Question

Factor the polynomial completely. 2x^2-15x+28

Answer 1

First step: try to factor out a common factor.
With coefficients 2, -15, 28 there isn't one
Next step: set up two sets of parentheses:
( )( )
Place the factorization of 2x^2 on left sides:
(2x )(x )

Answer 2

Next step, you need to factor 28 and place its factors in the right sides of the parentheses. Since the last term is _positive_ 28 and the middle term (-15x) is negative, the two factors of 28 must be negative. How can you factor 28 into two negative factors?
-1, -28
-2, -14
-4, -7
Those are the only choices.

Answer 3

Now you need to place the correct choice in the correct way in the parentheses.
Let's start with the last set of factors, -7, -4
They can be placed like this:
(2x - 7)(x -4)
or like this:
(2x - 4)(x - 7)
To see if one of them works, do teh OI of FOIL and see if you get the original midle term (-15x)
OI with the first one: (2x)(-4) + (-7)(x) = -8x - 7x = -15x
Since it works, that's the correct way of factoring
So the factorixation of 2x^2 - 15x + 28 is (2x - 7)(x - 4)