How Do You Factor and Solve Polynomial Equations to Find Box Dimensions?

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Discussion Overview

The discussion revolves around factoring and solving the polynomial equation \(x^3-7x^2+14x-8\) to determine the dimensions of a box that yields a volume of 12 cm³. Participants explore various methods of factoring and the implications of their findings on the dimensions of the box.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests moving 12 to the left side to form the equation \(x^3-7x^2+14x-20=0\) but expresses confusion about the next steps.
  • Another participant advises dividing the polynomial by the factor corresponding to the zero found (5) to reduce the polynomial's degree and find other roots, noting that the remaining roots may be imaginary.
  • Some participants argue against factoring the polynomial with the 12 included, insisting that it should be factored completely first into linear factors before considering the volume condition.
  • A later reply confirms that \(x=5\) is a solution and states that the other factor obtained from division has no real zeros, implying \(x=5\) is the only real solution.
  • One participant hints that \(x=1\) also satisfies the original polynomial equation, suggesting it could be relevant for finding dimensions.

Areas of Agreement / Disagreement

Participants exhibit disagreement on the approach to solving the problem, particularly regarding whether to factor with the volume condition included or to first find the linear factors of the polynomial. There is no consensus on the correct method to proceed.

Contextual Notes

Some participants express uncertainty about the correctness of the original problem statement, suggesting potential errors in the formulation. The discussion also highlights the challenge of finding real dimensions given the polynomial's roots.

thomasrules
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A box has dimensions that are linear factors of (x^3-7x^2+14x-8) cubic centimetres. What dimensions give a volume of 12 cm cubed?

I started it off by moving 12 to the left side giving the equation:

x^3-7x^2+14x-20=0

then from then I don't know...If I substitute 5, the equation is satisfied and is a factor but from there it doesn't work. Plus 5 won't really make it 12 so...i'm lost
 
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Once you know a zero of a polynomial, you can divide by the corresponding factor to reduce the degree of the polynomial. In this case, the zero you found is 5, so we will divide the polynomial by (x - 5). You can do this either with long or synthetic division.

Once you do that, you will have a reduced polynomial (a quadratic). You can then solve this to find the other roots. Having 5 as a factor does not mean you cannot get 12, since the other factors could be fractions or even irrational numbers.

However, now that I look at the problem closer, I see that, in fact, the other two roots will be imaginary. So, I suspect that either you wrote the problem incorrectly (a negative sign in the wrong place) or you were given a problem with to tangible solution.
 
Last edited:
thomasrules said:
I started it off by moving 12 to the left side

NO.

Factor [itex]x^3-7x^2+14x-8[/itex] completely into (x-a)(x-b)(x-c). Then find some combination of a, b, and c (with possible repitition) that multiply together to give 12. Don't try to factor it with the 12, that's not a part of the question.
 
Last edited:
CRGreathouse said:
NO.

Factor [itex]x^3-7x^2+14x-8[/itex] completely into (x-a)(x-b)(x-c). Then find some combination of a, b, and c (with possible repitition) that multiply together to give 12. Don't try to factor it with the 12, that's not a part of the question.
yes it is :biggrin:

you have to move it to the left side to get -20 then f(5)=0
 
You have correctly determined that a solution to [itex]x^3-7x^2+14x-8= 12[/itex] is x= 5. In fact, if you divide [itex]x^3-7x^2+14x-20[/itex] by x-5, you get that the other factor is [itex]x^2- 2x+ 4[/itex] which has no real zeroes so x= 5 is the only solution.

Now, CRGreathouse's point is that you need to factor [itex]x^3-7x^2+14x-8[/itex] into linear factors. Setting x= 5 in each of those linear factors gives the dimensions. (Hint: x= 1 satisfies [itex]x^3-7x^2+14x-8= 0[/itex].)
 

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