Advertisements
Advertisements
प्रश्न
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
a2y3 – 2aby2 + b2y
योग
Advertisements
उत्तर
We have,
a2y3 – 2aby2 + b2y
= y(a2y2 – 2aby + b2)
= y[(ay)2 – 2 × ay × b + b2]
= y(ay – b)2
= y(ay – b)(ay – b)
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Algebraic Expression, Identities and Factorisation - Exercise [पृष्ठ २३५]
APPEARS IN
संबंधित प्रश्न
Expand: (5x - 4)2
Factorise the following expressions
y2 – 10y + 25
The factors of x2 – 4x + 4 are __________
The factors of x2 – 6x + 9 are
a2 – b2 is equal to ______.
Square of 9x – 7xy is ______.
(a – b)2 = a2 – b2
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
`9y^2 - 4xy + (4x^2)/9`
Factorise the following.
x2 – 10x + 21
Factorise the following.
x2 + 4x – 77
