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  1. The area of any regular rectangle is found by multiplying its length by its width dimensions. Therefore, to calculate the missing length expression, we must divide the total area polynomial by the width expression. Factorizing the area polynomial 2x square + 7x + 3 by middle term splitting gives theRead more

    The area of any regular rectangle is found by multiplying its length by its width dimensions. Therefore, to calculate the missing length expression, we must divide the total area polynomial by the width expression. Factorizing the area polynomial 2x square + 7x + 3 by middle term splitting gives the paired factors (2x + 1) multiplied by (x + 3). Cancelling out the common width binomial factor (2x + 1) leaves the remaining linear factor (x + 3).

     

    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. Let the unknown target number be represented by the variable x. According to the wording of the problem, the algebraic equation is written as x plus 1/x equals 10/3. Multiplying the entire equation by 3x clears out the denominators, transforming it into the standard quadratic equation form 3x squareRead more

    Let the unknown target number be represented by the variable x. According to the wording of the problem, the algebraic equation is written as x plus 1/x equals 10/3. Multiplying the entire equation by 3x clears out the denominators, transforming it into the standard quadratic equation form 3x square – 10x + 3 = 0. Splitting the middle term results in the linear factors (3x – 1) and (x – 3), giving the solutions 3 or 1/3.

     

    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. The inner playground is a square with an area of 40 times 40, which equals 1600 square metres. When a path of width s is added around all sides, the new outer boundary forms a larger square with a total side length of 40 + 2s metres. The total outer area is the square of (40 + 2s), which expands toRead more

    The inner playground is a square with an area of 40 times 40, which equals 1600 square metres. When a path of width s is added around all sides, the new outer boundary forms a larger square with a total side length of 40 + 2s metres. The total outer area is the square of (40 + 2s), which expands to 1600 + 160s + 4s square. Subtracting the inner playground area leaves the path area.

     

    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. 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/

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  5. 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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