(i) p²+6p+8 = p²+(4+2)p+4X2 = p²+4p+2p+4x2 = p(p+4)+2(p+4) = (p+4)(p+2) (ii) q²-10q+21 = q²-(7+3)q+7x3 = q²-7q-3q+7x3 = q(q-7)-3(q-7) = (q-7)(q-3) Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14Read more
(i) a⁴-b⁴ = (a²)²-(b²)² = (a²-b²)(a²+b²) [∵ a²-b²-(a-b)(a+b)] = (a-b)(a+b)(a²+b²) [∵ a²-b²-(a-b)(a+b)] (ii) p⁴-81=(p²)²-(9²)² = (9²-9)(9²+9) [∵ a²-b²-(a-b)(a+b)] = (p²-3²)(p²+9) = (p-3)(p+3)(p²+9) [∵ a²-b²-(a-b)(a+b)] Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to:Read more
(i) ax²+bx = x(ax+b) (ii) 7p²+21q² = 7(p²+3q²) Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14/
(i) ax²+bx = x(ax+b)
(ii) 7p²+21q² = 7(p²+3q²)
Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video
(i) 4p²-9p² = (2p)²-(3q)² = (2p-3q)(2p+3q) [∵ a²-b²=(a-b)(a+b)] (ii) 63a²-112b² =7(9a²-16b²)= 7[(3a)²-(4b)²] = 7(3a-4b)(3a+4b) [∵ a²-b²=(a-b)(a+b)] Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-1Read more
(i) a²+8a+16 = a²+(4+4)a+4X4 Using identity x²+(a+b)x+ab=(x+a)(x+b), Here x=a,a=4 and b=4 a²+8a+16 = (a+4)(a+4)=(a+4)² (ii) p²-10p+25 = p²+(-5-5)p+(-5)(-5) Using identity x²+(a+b)x+ab=(x+a)(x+b), Here x=p,a=-5and b=-5 Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to:Read more
(i) a²+8a+16 = a²+(4+4)a+4X4
Using identity x²+(a+b)x+ab=(x+a)(x+b),
Here x=a,a=4 and b=4
a²+8a+16 = (a+4)(a+4)=(a+4)²
(ii) p²-10p+25 = p²+(-5-5)p+(-5)(-5)
Using identity x²+(a+b)x+ab=(x+a)(x+b),
Here x=p,a=-5and b=-5
Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video
Factorize the following expressions: (i) p²+6p+8 (ii) q²-10q+21
(i) p²+6p+8 = p²+(4+2)p+4X2 = p²+4p+2p+4x2 = p(p+4)+2(p+4) = (p+4)(p+2) (ii) q²-10q+21 = q²-(7+3)q+7x3 = q²-7q-3q+7x3 = q(q-7)-3(q-7) = (q-7)(q-3) Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14Read more
(i) p²+6p+8 = p²+(4+2)p+4X2
= p²+4p+2p+4×2
= p(p+4)+2(p+4)
= (p+4)(p+2)
(ii) q²-10q+21 = q²-(7+3)q+7×3
= q²-7q-3q+7×3
= q(q-7)-3(q-7)
= (q-7)(q-3)
Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14/
Factorize: (i) a⁴-b⁴ (ii) p⁴-81
(i) a⁴-b⁴ = (a²)²-(b²)² = (a²-b²)(a²+b²) [∵ a²-b²-(a-b)(a+b)] = (a-b)(a+b)(a²+b²) [∵ a²-b²-(a-b)(a+b)] (ii) p⁴-81=(p²)²-(9²)² = (9²-9)(9²+9) [∵ a²-b²-(a-b)(a+b)] = (p²-3²)(p²+9) = (p-3)(p+3)(p²+9) [∵ a²-b²-(a-b)(a+b)] Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to:Read more
(i) a⁴-b⁴ = (a²)²-(b²)²
= (a²-b²)(a²+b²) [∵ a²-b²-(a-b)(a+b)]
= (a-b)(a+b)(a²+b²) [∵ a²-b²-(a-b)(a+b)]
(ii) p⁴-81=(p²)²-(9²)²
= (9²-9)(9²+9) [∵ a²-b²-(a-b)(a+b)]
= (p²-3²)(p²+9)
= (p-3)(p+3)(p²+9) [∵ a²-b²-(a-b)(a+b)]
Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14/
Factorize the expressions: (i) ax²+bx (ii) 7p²+21q²
(i) ax²+bx = x(ax+b) (ii) 7p²+21q² = 7(p²+3q²) Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14/
(i) ax²+bx = x(ax+b)
(ii) 7p²+21q² = 7(p²+3q²)
Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14/
Factorize: (i) 4p²-9p² (ii) 63a²-112b²
(i) 4p²-9p² = (2p)²-(3q)² = (2p-3q)(2p+3q) [∵ a²-b²=(a-b)(a+b)] (ii) 63a²-112b² =7(9a²-16b²)= 7[(3a)²-(4b)²] = 7(3a-4b)(3a+4b) [∵ a²-b²=(a-b)(a+b)] Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-1Read more
(i) 4p²-9p² = (2p)²-(3q)²
= (2p-3q)(2p+3q) [∵ a²-b²=(a-b)(a+b)]
(ii) 63a²-112b² =7(9a²-16b²)= 7[(3a)²-(4b)²]
= 7(3a-4b)(3a+4b) [∵ a²-b²=(a-b)(a+b)]
Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14/
Factorize the following expressions: (i) a²+8a+16 (ii) p²-10p+25
(i) a²+8a+16 = a²+(4+4)a+4X4 Using identity x²+(a+b)x+ab=(x+a)(x+b), Here x=a,a=4 and b=4 a²+8a+16 = (a+4)(a+4)=(a+4)² (ii) p²-10p+25 = p²+(-5-5)p+(-5)(-5) Using identity x²+(a+b)x+ab=(x+a)(x+b), Here x=p,a=-5and b=-5 Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video for more answers vist to:Read more
(i) a²+8a+16 = a²+(4+4)a+4X4
Using identity x²+(a+b)x+ab=(x+a)(x+b),
Here x=a,a=4 and b=4
a²+8a+16 = (a+4)(a+4)=(a+4)²
(ii) p²-10p+25 = p²+(-5-5)p+(-5)(-5)
Using identity x²+(a+b)x+ab=(x+a)(x+b),
Here x=p,a=-5and b=-5
Class 8 Maths Chapter 11 Exercise 14.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-14/