Given: 33x^2-x-14=0
Since there is an x^2 we know there will be two factors for this equation so we can start setting up the factor brackets:
(x ...)(x ...) = 0
We then need to find factors of both 33 and 14 that will add or subtract to leave a value of -1 which is the multiplier of the middle or x term.
Factors of 33 are 33*1; 11*3; and that's it.
Factors of 14 are 14*1; 7*2; and that's it.
Addition or subtraction of factors 33*1 and 14*1 will not give us -1.
Addition or subtraction of factors 11*3 and 7*2 may give us -1.
Place the factors into the brackets:
(11x ...7)(3x ...2) = 0
We can see that 11*2=22 and 7*3=21 which is needed to give the -1 we need, if the first is negative and the second is positive.
We also know that one factor will contain a positive integer and the other will contain a negative integer because the 14 is negative.
Inserting the signs: (11x+7)(3x-2) = 0
Now we can solve for the two factors of x:
11x+7=0; 11x=-7; x=-7/11
3x-2=0; 3x=2; x=2/3
To check, substitute answers into the given equation:
33x^2-x-14=0
33(2/3)^2-2/3-14=0
33(4/9)-2/3-14=0
cancel(33)(4/cancel(9))^2-14 2/3=0
44/3-14 2/3=0
