To factor this six-term polynomial, we look at the signs of the products to find which base value carries the negative sign. The perfect square bases are 3a, b and 2c. We notice that the terms 6ab and 4bc are negative, while 12ac is positive. Since the negative components both contain the variable bRead more
To factor this six-term polynomial, we look at the signs of the products to find which base value carries the negative sign. The perfect square bases are 3a, b and 2c. We notice that the terms 6ab and 4bc are negative, while 12ac is positive. Since the negative components both contain the variable b, the base associated with b must be negative. This gives the grouped factor (3a – b + 2c) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
We factor this quadratic expression by recognizing the underlying perfect square trinomial pattern. The first term 16s square is the square of 4s and the second term 25t square is the square of 5t. The middle term has a minus sign and is exactly equal to minus two multiplied by 4s multiplied by 5t,Read more
We factor this quadratic expression by recognizing the underlying perfect square trinomial pattern. The first term 16s square is the square of 4s and the second term 25t square is the square of 5t. The middle term has a minus sign and is exactly equal to minus two multiplied by 4s multiplied by 5t, which equals minus 40st. This matches the subtraction identity, yielding the factor (4s – 5t) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
To factor this expression without tiles, we apply the middle term splitting method. We need two numbers that add up to minus 1 and multiply to minus 42. These target integers are minus 7 and positive 6. Rewriting the expression gives r square - 7r + 6r - 42. Factoring by grouping in pairs gives r(rRead more
To factor this expression without tiles, we apply the middle term splitting method. We need two numbers that add up to minus 1 and multiply to minus 42. These target integers are minus 7 and positive 6. Rewriting the expression gives r square – 7r + 6r – 42. Factoring by grouping in pairs gives r(r – 7) + 6(r – 7). Taking out the common binomial bracket leaves the final factors (r – 7)(r + 6).
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
This algebraic expression can be factored by identifying the perfect square components. The first part of the expression is 49g square, which is equal to (7g) square. The last part is h square, which is equal to (h) square. The middle value 14gh is exactly two times 7g times h. Since it matches theRead more
This algebraic expression can be factored by identifying the perfect square components. The first part of the expression is 49g square, which is equal to (7g) square. The last part is h square, which is equal to (h) square. The middle value 14gh is exactly two times 7g times h. Since it matches the standard addition identity completely, it condenses into the final perfect square binomial factor (7g + h) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
To factor this long polynomial, we examine the perfect square bases which are 8u, 11v and 2w. We observe that the cross-product terms 176uv and 32uw are negative, while 44vw is positive. Because the product of v and w becomes positive while their separate combinations with u are negative, both the 1Read more
To factor this long polynomial, we examine the perfect square bases which are 8u, 11v and 2w. We observe that the cross-product terms 176uv and 32uw are negative, while 44vw is positive. Because the product of v and w becomes positive while their separate combinations with u are negative, both the 11v and 2w bases must carry negative signs. This produces the complete factored solution (8u – 11v – 2w) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
Factor the following: 9a square + b square + 4c square – 6ab + 12ac – 4bc
To factor this six-term polynomial, we look at the signs of the products to find which base value carries the negative sign. The perfect square bases are 3a, b and 2c. We notice that the terms 6ab and 4bc are negative, while 12ac is positive. Since the negative components both contain the variable bRead more
To factor this six-term polynomial, we look at the signs of the products to find which base value carries the negative sign. The perfect square bases are 3a, b and 2c. We notice that the terms 6ab and 4bc are negative, while 12ac is positive. Since the negative components both contain the variable b, the base associated with b must be negative. This gives the grouped factor (3a – b + 2c) square.
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 lessFactor the following: 16s square + 25t square – 40st
We factor this quadratic expression by recognizing the underlying perfect square trinomial pattern. The first term 16s square is the square of 4s and the second term 25t square is the square of 5t. The middle term has a minus sign and is exactly equal to minus two multiplied by 4s multiplied by 5t,Read more
We factor this quadratic expression by recognizing the underlying perfect square trinomial pattern. The first term 16s square is the square of 4s and the second term 25t square is the square of 5t. The middle term has a minus sign and is exactly equal to minus two multiplied by 4s multiplied by 5t, which equals minus 40st. This matches the subtraction identity, yielding the factor (4s – 5t) square.
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 lessFactor the following: r square – r – 42
To factor this expression without tiles, we apply the middle term splitting method. We need two numbers that add up to minus 1 and multiply to minus 42. These target integers are minus 7 and positive 6. Rewriting the expression gives r square - 7r + 6r - 42. Factoring by grouping in pairs gives r(rRead more
To factor this expression without tiles, we apply the middle term splitting method. We need two numbers that add up to minus 1 and multiply to minus 42. These target integers are minus 7 and positive 6. Rewriting the expression gives r square – 7r + 6r – 42. Factoring by grouping in pairs gives r(r – 7) + 6(r – 7). Taking out the common binomial bracket leaves the final factors (r – 7)(r + 6).
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 lessFactor the following: 49g square + 14gh + h square
This algebraic expression can be factored by identifying the perfect square components. The first part of the expression is 49g square, which is equal to (7g) square. The last part is h square, which is equal to (h) square. The middle value 14gh is exactly two times 7g times h. Since it matches theRead more
This algebraic expression can be factored by identifying the perfect square components. The first part of the expression is 49g square, which is equal to (7g) square. The last part is h square, which is equal to (h) square. The middle value 14gh is exactly two times 7g times h. Since it matches the standard addition identity completely, it condenses into the final perfect square binomial factor (7g + h) square.
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 lessFactor the following: 64u square + 121v square + 4w square – 176uv – 32uw + 44vw
To factor this long polynomial, we examine the perfect square bases which are 8u, 11v and 2w. We observe that the cross-product terms 176uv and 32uw are negative, while 44vw is positive. Because the product of v and w becomes positive while their separate combinations with u are negative, both the 1Read more
To factor this long polynomial, we examine the perfect square bases which are 8u, 11v and 2w. We observe that the cross-product terms 176uv and 32uw are negative, while 44vw is positive. Because the product of v and w becomes positive while their separate combinations with u are negative, both the 11v and 2w bases must carry negative signs. This produces the complete factored solution (8u – 11v – 2w) square.
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