Question

Factor 60x2 – 155x + 100 completely

Answer 1

hello :

60x² – 155x + 100 = 5 ( 12 x² - 31 x + 20 )

delta of ( 12x² - 31 x + 20 ) is b² - 4ac when : a = 12 b = -31 c = 20

delta = ( - 31 )² - 4(12)(20) = 961 - 960 = 1

x1 = (31-1)/24 = 30/24 = 5/4

x2 = (31+1)/24 = 32/24 = 4/3

Factor 60x2 – 155x + 100 = 5 (12(x -5/4)(x- 4/3)) = 60 (x -5/4)(x- 4/3)

Answer 2

The factor of the provided expression is
`5[\left(4x-5\right)\left(3x-4\right)]`.

Explanation:

Consider the provided expression.

`60x^2-155x+100`

Factor out common term 5.

`5(12x^2-31x+20)`

`5(12x^2-15x-16x+20)`

`5[\left(12x^2-15x\right)+\left(-16x+20\right)]`

`5[3x\left(4x-5\right)-4\left(4x-5\right)]`

`5[\left(4x-5\right)\left(3x-4\right)]`

Hence, the factor of the provided expression is
`5[\left(4x-5\right)\left(3x-4\right)]`.