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Factor 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 lessFactor using suitable identities: p square/16 – 2 + 16/p square
We factor this variable fraction expression by recognizing the underlying square identity. The first term is the square of p/4, and the final term 16/p square is the square of 4/p. The middle term is negative, and when we calculate two times p/4 times 4/p, the variables cancel out perfectly to leaveRead more
We factor this variable fraction expression by recognizing the underlying square identity. The first term is the square of p/4, and the final term 16/p square is the square of 4/p. The middle term is negative, and when we calculate two times p/4 times 4/p, the variables cancel out perfectly to leave just the number two. This gives us the final factored expression (p/4 – 4/p) 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 using suitable identities: m square/9 + mk/3 + k square/4 + 3nk + 2mn + 9n square
We observe that this long expression has six separate terms, which indicates it comes from a three-term square identity. The three perfect square terms are m square/9, k square/4 and 9n square, which are the squares of m/3, k/2 and 3n respectively. The remaining three terms perfectly match the doublRead more
We observe that this long expression has six separate terms, which indicates it comes from a three-term square identity. The three perfect square terms are m square/9, k square/4 and 9n square, which are the squares of m/3, k/2 and 3n respectively. The remaining three terms perfectly match the double cross-products of these bases. Therefore, the complete factored form is (m/3 + k/2 + 3n) 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/
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