Question 1181482: The volume of a rectangular box is modelled by the polynomial V(x)=x^3-14x^2+61x, and the volume of the box is 84 cm^3. Give the possible values of x
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850)   (Show Source):
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
The cubic equation to be solved is this:
Tutor @MathLover1 likes to show factoring cubic polynomials by splitting the terms into parts. For example, she starts this one by splitting the -14x^2 into -7x^2 and -7x^2; then she splits other terms in a convenient way that eventually allows you to use grouping to factor the whole cubic into three linear factors.
The trouble with that method is that the factorization only works if you already know the factorization.
In this example, since the roots turn out to be 3, 4, and 7, splitting the -14x^2 into -7x^2 and -7x^2 works, because one of the roots is 7.
Since the other roots are 3 and 4, you could also do the factoring by her method by splitting the -14x^ into -4x^2 and -10x^2, or into -3x^2 and -11x^2.
But the method wouldn't get you anywhere if you started by splitting the -14x^2 into -6x^2 and -8x^2, because neither 6 nor 8 is a root.
So her response teaches the student nothing....
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The standard method for factoring cubic polynomials is to use the rational roots theorem to find the possible rational roots and then use substitution, or synthetic division, or long division to find the first actual root. Then dividing out the linear factor corresponding to that root reduces the cubic to a quadratic, which can be solved either by factoring or using the quadratic formula.
The possible rational roots are the integer factors of 84: plus or minus 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.
Using substitution or synthetic division to identify the first root will still involve a lot of trial and error.
Note that the method used by the other tutor also involves trial and error; it will lead to the solution only if you split the -14x^2 term into two parts in certain ways.
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So here is what I think is the fastest way to do the factorization using formal math and a little bit of trial and error. Yes, it still uses trial and error; but the work involved is much easier than using substitution or division to identify the roots. In fact, this method makes the work MUCH easier by identifying all three roots at the same time.
Descartes' rule of signs tells us that, because of the alternating signs of the terms, the roots are all positive. So if we call the three roots p, q, and r, then we know the product pqr is 84. So the three roots are three numbers from the above list of the integer factors of 84 whose product is 84.
There are several combinations of three of those factors of 84 whose product is 84. To narrow the number of possibilities down to one, we use two other facts: the coefficients of the quadratic and linear terms tell us that (1) the sum of the roots p+q+r is 14; and the sum pq+pr+qr is 61.
There are several combinations (allowing repetition) of three integer factors of 84 that give the correct sum p+q+r=14, but only one of them gives pq+pr+qr=61 -- 3, 4, and 7.
So the factorization of the cubic polynomial is
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So here is a recap of what is probably the fastest way to find the factorization algebraically....
(1) make a list of the positive integer factors of the constant term, 84 (the negative sign of the constant can be ignored);
(2) find three numbers, possibly with repetition, from that list for which
(a) pqr=84;
(b) p+q+r=14; and
(c) pq+pr+qr=61
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