Question 1163225: 'What is the length of the edge of a cube if its volume could be doubled by an increase of 6 centimeters in one edge, an increase of 12 centimeters in a second edge, and a decrease of 4 centimeters in the third edge?'
I've seen the answer of this question on this website. However, I wish to understand the concepts. Is it alright if a list of concepts may be provided? I'm fine with understanding it on my own, but I don't know where to start. Thank you!
Found 3 solutions by ankor@dixie-net.com, MathTherapy, ikleyn: Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! What is the length of the edge of a cube if its volume could be doubled by an increase of 6 centimeters in one edge, an increase of 12 centimeters in a second edge, and a decrease of 4 centimeters in the third edge?'
:
Let's just see what we have, and follow it algebraically to the bitter end!
:
let s = side of the original cube
then s^3 = the vol of original cube
:
the problem:
(s+6)(s+12)(s-4) = 2s^3
Foil the last two expressions
(s+6)(s^2 - 4x + 12x - 48) = 2s^3
(s+6)(s^2 + 8x - 48) = 2s^3
old fashioned multiplication
s^3 + 14s^2 + 0s - 288 = 2s^3
s^3 - 2s^3 + 14s^2 + 0s - 288 = 0
-s^3 + 14s^2 + 0s - 288 = 0
Use synthetic division, the divisor, can't be 4, try 6
:
... ______________________
6 |-1 + 14 + 0 - 288
Works!
:
s = 6 cm is the original side length
:
Check this: 6^3 = 216 cu/cm
Using the new dimensions
12 * 18 * 2 = 432 twice the volume
Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!
What is the length of the edge of a cube if its volume could be doubled by an increase of 6 centimeters in one edge, an increase of 12 centimeters in a second edge, and a decrease of 4 centimeters in the third edge?'
I've seen the answer of this question on this website. However, I wish to understand the concepts. Is it alright if a list of concepts may be provided? I'm fine with understanding it on my own, but I don't know where to start. Thank you!
Let length of one of the edges, be L
Then original volume = L3
We then get: 
Using the Rational Root Theorem (this I prefer), or Synthetic Division, we get one of the roots of the above equation as 6, or the factor, L - 6.
Using the root, 6, a and the Rational Root Theorem again, or Synthetic Division, we get another root, 12.
Or, using the factor, L - 6 and Long Division of Polynomials, we get the resulting quadratic: , which factors to (L - 12)(L + 4),
where we see the other root, 12.
Therefore, we can say that: 
This means that the length of the edge of the cube, or
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Since you want to understand concepts, I want to share my vision.
(1) The problem is about the size of the cube edge.
So, it is about real number solutions, and we have no any reasons (bases) to assume from the very beginning
that the answer is in integer numbers.
Therefore, after you obtained this basic polynomial equation of the third degree,
you have no a reason to apply the Rational root theorem to seek the integer roots.
In such cases, it is much more productive to make a plot of the polynomial and to get an information
about the roots from the plot.
If from the plot you see that one (or several) roots are (or are expected to be) integer numbers,
then you can simply CHECK if these candidates / (these integer numbers) are really the roots.
If at least one root is integer and you just found it, then you can move further by dividing the polynomial to
an appropriate linear binomial and reduce the degree.
Good tool to make plots quickly and easily in seconds is this free of charge Internet site
www.desmos.com.
And THEN, when the roots (the integer roots, in this case) are just found, you may apply the Rational root theorem
to demonstrate to your teacher, how smart you are.
Do not think, PLEASE, that I discreditate the Rational root theorem.
It is a nice tool in Algebra.
But, as I joke from time to time, in complicated cases it works especially good, when the answer / (the solution) is known in advance.
(2) A person experienced in Algebra may drive his (or her) thoughts by different way.
He (or she) notices that the polynimial p(x) = was obtained as a product of the three linear binomials
(x+6)*(x+12)*(x-4)
Then he needs to solve a modified equation = 0.
He (or she) notices that the modified polinomial q(x) = has the coefficients that are opposed
to the coefficients of the polynomial p(x).
More precisely, he notices that the modified polynomial has opposite coefficients at the term with x^2 and the constant term,
while the coefficient at x is zero in both polynomials.
Then the person recalls the Vieta's theorem, which tells him (or her), that IT MEANS THAT the roots of the polynomial q(x)
are opposed numbers to the roots of the polynomial q(x).
It means that the roots of the equation = 0 are 6, 12 and -4.
Probably, this explanation is slightly over than the standard/regular school level; but for an advanced student it may work.
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Happy learning (!)
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