Advertisements
Advertisements
प्रश्न
Factorise: 1 - 25 (a+ b)2
योग
Advertisements
उत्तर
1 - 25 (a+ b)2
= (1)2 - [5(a+b)]2
= [1 + 5(a+b)] [1 - 5(a+b)]
= (1 + 5a + 5b) (1 - 5a - 5b)
shaalaa.com
Factorisation by Difference of Two Squares
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
Factorise: `4/25 - 25b^2`
Factorise: (a+ 2b)2 - a2
Factorise: 25(2x + y)2 - 16(x - y)2
Factorise: a2 + 4a + 4 - b2
Factorise: a2 - b2 - 2b - 1
factorise : 25(2x - y)2 - 16(x - 2y)2
Factorise: `9x^2 - 1/16`
Factorise: (a - 3b)2 - 36 b2
Factorise: 25(a - 5b)2 - 4(a - 3b)2
Factorise a2b − b3. Using this result, find the value of 1012 × 100 − 1003.
