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Find the product without multiplying directly: (18 x 29)
To find this product using an identity approach, we can break down both numbers relative to base tens, writing them as the binomial product (20 - 2)(30 - 1). Applying the distributive expansion gives 20 times 30 minus 20 times 1 minus 2 times 30 plus 2 times 1. This evaluates to 600 minus 20 minus 6Read more
To find this product using an identity approach, we can break down both numbers relative to base tens, writing them as the binomial product (20 – 2)(30 – 1). Applying the distributive expansion gives 20 times 30 minus 20 times 1 minus 2 times 30 plus 2 times 1. This evaluates to 600 minus 20 minus 60 plus 2. Solving this basic arithmetic sequence yields 522.
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 the product without multiplying directly: (34 x 43)
We solve this multiplication by centering the numbers around the base value 40, transforming the problem into (40 - 6) multiplied by (40 + 3). This fits the identity formula matching the template (x + a)(x + b) = x square + (a + b)x + ab. Here, x is 40, a is minus 6 and b is 3. This gives 1600 minusRead more
We solve this multiplication by centering the numbers around the base value 40, transforming the problem into (40 – 6) multiplied by (40 + 3). This fits the identity formula matching the template (x + a)(x + b) = x square + (a + b)x + ab. Here, x is 40, a is minus 6 and b is 3. This gives 1600 minus 120 minus 18, resulting in 1462.
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 the product without multiplying directly: (205) square
To find the value using an algebraic shortcut, we rewrite 205 as the sum of 200 and 5. This lets us use the regular binomial identity where a equals 200 and b equals 5. The square of 200 is 40000 and the square of 5 is 25. The intermediate term is two times 200 times 5, which equals 2000. Adding 400Read more
To find the value using an algebraic shortcut, we rewrite 205 as the sum of 200 and 5. This lets us use the regular binomial identity where a equals 200 and b equals 5. The square of 200 is 40000 and the square of 5 is 25. The intermediate term is two times 200 times 5, which equals 2000. Adding 40000, 2000 and 25 together gives 42025.
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: 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/
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