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

[thomasrules]

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

 

[BSMSMSTMSPHD]

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.

 

[CRGreathouse]

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.

 

[thomasrules]

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

you have to move it to the left side to get -20 then f(5)=0

 

[HallsofIvy]

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].)