Question

How to factor 6n2 - 6n - 12

Answer 1

We see that in these 3 terms, 6 is a common factor. So, let's factor out a 6:

`6n^2-6n-12 = 6(n^2-n-2)`

We can set this equal to 0 and factor by using the quadratic formula which is:

`x = (-b \pm √(b^2-4ac) )/(2a)`

So, let's do just that:

`6(n^2-n-2) = 0`

Note that the 6 goes away if you divide both sides by 6. In this case, a = 1, b = -1, and c = -2. Let's plug that into the quadratic equation:

`(1 \pm √((-1)^2-4(1)(-2)) )/(2(1)) = (-1 \pm √(1-(-8)) )/(2)`

`(-1 \pm √(9) )/(2) = {1, -2}`

So, we can write this as:

`6n^2 - 6n - 12 = 6(n-1)(n+2)`

Notice that the 6 comes back because it was only temporarily mad. And that the roots have opposite signs in the parentheses because to find the roots, you need to set each parentheses equal to 0 and solve for n. With that in mind, your final answer is 6(n-1)(n+2). Hope I could help you!