Factor Completely: 6n^2-n-35?

Question

I know the answer, but it's the steps that's been confusing me and I don't even know if my answer is right. When I do problems like these I usually come up with one line, skipping all the steps. Can someone guide me through this step by step?

Answer 1

First write down possible factors of 6n²
Try:-
(3n"""""")(2n""""""")
Now require factors of - 35 that combine with 3n and 2n to give - n
Try 7 and 5:-
(3n"""""7)(2n"""""5)
Affix signs:-
(3n + 7) (2n - 5)

Answer 2

standard form of quadratic equation
ax^2 + bx + c
to factorize, you have to split b into sum of two numbers so that their product is equal to (ac). So factorize (ac) ino two numbers in such a way that their sum is equal to b

6n^2 - n - 35
here b = -n
ac = 6(-35) = - 210
factorize 210
2*3*5*7 = (5* 3)(2*7) = 15(14) =(-15)(+14)
if you add -15 and 14, you will get - 1 and the product of (-15)and (14) is -210, satisfying the requirement
so 6n^2 - n - 35 = 6n^2 - 15n + 14n - 35
take common factors
3n(2n - 5) + 7(2n - 5)
take out common factor (2n - 5)
(2n-5)(3n + 7)

Answer 3

Use the quadratic formula where

(I can't write it here, the notations suck)

a = 6, b = -1, and c = -35

http://en.wikipedia.org/wiki/Quadratic_equation

Your answers should be 28/12 = 4 or 30/12 = 4 1/6

Answer 4

6n^2 - n -35
The answer is in the form (ax+b)(cx+d)
a*c has to equal 6
a*d + b*c =-1
b*d = -35

So first try out possibilities of a, c, b, and d using the first and third of the equations listed above.
By trial you know the answer is:
(3n+7 )(2n -5 )