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Find possible expressions for the length, breadth and height of the cuboid whose volume is 3ps square – 15ps + 12p
We find the dimensions by factoring the given cubic volume polynomial step by step. First, we pull out the common term 3p from all three components, yielding 3p times the trinomial s square - 5s + 4. Next, we apply the middle term splitting technique to this remaining quadratic expression. The middlRead more
We find the dimensions by factoring the given cubic volume polynomial step by step. First, we pull out the common term 3p from all three components, yielding 3p times the trinomial s square – 5s + 4. Next, we apply the middle term splitting technique to this remaining quadratic expression. The middle coefficient minus 5 splits into minus 1 and minus 4, which solves into the final three dimensional factors.
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 lessFind possible expressions for the length, breadth and height of the cuboid whose volume is 6a square – 24b square
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/
See lessFind possible expressions for the length and breadth of the rectangle whose area is 36s square – 49t square
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/
See lessFind possible expressions for the length and breadth of the rectangle whose area is 25a square – 30ab + 9b square
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/
See lessFactor using suitable identities: 9a square + 4b square + c square – 12ab + 6ac – 4bc
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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