(i) (x²-5)(x+5)+25 = x²(x+5)-(x+5)+25 = x²X+x²x5-5Xx-5X5+25 = x³+5x²-5x-25+25 = x³+5x²-5x (ii) (a²+5)(b²+3)+5 = a²(b³+3)+5(b³+3)+5 = a²xb³+a²x3+5xb³+5x3+5 = a²b³+3a²++5b³+15+5 = a²b³+3a²+5b³+20 Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video for more answers vist to: https://www.tiwariacademyRead more
(i) (5-2x)(3+2)=5x(3+x)-2x(3+x)= 5x3+5xX-2xX3-2xx3-2xXx = 15+5x-6x-2x²=15-x-2x² (ii) (x+7y)(7x-y)= x(7x-y)+7yx(7x-y) = xx7x-Xxy+7yx7x-7yxy = 7x²-xy+49xy-7y² = 7x²+48xy-7y² Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/cRead more
(i) (2x+5)and(4x-3) =2x(4x-3)+5(4x-3) = 2xX4x-2xX3+5X4x-5X3 = 8x²-6x+20x-15 = 8x² + 14x-15 (ii) (y-8) and (3y-4) =y(3y-4)-8(3y-4) = yX3y-yx4-8x3y-8x-4 = 3y²-4y-24y+12 = 3y²-28y+12 Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solRead more
p(p-q)+q(q-r)+r(r-p)=p²-pq+q²-qr²+r²-rp = p²+q²+r²-pq-qr-rp Class 8 Maths Chapter 9 Exercise 9.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
3x(4x-5) +3= 3x X 4x X 3x X 5+3 = 12x²-15x+3 (i) for x=3, 12x²+15x+3=12(3)²-15x3+3 = 12x9–45+3=108–45+3=66 (ii) for x=1/2, 12x² -15x+3=12(1/2)²-15x1/2+3=12x1/4-15/2+3 = 6-15/2=12-15/2=-3/2 Class 8 Maths Chapter 9 Exercise 9.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/Read more
3x(4x-5) +3= 3x X 4x X 3x X 5+3 = 12x²-15x+3
(i) for x=3, 12x²+15x+3=12(3)²-15×3+3
= 12×9–45+3=108–45+3=66
(ii) for x=1/2, 12x² -15x+3=12(1/2)²-15×1/2+3=12×1/4-15/2+3
= 6-15/2=12-15/2=-3/2
Class 8 Maths Chapter 9 Exercise 9.3 Solution in Video
Simplify: (i) (x²-5)(x+5)+25 (ii) (a²+5)(b²+3)+5
(i) (x²-5)(x+5)+25 = x²(x+5)-(x+5)+25 = x²X+x²x5-5Xx-5X5+25 = x³+5x²-5x-25+25 = x³+5x²-5x (ii) (a²+5)(b²+3)+5 = a²(b³+3)+5(b³+3)+5 = a²xb³+a²x3+5xb³+5x3+5 = a²b³+3a²++5b³+15+5 = a²b³+3a²+5b³+20 Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video for more answers vist to: https://www.tiwariacademyRead more
(i) (x²-5)(x+5)+25 = x²(x+5)-(x+5)+25
= x²X+x²x5-5Xx-5X5+25
= x³+5x²-5x-25+25
= x³+5x²-5x
(ii) (a²+5)(b²+3)+5 = a²(b³+3)+5(b³+3)+5
= a²xb³+a²x3+5xb³+5×3+5
= a²b³+3a²++5b³+15+5
= a²b³+3a²+5b³+20
Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
Find the product: (i) (5-2x) (3+2) (ii) (x+7y)(7x-y)
(i) (5-2x)(3+2)=5x(3+x)-2x(3+x)= 5x3+5xX-2xX3-2xx3-2xXx = 15+5x-6x-2x²=15-x-2x² (ii) (x+7y)(7x-y)= x(7x-y)+7yx(7x-y) = xx7x-Xxy+7yx7x-7yxy = 7x²-xy+49xy-7y² = 7x²+48xy-7y² Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/cRead more
(i) (5-2x)(3+2)=5x(3+x)-2x(3+x)= 5×3+5xX-2xX3-2xx3-2xXx
= 15+5x-6x-2x²=15-x-2x²
(ii) (x+7y)(7x-y)= x(7x-y)+7yx(7x-y)
= xx7x-Xxy+7yx7x-7yxy
= 7x²-xy+49xy-7y²
= 7x²+48xy-7y²
Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
Multiply the binomials: (i) (2x+5) and (4x-3) (ii) (y-8) and (3y-4)
(i) (2x+5)and(4x-3) =2x(4x-3)+5(4x-3) = 2xX4x-2xX3+5X4x-5X3 = 8x²-6x+20x-15 = 8x² + 14x-15 (ii) (y-8) and (3y-4) =y(3y-4)-8(3y-4) = yX3y-yx4-8x3y-8x-4 = 3y²-4y-24y+12 = 3y²-28y+12 Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solRead more
(i) (2x+5)and(4x-3)
=2x(4x-3)+5(4x-3)
= 2xX4x-2xX3+5X4x-5X3
= 8x²-6x+20x-15
= 8x² + 14x-15
(ii) (y-8) and (3y-4) =y(3y-4)-8(3y-4)
= yX3y-yx4-8x3y-8x-4
= 3y²-4y-24y+12
= 3y²-28y+12
Class 8 Maths Chapter 9 Exercise 9.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
add: p(p-q), q(q-r) and r(r-p)
p(p-q)+q(q-r)+r(r-p)=p²-pq+q²-qr²+r²-rp = p²+q²+r²-pq-qr-rp Class 8 Maths Chapter 9 Exercise 9.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
p(p-q)+q(q-r)+r(r-p)=p²-pq+q²-qr²+r²-rp
= p²+q²+r²-pq-qr-rp
Class 8 Maths Chapter 9 Exercise 9.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
Simplify: and find values for 3x(4x-5) +3 and find values for (i) x=3 (ii) x=1/2.
3x(4x-5) +3= 3x X 4x X 3x X 5+3 = 12x²-15x+3 (i) for x=3, 12x²+15x+3=12(3)²-15x3+3 = 12x9–45+3=108–45+3=66 (ii) for x=1/2, 12x² -15x+3=12(1/2)²-15x1/2+3=12x1/4-15/2+3 = 6-15/2=12-15/2=-3/2 Class 8 Maths Chapter 9 Exercise 9.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/Read more
3x(4x-5) +3= 3x X 4x X 3x X 5+3 = 12x²-15x+3
(i) for x=3, 12x²+15x+3=12(3)²-15×3+3
= 12×9–45+3=108–45+3=66
(ii) for x=1/2, 12x² -15x+3=12(1/2)²-15×1/2+3=12×1/4-15/2+3
= 6-15/2=12-15/2=-3/2
Class 8 Maths Chapter 9 Exercise 9.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/