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  1. To find three dimensional expressions for a cuboid volume, we must fully factorize the given expression into three parts. First, we isolate the highest common integer factor, which is 6. This gives 6 multiplied by the binomial quantity a square - 4b square. The bracketed part is a difference of squaRead more

    To find three dimensional expressions for a cuboid volume, we must fully factorize the given expression into three parts. First, we isolate the highest common integer factor, which is 6. This gives 6 multiplied by the binomial quantity a square – 4b square. The bracketed part is a difference of squares that expands into the linear components (a + 2b) and (a – 2b). These three distinct factors constitute the dimensions.

     

    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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  2. The area of the rectangle is given as a binomial with two perfect squares separated by a minus sign. We utilize the difference of squares algebraic identity to factor this expression. Here, the term 36s square is the square of 6s and the term 49t square is the square of 7t. Factoring gives the produRead more

    The area of the rectangle is given as a binomial with two perfect squares separated by a minus sign. We utilize the difference of squares algebraic identity to factor this expression. Here, the term 36s square is the square of 6s and the term 49t square is the square of 7t. Factoring gives the product of the sum and difference of these bases, yielding the length and breadth expressions.

     

    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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  3. To find the dimensions of the rectangle, we look for the factors of the given quadratic area polynomial. The expression matches the perfect square subtraction identity. The first term 25a square is the square of 5a and the last term 9b square is the square of 3b. The middle term minus 30ab representRead more

    To find the dimensions of the rectangle, we look for the factors of the given quadratic area polynomial. The expression matches the perfect square subtraction identity. The first term 25a square is the square of 5a and the last term 9b square is the square of 3b. The middle term minus 30ab represents minus two times 5a times 3b. Therefore, it factors completely into (5a – 3b) square, giving the possible side lengths.

     

    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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  4. To factor this six-term polynomial, we look at the signs of the products to determine which base is negative. The perfect square bases are 3a, 2b and c. We notice that the terms 12ab and 4bc are negative, while 6ac is positive. Since the negative terms both contain the variable b, the base associateRead more

    To factor this six-term polynomial, we look at the signs of the products to determine which base is negative. The perfect square bases are 3a, 2b and c. We notice that the terms 12ab and 4bc are negative, while 6ac is positive. Since the negative terms both contain the variable b, the base associated with b must carry the negative sign. This results in the factor (3a – 2b + c) 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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  5. We use the three-term square identity to expand the given expression. By setting a equal to p, b equal to 3q and c equal to 7r, we square each term to get p square, 9q square and 49r square. Then we calculate the three double cross-product terms: two times p times 3q gives 6pq; two times 3q times 7rRead more

    We use the three-term square identity to expand the given expression. By setting a equal to p, b equal to 3q and c equal to 7r, we square each term to get p square, 9q square and 49r square. Then we calculate the three double cross-product terms: two times p times 3q gives 6pq; two times 3q times 7r gives 42qr; two times 7r times p gives 14rp. Combining them gives the full expansion.

     

    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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