To find the missing binomial factor, we need to factorize the quadratic expression on the right side. We split the middle coefficient minus 4 into minus 7 and positive 3 because their product matches the product of 3 and minus 7. Rewriting the expression gives 3x square + 3x - 7x - 7. Grouping the tRead more
To find the missing binomial factor, we need to factorize the quadratic expression on the right side. We split the middle coefficient minus 4 into minus 7 and positive 3 because their product matches the product of 3 and minus 7. Rewriting the expression gives 3x square + 3x – 7x – 7. Grouping the terms gives 3x(x + 1) – 7(x + 1). Factoring out the common binomial yields the missing part, (3x – 7).
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
We solve for the blanks by splitting the middle term of the quadratic expression. We need two numbers that add up to minus 11 and multiply to minus 60. These numbers are minus 15 and positive 4. This transforms the polynomial into 10x square - 15x + 4x - 6. Grouping into pairs gives 5x(2x - 3) + 2(2Read more
We solve for the blanks by splitting the middle term of the quadratic expression. We need two numbers that add up to minus 11 and multiply to minus 60. These numbers are minus 15 and positive 4. This transforms the polynomial into 10x square – 15x + 4x – 6. Grouping into pairs gives 5x(2x – 3) + 2(2x – 3). This gives the complete factored form matching the blanks perfectly.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
To factorize this quadratic equation, we split the middle coefficient 7 into two parts that add up to 7 and multiply to 12. These numbers are 3 and 4. We rewrite the quadratic expression as 6x square + 3x + 4x + 2. Factoring by grouping gives 3x(2x + 1) + 2(2x + 1). Taking out the common bracket resRead more
To factorize this quadratic equation, we split the middle coefficient 7 into two parts that add up to 7 and multiply to 12. These numbers are 3 and 4. We rewrite the quadratic expression as 6x square + 3x + 4x + 2. Factoring by grouping gives 3x(2x + 1) + 2(2x + 1). Taking out the common bracket results in the final factors of (2x + 1) and (3x + 2).
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
To find this square value using an identity shortcut, we express 41 as the sum of 40 and 1. We then apply the binomial identity where a is 40 and b is 1. The square of 40 is 1600 and the square of 1 is 1. The middle term is two times 40 times 1, which equals 80. Adding 1600, 80 and 1 together givesRead more
To find this square value using an identity shortcut, we express 41 as the sum of 40 and 1. We then apply the binomial identity where a is 40 and b is 1. The square of 40 is 1600 and the square of 1 is 1. The middle term is two times 40 times 1, which equals 80. Adding 1600, 80 and 1 together gives 1681.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
We evaluate this square by rewriting 27 as a subtraction expression, 30 - 3. This matches our subtraction square identity with a equal to 30 and b equal to 3. The square of 30 is 900 and the square of 3 is 9. The middle product term to subtract is two times 30 times 3, which equals 180. Computing 90Read more
We evaluate this square by rewriting 27 as a subtraction expression, 30 – 3. This matches our subtraction square identity with a equal to 30 and b equal to 3. The square of 30 is 900 and the square of 3 is 9. The middle product term to subtract is two times 30 times 3, which equals 180. Computing 900 minus 180 plus 9 yields 729.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
Fill in the blanks to complete the following identities: (_) (x + 1) = (3x square – 4x – 7)
To find the missing binomial factor, we need to factorize the quadratic expression on the right side. We split the middle coefficient minus 4 into minus 7 and positive 3 because their product matches the product of 3 and minus 7. Rewriting the expression gives 3x square + 3x - 7x - 7. Grouping the tRead more
To find the missing binomial factor, we need to factorize the quadratic expression on the right side. We split the middle coefficient minus 4 into minus 7 and positive 3 because their product matches the product of 3 and minus 7. Rewriting the expression gives 3x square + 3x – 7x – 7. Grouping the terms gives 3x(x + 1) – 7(x + 1). Factoring out the common binomial yields the missing part, (3x – 7).
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessFill in the blanks to complete the following identities: 10x square – 11x – 6 = (2x – ) ( + 2)
We solve for the blanks by splitting the middle term of the quadratic expression. We need two numbers that add up to minus 11 and multiply to minus 60. These numbers are minus 15 and positive 4. This transforms the polynomial into 10x square - 15x + 4x - 6. Grouping into pairs gives 5x(2x - 3) + 2(2Read more
We solve for the blanks by splitting the middle term of the quadratic expression. We need two numbers that add up to minus 11 and multiply to minus 60. These numbers are minus 15 and positive 4. This transforms the polynomial into 10x square – 15x + 4x – 6. Grouping into pairs gives 5x(2x – 3) + 2(2x – 3). This gives the complete factored form matching the blanks perfectly.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessFill in the blanks to complete the following identities: 6×2 + 7x + 2 = ( _ ) ( _ )
To factorize this quadratic equation, we split the middle coefficient 7 into two parts that add up to 7 and multiply to 12. These numbers are 3 and 4. We rewrite the quadratic expression as 6x square + 3x + 4x + 2. Factoring by grouping gives 3x(2x + 1) + 2(2x + 1). Taking out the common bracket resRead more
To factorize this quadratic equation, we split the middle coefficient 7 into two parts that add up to 7 and multiply to 12. These numbers are 3 and 4. We rewrite the quadratic expression as 6x square + 3x + 4x + 2. Factoring by grouping gives 3x(2x + 1) + 2(2x + 1). Taking out the common bracket results in the final factors of (2x + 1) and (3x + 2).
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessSelect and use the identity that will help you to find the product without multiplying directly: (41) square
To find this square value using an identity shortcut, we express 41 as the sum of 40 and 1. We then apply the binomial identity where a is 40 and b is 1. The square of 40 is 1600 and the square of 1 is 1. The middle term is two times 40 times 1, which equals 80. Adding 1600, 80 and 1 together givesRead more
To find this square value using an identity shortcut, we express 41 as the sum of 40 and 1. We then apply the binomial identity where a is 40 and b is 1. The square of 40 is 1600 and the square of 1 is 1. The middle term is two times 40 times 1, which equals 80. Adding 1600, 80 and 1 together gives 1681.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessSelect and use the identity that will help you to find the product without multiplying directly: (27) square
We evaluate this square by rewriting 27 as a subtraction expression, 30 - 3. This matches our subtraction square identity with a equal to 30 and b equal to 3. The square of 30 is 900 and the square of 3 is 9. The middle product term to subtract is two times 30 times 3, which equals 180. Computing 90Read more
We evaluate this square by rewriting 27 as a subtraction expression, 30 – 3. This matches our subtraction square identity with a equal to 30 and b equal to 3. The square of 30 is 900 and the square of 3 is 9. The middle product term to subtract is two times 30 times 3, which equals 180. Computing 900 minus 180 plus 9 yields 729.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See less