Advertisements
Advertisements
प्रश्न
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
a2y2 – 2aby + b2
योग
Advertisements
उत्तर
We have,
a2y2 – 2aby + b2
= (ay)2 – 2 × ay × b + b2
= (ay – b)2
= (ay – b)(ay – b)
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Algebraic Expression, Identities and Factorisation - Exercise [पृष्ठ २३४]
APPEARS IN
संबंधित प्रश्न
Expand (2)2p − 3q
Expand the following square, using suitable identities
(xyz – 1)2
Evaluate the following, using suitable identity
472
Square of 9x – 7xy is ______.
(a – b) ______ = a2 – 2ab + b2
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
4a2 – 4ab + b2
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
p2y2 – 2py + 1
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
`x^2/4 - 2x + 4`
If m – n = 16 and m2 + n2 = 400, then find mn.
Subtract b(b2 + b – 7) + 5 from 3b2 – 8 and find the value of expression obtained for b = – 3.
