Question

Factor 60x2 – 155x + 100 completely

Answer 1

Answer:

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

Step-by-step 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 $5[\left(4x-5\right)\left(3x-4\right)]$.

Answer 2

Answer:

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)