors as (x + y + 3)(x − y − 3). Note that the minus sign distributes over the whole bracket, so the second factor is x − y − 3, not x − y + 3.

Explanation: The big square has side a + b + c, and each side is divided into segments of lengths a, b and c, cutting the figure into a 3 by 3 grid of nine pieces. The three pieces on the diagonal are squares: an a x a square (area a²), a b x b square (area b²) and a c x c square (area c²). The remaining six pieces are rectangles that come in equal pairs: two a x b rectangles (area ab each), two b x c rectangles (area bc each) and two c x a rectangles (area ca each). The whole figure is a square of side (a + b + c), so its area (a + b + c)² equals the sum of the nine labelled pieces, giving (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

Explanation: The top row shows four squares of sides (a + b + c), (a + b - c), (a - b + c) and (a - b - c), while the bottom row shows three squares of sides 2a, 2b and 2c. In the figure, the (a + b + c) square is dissected into red pieces plus a 2b x 2b square and a 2c x 2c square, and the same red pieces together with the three smaller squares of the top row exactly fill the 2a x 2a square, so both rows are made of the same pieces and have equal total area. Algebra confirms this: each of the four squares expands to a² + b² + c² with cross terms +/-2ab, +/-2bc, +/-2ca, and on adding the four expansions every cross term cancels. This leaves 4a² + 4b² + 4c², which is exactly (2a)² + (2b)² + (2c)².