Advertisements
Advertisements
Question
Factorize `6ab - b^2 + 12ac - 2bc`
Advertisements
Solution
Taking common b in (6ab - b2 )and 2c in (12ac - 2bc)
= b (6a - b) + 2c (6a - b)
Taking (6a - b) common in both terms
= (6a - b)(b + 2c)
∴ 6ab - b2 +12ac - 2bc = (6a - b)(b + 2c)
APPEARS IN
RELATED QUESTIONS
Get the algebraic expression in the following case using variables, constants and arithmetic operations.
Numbers x and y both squared and added.
Factorize `5sqrt5x^2 + 20x + 3sqrt5`
Given possible expressions for the length and breadth of the rectangle having 35y2 + 13y – 12 as its area.
Factorize 125x3 - 27 y3 - 225x2 y +135xy2
Factorize x3 + 8y3 + 6x2 y +12xy2
Multiply: x2 + 4y2 + z3 + 2xy + xz − 2yz by x − 2y − z
If (x + y)3 − (x − y)3 − 6y(x2 − y2) = ky2, then k =
Multiply: (1 + 6x2 - 4x3)(-1 + 3x - 3x2)
Divide: 8a2 + 4a - 60 by 2a - 5
Rohan’s mother gave him ₹ 3xy2 and his father gave him ₹ 5(xy2 + 2). Out of this total money he spent ₹ (10 – 3xy2) on his birthday party. How much money is left with him?
