ICSE-VIII-Mathematics
12: Algebraic Identities Class 8 Maths
- #1Use direct method to evaluate the following products : (i) (x + 8)(x + 3) (ii) (y + 5)(y - 3) (iii) (a - 8)(a + 2) (iv) (b - 3)(b - 5) (v) (3x - 2y)(2x + y) (vi) (5a + 16)(3a - 7) (vii) (8 - b) (3 + b)Ans : (i) (x + 8)(x + 3) = (x × x) + (x × 3) + (8 × x) + (8 × 3)
= x2 + 3x + 8x + 24
= x2 + 11x + 24 (ii) (y + 5)(y - 3) = (y × y) + (y × -3) + (5 × y) + (5 × -3)
= y2 + (-3y) + (5y) - 15
= y2 - 3y + 5y - 15
= y2 + 2y - 15 (iii) (a - 8)(a + 2) = (a × a) + (a × 2) + (-8) × a + (-8)(2)
= a2 + 2a - 8a - 16
= a2 - 6a - 16 (iv) (b - 3)(b - 5) = (b × b) + (b × -5) + (-3) × b + (-3)(-5)
= b2 - 5b - 3b + 15
= b2 - 8b + 15 (v) (3x - 2y)(2x + y) = (3x × 2x) + (3x × y) + (-2y × 2x) + (-2y × y)
= 6x2 + 3xy - 4xy -2y2
= 6x2 - xy - 2y2 (vi) (5a + 16)(3a - 7) = (5a × 3a) + (5a × -7) + (16 × 3a) + 16 × -7
= 15a2 + (-35a) + 48a + (-112)
= 15a2 - 35a + 48a - 112
= 15a2 + 13a - 112 (vii) (8 - b)(3 + b) = (8 × 3) + (8 × b) + (-b × 3) + (-b × b)
= 24 + 8b - 3b - b2
= 24 + 5b - b2
- #1-i(x + 8)(x + 3)Ans : (x + 8)(x + 3) = (x × x) + (x × 3) + (8 × x) + (8 × 3)
= x2 + 3x + 8x + 24
= x2 + 11x + 24
- #1-ii(y + 5)(y - 3)Ans : (y + 5)(y - 3) = (y × y) + (y × -3) + (5 × y) + (5 × -3)
= y2 + (-3y) + (5y) - 15
= y2 - 3y + 5y - 15
= y2 + 2y - 15
- #1-iii(a - 8)(a + 2)Ans : (a - 8)(a + 2) = (a × a) + (a × 2) + (-8) × a + (-8)(2)
= a2 + 2a - 8a - 16
= a2 - 6a - 16
- #1-iv(b - 3)(b - 5)Ans : (b - 3)(b - 5) = (b × b) + (b × -5) + (-3) × b + (-3)(-5)
= b2 - 5b - 3b + 15
= b2 - 8b + 15
- #1-v(3x - 2y)(2x + y)Ans : (3x - 2y)(2x + y) = (3x × 2x) + (3x × y) + (-2y × 2x) + (-2y × y)
= 6x2 + 3xy - 4xy -2y2
= 6x2 - xy - 2y2
- #1-vi(5a + 16)(3a - 7)Ans : (5a + 16)(3a - 7) = (5a × 3a) + (5a × -7) + (16 × 3a) + 16 × -7
= 15a2 + (-35a) + 48a + (-112)
= 15a2 - 35a + 48a - 112
= 15a2 + 13a - 112
- #1-vii(8 - b) (3 + b)Ans : (8 - b)(3 + b) = (8 × 3) + (8 × b) + (-b × 3) + (-b × b)
= 24 + 8b - 3b - b2
= 24 + 5b - b2