5y3 - 125y = 5y(y2-25) = 5y (y+5)(y-5)
3(3x - 5y)(3x + 5y)
x2 + 8xy + 15y2 = (x + 3y) (x + 5y)
(8x + 5y)(64x^2 - 40xy + 25y^2)
One such expression is 5n
5y3 - 125y = 5y(y2-25) = 5y (y+5)(y-5)
The cube of a binomial is the cube of two terms separated by an addition or subtraction sign, such as (2a + 3b) or (ab - cd).For example, (2x - 5y)3 = 8x3 - 40x2y + 50xy2 - 20x2y + 100xy2 - 125y3.The detailed method of expanding this binomial is : (2x - 5y)3 = (2x - 5y)(2x - 5y)(2x - 5y) = (4x2 - 20xy + 25y2)(2x - 5y) = 8x3 - 40x2y + 50xy2 - 20x2y + 100xy2 - 125y3
12y3 - 30y2 + 12yFactor out a y:y(12y2 - 30y + 12)Factor out a 6:6y(2y2 -5y + 2)Factor 2y2 - 5y + 26y(2y - 1)(y - 2)
(5y - 2)(5y + 2)
5y - 65xy '5' is a common factor both numbers. Hence 'take it out'. 5(y - 13xy) 'y' is a common factor to both terms, so take it out too!!!! 5y(1 - 13x) Done Factored!!!!!
The GCF is 5y.20y3 = (5y)(4y2)5y2 = (5y)(y)45y = (5y)(9)
(5y + 7w)(5y - 7w)
No
2(x - 5y)
Remember both 16 & 25 are squared numbers. 16 = 4^2 & 25 = 5^2 Hence we can write (4x)^2 - (5y)^2 Remember two squared terms with a NEGATIVE Between them will factor. ( 4x - 5y)(4x + 5y) Note the difference signs. NNB Two squared terms with a positive (+) between them DOES NOT factor.
3(3x - 5y)(3x + 5y)
y to the six is common, so it becomes y^3(y^3 - 125). Then we use difference of cubes ((a-b)(a^2 - 2ab + b^2)), so it becomes y^3(y-5)(y^2 + 5y + 25). Since we cannot factor out that last quadratic part, this concludes the factoring process and we end up with that last line as our answer.