myhelper: RD Sharma solution class 8 chapter 7 Factorization Exercise 7.8

Exercise 7.8

Page-7.30

Question 1:

Resolve each of the following quadratic trinomial into factor:
2x² + 5x + 3

Answer 1:

$The given expression is 2 x^{2} + 5 x + 3 . (Coefficient of x^{2} = 2, coefficient of x = 5 and constant term = 3) We will split the coefficient of x into two parts such that their sum is 5 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 2 \times 3 = 6 . Now, 2 + 3 = 5 and 2 \times 3 = 6 Replacing the middle term 5 x by 2 x + 3 x, we have : 2 x^{2} + 5 x + 3 = 2 x^{2} + 2 x + 3 x + 3 = (2 x^{2} + 2 x) + (3 x + 3) = 2 x (x + 1) + 3 (x + 1) = (x + 1) (2 x + 3)$

Question 2:

Resolve each of the following quadratic trinomial into factor:
2x² − 3x − 2

Answer 2:

$The given expression is 2 x^{2} - 3 x - 2 . (Coefficient of x^{2} = 2, coefficient of x = - 3 and constant term = - 2) We will split the coefficient of x into two parts such that their sum is - 3 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 2 \times (- 2) = - 4 . Now, (- 4) + 1 = - 3 and (- 4) \times 1 = - 4 Replacing the middle term 3 x by - 4 x + x, we have : 2 x^{2} - 3 x - 2 = 2 x^{2} - 4 x + x - 2 = (2 x^{2} - 4 x) + (x - 2) = 2 x (x - 2) + (x - 2) = (2 x + 1) (x - 2)$

Question 3:

Resolve each of the following quadratic trinomial into factor:
3x² + 10x + 3

Answer 3:

$The given expression is 3 x^{2} + 10 x + 3 . (Coefficient of x^{2} = 3, coefficient of x = 10 and constant term = 3) We will split the coefficient of x into two parts such that their sum is 10 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 3 \times 3 = 9 . Now, 9 + 1 = 10 and 9 \times 1 = 9 Replacing the middle term 10 x by 9 x + x, we have : 3 x^{2} + 10 x + 3 = 3 x^{2} + 9 x + x + 3 = (3 x^{2} + 9 x) + (x + 3) = 3 x (x + 3) + (x + 3) = (3 x + 1) (x + 3)$

Question 4:

Resolve each of the following quadratic trinomial into factor:
7x − 6 − 2x²

Answer 4:

$The given expression is 7 x - 6 - 2 x^{2} . (Coefficient of x^{2} = - 2, coefficient of x = 7 and constant term = - 6) We will split the coefficient of x into two parts such that their sum is 7 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., (- 2) \times (- 6) = 12 . Now, 4 + 3 = 7 and 4 \times 3 = 12 Replacing the middle term 7 x by 4 x + 3 x, we have : 7 x - 6 - 2 x^{2} = - 2 x^{2} + 4 x + 3 x - 6 = (- 2 x^{2} + 4 x) + (3 x - 6) = 2 x (2 - x) - 3 (2 - x) = (2 x - 3) (2 - x)$

Question 5:

Resolve each of the following quadratic trinomial into factor:
7x² − 19x − 6

Answer 5:

$The given expression is 7 x^{2} - 19 x - 6 . (Coefficient of x^{2} = 7, coefficient of x = - 19 and constant term = - 6) We will split the coefficient of x into two parts such that their sum is - 19 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 7 \times (- 6) = - 42 . Now, (- 21) + 2 = - 19 and (- 21) \times 2 = - 42 Replacing the middle term - 19 x by - 21 x + 2 x, we have : 7 x^{2} - 19 x - 6 = 7 x^{2} - 21 x + 2 x - 6 = (7 x^{2} - 21 x) + (2 x - 6) = 7 x (x - 3) + 2 (x - 3) = (7 x + 2) (x - 3)$

Question 6:

Resolve each of the following quadratic trinomial into factor:
28 − 31x − 5x²

Answer 6:

$The given expression is 28 - 31 x - 5 x^{2} . (Coefficient of x^{2} = - 5, coefficient of x = - 31 and constant term = 28) We will split the coefficient of x into two parts such that their sum is - 31 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., (- 5) \times (28) = - 140 . Now, (- 35) + 4 = - 31 and (- 35) \times 4 = - 140 Replacing the middle term - 31 x by - 35 x + 4 x, we have : - 5 x^{2} - 31 x + 28 = - 5 x^{2} - 35 x + 4 x + 28 = (- 5 x^{2} - 35 x) + (4 x + 28) = - 5 x (x + 7) + 4 (x + 7) = (4 - 5 x) (x + 7)$

Question 7:

Resolve each of the following quadratic trinomial into factor:
3 + 23y − 8y²

Answer 7:

$The given expression is 3 + 23 y - 8 y^{2} . (Coefficient of y^{2} = - 8, coefficient of y = 23 and constant term = 3) We will split the coefficient of y into two parts such that their sum is 23 and their product equals the product of the coefficient of y^{2} and the constant term, i . e ., (- 8) \times 3 = - 24 . Now, (- 1) + 24 = 23 and (- 1) \times 24 = - 24 Replacing the middle term 23 y by - y + 24 y, we have : 3 + 23 y - 8 y^{2} = - 8 y^{2} + 23 y + 3 = - 8 y^{2} - y + 24 y + 3 = (- 8 y^{2} - y) + (24 y + 3) = - y (8 y + 1) + 3 (8 y + 1) = (3 - y) (8 y + 1)$

Question 8:

Resolve each of the following quadratic trinomial into factor:
11x² − 54x + 63

Answer 8:

$The given expression is 11 x^{2} - 54 x + 63 . (Coefficient of x^{2} = 11, coefficient of x = - 54 and constant term = 63) We will split the coefficient of x into two parts such that their sum is - 54 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 11 \times 63 = 693 . Now, (- 33) + (- 21) = - 54 and (- 33) \times (- 21) = 693 Replacing the middle term - 54 x by - 33 x - 21 x, we have : 11 x^{2} - 54 x + 63 = 11 x^{2} - 33 x - 21 x + 63 = (11 x^{2} - 33 x) + (- 21 x + 63) = 11 x (x - 3) - 21 (x - 3) = (11 x - 21) (x - 3)$

Question 9:

Resolve each of the following quadratic trinomial into factor:
7x − 6x² + 20

Answer 9:

$The given expression is 7 x - 6 x^{2} + 20 . (Coefficient of x^{2} = - 6, coefficient of x = 7 and constant term = 20) We will split the coefficient of x into two parts such that their sum is 7 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., (- 6) \times 20 = - 120 . Now, 15 + (- 8) = 7 and 15 \times (- 8) = - 120 Replacing the middle term 7 x by 15 x - 8 x, we get : 7 x - 6 x^{2} + 20 = - 6 x^{2} + 7 x + 20 = - 6 x^{2} + 15 x - 8 x + 20 = (- 6 x^{2} + 15 x) + (- 8 x + 20) = 3 x (- 2 x + 5) + 4 (- 2 x + 5) = (3 x + 4) (- 2 x + 5)$

Question 10:

Resolve each of the following quadratic trinomial into factor:
3x² + 22x + 35

Answer 10:

$The given expression is 3 x^{2} + 22 x + 35 . (Coefficient of x^{2} = 3, coefficient of x = 22 and constant term = 35) We will split the coefficient of x into two parts such that their sum is 22 and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 3 \times 35 = 105 . Now, 15 + 7 = 22 and 15 \times 7 = 105 Replacing the middle term 22 x by 15 x + 7 x, we get : 3 x^{2} + 22 x + 35 = 3 x^{2} + 15 x + 7 x + 35 = (3 x^{2} + 15 x) + (7 x + 35) = 3 x (x + 5) + 7 (x + 5) = (3 x + 7) (x + 5)$

Question 11:

Resolve each of the following quadratic trinomial into factor:
12x² − 17xy + 6y²

Answer 11:

$The given expression is 12 x^{2} - 17 xy + 6 y^{2} . (Coefficient of x^{2} = 12, coefficient of x = - 17 y and constant term = 6 y^{2}) We willsplit the coefficient of x into two parts such that their sum is - 17 y and their product equals the product of the coefficient of x^{2} and the constant term i . e ., 12 \times 6 y^{2} = 72 y^{2} . Now, (- 9 y) + (- 8 y) = - 17 y and (- 9 y) \times (- 8 y) = 72 y^{2} Replacing the middle term - 17 xy by - 9 xy - 8 xy, we get : 12 x^{2} - 17 xy + 6 y^{2} = 12 x^{2} - 9 xy - 8 xy + 6 y^{2} = (12 x^{2} - 9 xy) - (8 xy - 6 y^{2}) = 3 x (4 x - 3 y) - 2 y (4 x - 3 y) = (3 x - 2 y) (4 x - 3 y)$

Question 12:

Resolve each of the following quadratic trinomial into factor:
6x² − 5xy − 6y²

Answer 12:

$The given expression is 6 x^{2} - 5 xy - 6 y^{2} . (Coefficient of x^{2} = 6, coefficient of x = - 5 y and constant term = - 6 y^{2}) We will split the coefficient of x into two parts such that their sum is - 5 y and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 6 \times (- 6 y^{2}) = - 36 y^{2} . Now, (- 9 y) + 4 y = - 5 y and (- 9 y) \times 4 y = - 36 y^{2} Replacing the middle term - 5 xy by - 9 xy + 4 xy, we get : 6 x^{2} - 5 xy - 6 y^{2} = 6 x^{2} - 9 xy + 4 xy - 6 y^{2} = (6 x^{2} - 9 xy) + (4 xy - 6 y^{2}) = 3 x (2 x - 3 y) + 2 y (2 x - 3 y) = (3 x + 2 y) (2 x - 3 y)$

Question 13:

Resolve each of the following quadratic trinomial into factor:
6x² − 13xy + 2y²

Answer 13:

$The given expression is 6 x^{2} - 13 xy + 2 y^{2} . (Coefficient of x^{2} = 6, coefficient of x = - 13 y and constant term = 2 y^{2}) We will split the coefficient of x into two parts such that their sum is - 13 y and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 6 \times (2 y^{2}) = 12 y^{2} . Now, (- 12 y) + (- y) = - 13 y and (- 12 y) \times (- y) = 12 y^{2} Replacing the middle term - 13 xy by - 12 xy - xy, we get : 6 x^{2} - 13 xy + 2 y^{2} = 6 x^{2} - 12 xy - xy + 2 y^{2} = (6 x^{2} - 12 xy) - (xy - 2 y^{2}) = 6 x (x - 2 y) - y (x - 2 y) = (6 x - y) (x - 2 y)$

Question 14:

Resolve each of the following quadratic trinomial into factor:
14x² + 11xy − 15y²

Answer 14:

$The given expression is 14 x^{2} + 11 xy - 15 y^{2} . (Coefficient of x^{2} = 14, coefficient of x = 11 y and constant term = - 15 y^{2}) Now, we will split the coefficient of x into two parts such that their sum is 11 y and their product equals the product of the coefficient of x^{2} and the constant term, i . e ., 14 \times (- 15 y^{2}) = - 210 y^{2} . Now, 21 y + (- 10 y) = 11 y and 21 y \times (- 10 y) = - 210 y^{2} Replacing the middle term - 11 xy by - 10 xy + 21 xy, we get : 14 x^{2} + 11 xy - 15 y^{2} = 14 x^{2} - 10 xy + 21 xy - 15 y^{2} = (14 x^{2} - 10 xy) + (21 xy - 15 y^{2}) = 2 x (7 x - 5 y) + 3 y (7 x - 5 y) = (2 x + 3 y) (7 x - 5 y)$

Question 15:

Resolve each of the following quadratic trinomial into factor:
6a² + 17ab − 3b²

Answer 15:

$The given expression is 6 a^{2} + 17 ab - 3 b^{2} . (Coefficient of a^{2} = 6, coefficient of a = 17 b and constant term = - 3 b^{2}) Now, we will split the coefficient of a into two parts such that their sum is 17 b and their product equals the product of the coefficient of a^{2} and the constant term, i . e ., 6 \times (- 3 b^{2}) = - 18 b^{2} . Now, 18 b + (- b) = 17 b and 18 b \times (- b) = - 18 b^{2} Replacing the middle term 17 ab by - ab + 18 ab, we get : 16 a^{2} + 17 ab - 3 b^{2} = 6 a^{2} + - ab + 18 ab - 3 b^{2} = (6 a^{2} - ab) + (18 ab - 3 b^{2}) = a (6 a - b) + 3 b (6 a - b) = (a + 3 b) (6 a - b)$

Question 16:

Resolve each of the following quadratic trinomial into factor:
36a² + 12abc − 15b²c²

Answer 16:

$The given expression is 36 a^{2} + 12 abc - 15 b^{2} c^{2} . (Coefficient of a^{2} = 36, coefficient of a = 12 bc and constant term = - 15 b^{2} c^{2}) Now, we will split the coefficient of a into two parts such that their sum is 12 bc and their product equals the product of the coefficient of a^{2} and the constant term, i . e ., 36 \times (- 15 b^{2} c^{2}) = - 540 b^{2} c^{2} . Now, (- 18 bc) + 30 bc = 12 bc and (- 18 bc) \times 30 bc = - 540 b^{2} c^{2} Replacing the middle term 12 abc by - 18 abc + 30 abc, we get : 36 a^{2} + 12 abc - 15 b^{2} c^{2} = 36 a^{2} - 18 abc + 30 abc - 15 b^{2} c^{2} = (36 a^{2} - 18 abc) + (30 abc - 15 b^{2} c^{2}) = 18 a (2 a - bc) + 15 bc (2 a - bc) = (18 a + 15 bc) (2 a - bc) = 3 (6 a + 5 bc) (2 a - bc)$

Question 17:

Resolve each of the following quadratic trinomial into factor:
15x² − 16xyz − 15y²z²

Answer 17:

$T h e g i v e n e x p r e s s i o n i s 15 x^{2} - 16 x y z - 15 y^{2} z^{2} . (C o e f f i c i e n t o f x^{2} = 15, c o e f f i c i e n t o f x = - 16 y z a n d c o n s t a n t t e r m = - 15 y^{2} z^{2}) N o w, w e w i l l s p l i t t h e c o e f f i c i e n t o f x i n t o t w o p a r t s s u c h t h a t t h e i r s u m i s - 16 y z a n d t h e i r p r o d u c t e q u a l s t h e p r o d u c t o f t h e c o e f f i c i e n t o f x^{2} a n d t h e c o n s \tan t t e r m, i . e ., 15 \times (- 15 y^{2} z^{2}) = - 225 y^{2} z^{2} . N o w, (- 25 y z) + 9 y z = - 16 y x a n d (- 25 y z) \times 9 y z = - 225 y^{2} z^{2} R e p l a c i n g t h e m i d d l e t e r m - 16 x y z b y - 25 x y z + 9 x y z, w e h a v e : 15 x^{2} - 16 x y z - 15 y^{2} z^{2} = 15 x^{2} - 25 x y z + 9 x y z - 15 y^{2} z^{2} = (15 x^{2} - 25 x y z) + (9 x y z - 15 y^{2} z^{2}) = 5 x (3 x - 5 y z) + 3 y z (3 x - 5 y z) = (5 x + 3 y z) (3 x - 5 y z)$

Question 18:

Resolve each of the following quadratic trinomial into factor:
(x − 2y)² − 5(x − 2y) + 6

Answer 18:

$The given expression is a^{2} - 5 a + 6 . Assuming a = x - 2 y, we have : (x - 2 y)^{2} - 5 (x - 2 y) + 6 = a^{2} - 5 a + 6 (Coefficient of a^{2} = 1, coefficient of a = - 5 and constant term = 6) Now, we will split the coefficient of a into two parts such that their sum is - 5 and their product equals the product of the coefficient of a^{2} and the constant term, i . e ., 1 \times 6 = 6 . Clearly, (- 2) + (- 3) = - 5 and (- 2) \times (- 3) = 6 Replacing the middle term - 5 a by - 2 a - 3 a, we have : a^{2} - 5 a + 6 = a^{2} - 2 a - 3 a + 6 = (a^{2} - 2 a) - (3 a - 6) = a (a - 2) - 3 (a - 2) = (a - 3) (a - 2) Replacing a by (x - 2 y), we get : (a - 3) (a - 2) = (x - 2 y - 3) (x - 2 y - 2)$

Question 19:

Resolve each of the following quadratic trinomial into factor:
(2a − b)² + 2(2a − b) − 8

Answer 19:

$A s s u m i n g x = 2 a - b, w e h a v e : (2 a - b)^{2} + 2 (2 a - b) - 8 = x^{2} + 2 x - 8 T h e g i v e n e x p r e s s i o n b e c o m e s x^{2} + 2 x - 8 . (C o e f f i c i e n t o f x^{2} = 1 a n d t h a t o f x = 2; c o n s t a n t t e r m = - 8) N o w, w e w i l l s p l i t t h e c o e f f i c i e n t o f x i n t o t w o p a r t s s u c h t h a t t h e i r s u m i s 2 a n d t h e i r p r o d u c t e q u a l s t h e p r o d u c t o f t h e c o e f f i c i e n t o f x^{2} a n d t h e c o n s \tan t t e r m, i . e ., 1 \times (- 8) = - 8 . C l e a r l y, (- 2) + 4 = 2 a n d (- 2) \times 4 = - 8 R e p l a c i n g t h e m i d d l e t e r m 2 x b y - 2 x + 4 x, w e g e t : x^{2} + 2 x - 8 = x^{2} - 2 x + 4 x - 8 = (x^{2} - 2 x) + (4 x - 8) = x (x - 2) + 4 (x - 2) = (x + 4) (x - 2) R e l a c i n g x b y 2 a - b, w e g e t : (x + 4) (x - 2) = (2 a - b + 4) (2 a - b - 2)$

RD Sharma solution class 8 chapter 7 Factorization Exercise 7.8