Question 1176836
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f = first number
s = second number
t = third number
h = fourth number
the letter f was already taken, so I picked the last letter in "fourth"
All of these values are some rational number in the form a/b, where a,b are integers and b is nonzero.


We are given these four facts<ol><li>The product of the first, third, and fourth numbers is -6.</li><li>The second number is 3 less than the first number</li><li>The third is 2 less than the second</li><li>The fourth is 2 less than the third</li></ol>
Those facts lead to the corresponding equations (in the order shown)
f*t*h = -6
s = f-3
t = s-2
h = t-2


I'll refer to them as equations (1) through (4)


Let's start with equation (4)
h = t-2
Now replace t with s-2; this is valid due to equation (3)
h = t-2
h = (t)-2
h = (s-2)-2
h = s-4
This tells us the fourth value (h) is 4 less than the second value (variable s).


We can add 4 to both sides getting
s = h+4
Which then can be plugged into equation (2)
s = f-3
h+4 = f-3
h = f-3-4
h = f-7
The fourth value is 7 less than the first value


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We have these equations all with h to start off with
h = f-7
h = s-4
h = t-2
I sorted them so that the one with 'f' goes first, then 's' second, and 't' third


For each of those three equations shown, isolate the other variable. Doing so leads to these three results
f = h+7
s = h+4
t = h+2


What this allows us to do is replace the f and t values in equation (1) with something in terms of h. That way we can set up a single variable equation in which we can solve for h


f*t*h = -6
(f)*(t)*h = -6
(h+7)*(h+2)*h = -6
(h^2+9h+14)*h = -6 ..... FOIL rule
h^3+9h^2+14h = -6 .... distribute
h^3+9h^2+14h+6 = 0


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Recall that each f,s,t,h are all rational numbers. 
That must mean the last equation we arrived at has at least one rational root; or else, all the roots of that cubic equation are irrational and that contradicts the instructions. 


The rational root theorem says that we divide the factors of the last term (6) over the factors of the leading coefficient (1)
List out the plus and minus versions of each ratio


Luckily that list isn't too big
If the numerator is 6, then we have the ratios: -6/1, 6/1
That leads to -6 and 6 respectively.


If the numerator is 3, then we get -3/1 = -3 and 3/1 = 3
If the numerator is 2, then we get -2/1 = -2 and 2/1 = 2
If the numerator is 1, then we get -1/1 = -1 and 1/1 = 1


The possible rational roots are: 
-6,6
-3,3
-2,2
-1,1


As you can see, all we've done really is listed all the factors of 6 (positive and negative versions). This is due to the leading coefficient being 1.


From here, we plug each of those 8 values into the cubic equation
h^3+9h^2+14h+6 = 0
to see if the left side becomes 0 or not


Let's try out h = -6
h^3+9h^2+14h+6 = (-6)^3+9*(-6)^2+14*(-6)+6 = 30
That doesn't work. 


Now try h = 6
h^3+9h^2+14h+6 = (6)^3+9*(6)^2+14*(6)+6 = 630
That doesn't work either.


Through trial and error, you should find that only h = -1 works since,
h^3+9h^2+14h+6 = (-1)^3+9*(-1)^2+14*(-1)+6 = 0


So h = -1 is a rational root solution to h^3+9h^2+14h+6 = 0


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With that in mind, we can find the other values
f = h+7 = -1+7 = 6
s = h+4 = -1+4 = 3
t = h+2 = -1+2 = 1


To summarize, we have these values:
f = 6
s = 3
t = 1
h = -1
as the first through fourth values in that order.


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Check:


We'll plug in the values we found for each equation formed at the top of this page.


f*t*h = -6
6*1*(-1) = -6
-6 = -6
First equation works out


s = f-3
3 = 6-3
3 = 3
That works out also


t = s-2
1 = 3-2
1 = 1
Works too


h = t-2
-1 = 1-2
-1 = -1
We can see that all four original equations work; therefore, the answers have been confirmed.


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Answers:
first = 6
second = 3
third = 1
fourth = -1
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