What is the completely factored form of 6x2 - 13x - 5?

(2x - 5)(3x + 1)

Step-by-step explanation:

6x² - 13x - 5

you can solve this by factoring method by observing the factors of the x² term or the constant term. But because in this case the x² term is 6 (i.e ≠ 1), it will probably be easier to find the roots of the equation.

We'll use completing the square:

let 6x² - 13x - 5 = 0 (divide both sides by 6, to make the x² term = 1)

(6x² - 13x) / 6 = 5 /6

x² - (13/6)x = 5 /6  (complete the square by adding [(13/6)÷2 ]²= (13/12)² to both sides)

x² - (13/6)x + (13/12)²= 5 /6 + (13/12)²

[x - (13/12) ]²  = 289/144

x - (13/12) = ±√(289/144)

x - (13/12) = ± (17/12)

x = (13/12) ± (17/12)

hence

x = (13/12) + (17/12)

x = 5/2

2x = 5

2x-5 = 0 ---> hence (2x - 5) is a factor

OR

x = (13/12) - (17/12)

x = -1/3

3x = -1

3x+1 = 0 ---> hence (3x + 1) is a factor

Since we know that a quadratic equation can have at most 2 real roots, we can conclude that these are the only 2 real factors

Combining the 2 factors we have found,

6x² - 13x - 5 = (2x - 5)(3x + 1)

answer: check all that apply, where the options?

step-by-step explanation: