What are the dimensions of a small box given a specific volume increase?

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In summary, the dimensions of the original small box are x = 5 cm, x+1 = 6 cm, and x+2 = 7 cm. The volume of the box is 120 cm^3, and the formula for volume is v = lwh. To solve for x, the factor theorem or factoring can be used, and it is important to note that negative values for x are not physically valid.
  • #1
sin_city_stunner
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1. The height,length and width of a small box are consecutive intergers with the height being the smallest of the three dimensions. If the length and width are increased by 1 cm each and the height is doubled, then the volume is increased by 120 cm^3. Find the dimensions of he original small box



2. v=lwh



3. I tried to appoach this two different ways but couldn't get either to work:

1. h = x
w = x+1
l= x+2
therefore (2x)(x+2)(x+3) = 120
2x^3 + 10x^2+12x-120 = 0
i tried to use the factor theroem to solve the equation but could not get it to work.
p{+-1,+-2,+-3...}
q{+-1,+-2}
p/q {+-1,+-1/2, +-2, +-3, +-3/2...}

I knowthat in order to solve the question, I am going to need to need to find the roots of the polynomial and therefore have to solve for x. The factor theroem indicates that if i plug in p/q into the above given equation, it will equal zero, and i can finish the question. The only problem i am having is determining which p/q value satisfies the equation. Another solution would be to factor, but i tried that and it ended up being..

2x^2(x+5) + 12(x-10) and since there are no more common factors, it cannot be taken any further. As a result, i am pretty sure i have to use the factor theorem, but i can't figure out what p/q value satisfies the equation

So then i tried (2x)(x+2)(x+3)= 120 + (x)(x+1)(x+2)
and got..
x^3+7x^2-10x-120 = 0
and tried to use factor theorem again but couldn't get it to work.


Thanks in advance for the help.
 
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  • #2
That second equation is what you need to solve. You made some sort of algebra mistake when you expanded, the answer is a nice integer
 
  • #3
sin_city_stunner said:
3. I tried to appoach this two different ways but couldn't get either to work:

1. h = x
w = x+1
l= x+2
therefore (2x)(x+2)(x+3) = 120
2x^3 + 10x^2+12x-120 = 0
i tried to use the factor theroem to solve the equation but could not get it to work.
p{+-1,+-2,+-3...}
q{+-1,+-2}

So then i tried (2x)(x+2)(x+3)= 120 + (x)(x+1)(x+2)
and got..
x^3+7x^2-10x-120 = 0
and tried to use factor theorem again but couldn't get it to work.


Thanks in advance for the help.
Your second method is correct. You have made a mistake in your last line though; it should read x^3+7x^2+10x-120 = 0.

Try and find a factor here (it shouldn't take too many tries!) Note that you do not need to try negative values, since x is a length, and a negative length is not physically valid.
 
  • #4
ok i see it now, can't believe i had negative 10x..thanks for the help
 
  • #5
sin_city_stunner said:
So then i tried (2x)(x+2)(x+3)= 120 + (x)(x+1)(x+2)

That's correct.

and got..
x^3+7x^2-10x-120 = 0

Check your algebra. The "-10x" is wrong.
 

1. What is a polynomial equation?

A polynomial equation is an algebraic equation that contains one or more terms, where each term consists of a variable raised to a non-negative integer power, multiplied by a coefficient. The highest power of the variable is called the degree of the polynomial.

2. How do you solve a polynomial equation?

To solve a polynomial equation, you must find the values of the variable that make the equation true. This can be done by using algebraic techniques such as factoring, completing the square, or using the quadratic formula for quadratic equations. Other techniques for higher-degree polynomial equations include using the rational root theorem or the synthetic division method.

3. What is the difference between real and complex solutions to a polynomial equation?

Real solutions to a polynomial equation are values of the variable that result in real numbers when substituted into the equation. Complex solutions, on the other hand, involve imaginary numbers and are typically written in the form a + bi, where a and b are real numbers and i is the imaginary unit.

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To check if a given value is a solution to a polynomial equation, you can substitute the value into the equation and see if it results in a true statement. Alternatively, you can plug in the value into a graphing calculator or software and see if it lies on the graph of the equation.

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