Factorise :

$(i) a^{2} - 23 a + 42. (i i) a^{2} - 23 a - 108 (i i i) 1 - 18 x - 63 x^{2} (i v) 5 x^{2} - 4 x y - 12 y^{2} (v) x (3 x + 14) + 8 (v i) 5 - 4 x (1 + 3 x) (v i i) x^{2} y^{2} - 3 x y - 40 (v i i i) (3 x - 2 y)^{2} - 5 (3 x - 2 y) - 24 (i x) 12 (a + b)^{2} - (a + b) - 35$

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Solution

$(i) a^{2} - 23 a + 42. [42 = 21 \times 2 a n d 21 + 2 = 23] = a^{2} - 21 a - 2 a + 42 = a (a - 21) - 2 (a - 21) = (a - 21) (a - 2) (i i) a^{2} - 23 a - 108 = a^{2} - 27 a + 4 a - 108 [27 \times 4 = 108 a n d 27 - 4 = 23] = a (a - 27) + 4 (a - 27) = (a - 27) (a + 4) (i i i) 1 - 18 x - 63 x^{2} = 1 - 21 x + 3 x - 63 x^{2} = 1 (1 - 21 x) + 3 x (1 - 21 x) = (1 - 21 x) (1 + 3 x) (i v) 5 x^{2} - 4 x y - 12 y^{2} = 5 x^{2} - 10 x y + 6 x y - 12 y^{2} = 5 x (x - 2 y) + 6 y (x - 2 y) = (x - 2 y) (5 x + 6 y) (v) x (3 x + 14) + 8 = 3 x^{2} + 14 x + 8 = 3 x^{2} + 12 x + 2 x + 8 = 3 x (x + 4) + 2 (x + 4) = (x + 4) (3 x + 2) (v i) 5 - 4 x (1 + 3 x) = 5 - 4 x - 12 x^{2} = 5 - 10 x + 6 x - 12 x^{2} = 5 (1 - 2 x) + 6 x (1 - 2 x) = (1 - 2 x) (5 + 6 x) (v i i) x^{2} y^{2} - 3 x y - 40 = x^{2} y^{2} - 8 x y + 5 x y - 40 = x y (x y - 8) + 5 (x y - 8) = (x y - 8) (x y + 5) (v i i i) (3 x - 2 y)^{2} - 5 (3 x - 2 y) - 24 = (3 x - 2 y)^{2} - 8 (3 x - 2 y) + 3 (3 x - 2 y) - 24 = (3 x - 2 y) (3 x - 2 y - 8) + 3 (3 x - 2 y - 8) = (3 x - 2 y - 8) (3 x - 2 y + 3) (i x) 12 (a + b)^{2} - (a + b) - 35 = 12 (a + b)^{2} - 21 (a + b) + 20 (a + b) - 35 = 3 (a + b) [4 (a + b) - 7] + 5 [4 (a + b) - 7] = (4 a + 4 b - 7) (3 a + 3 b + 5)$

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Similar questions

Q. State whether True or False.

Factorization of

a^{2} - 23 a + 42

(a - 21) (a - 2)

Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5)

(iii) (2a − 7) (2a − 7) (iv)

(v) (1.1m − 0.4) (1.1 m + 0.4) (vi) (a² + b²) (− a² + b²)

(vii) (6x − 7) (6x + 7) (viii) (− a + c) (− a + c)

(ix) (x) (7a − 9b) (7a − 9b)

Factorise :

$(i) a^{2} - (b - c)^{2} (i i) 25 (2 x - y)^{2} - 16 (x - 2 y)^{2} (i i i) 16 (5 x + 4)^{2} - 9 (3 x - 2)^{2} (i v) 9 x^{2} - \frac{1}{16} (v) 25 (x - 2 y)^{2} - 4$

Q. Use the identity

(a + b) (a - b) = a^{2} - b^{2}

to find the product of

(3 x + 5) (3 x - 5)

State whether True or False.

Factorization of

a^{2} - 23 a - 108

(a - 27) (a + 4)

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Factorise : (i) a2−23a+42.(ii) a2−23a−108(iii) 1−18x−63x2(iv) 5x2−4xy−12y2(v) x(3x+14)+8(vi) 5−4x(1+3x)(vii) x2y2−3xy−40(viii) (3x−2y)2−5(3x−2y)−24(ix) 12(a+b)2−(a+b)−35

Factorise :

$(i) a^{2} - 23 a + 42. (i i) a^{2} - 23 a - 108 (i i i) 1 - 18 x - 63 x^{2} (i v) 5 x^{2} - 4 x y - 12 y^{2} (v) x (3 x + 14) + 8 (v i) 5 - 4 x (1 + 3 x) (v i i) x^{2} y^{2} - 3 x y - 40 (v i i i) (3 x - 2 y)^{2} - 5 (3 x - 2 y) - 24 (i x) 12 (a + b)^{2} - (a + b) - 35$