# word problem using polynomials

• Dec 16th 2009, 11:23 AM
extraordinarymachine
word problem using polynomials
hi i need help with this question, i got some of it...

The volume of a cardboard box is $\displaystyle x^3 - 8x^2 + 17x - 10$ Factor this polynomial.

i got (x-1)(x-5)(x-2)

if these factors represent the length, width, and height, find the demensions of a box that has a volume of $\displaystyle 20cm^3$

i got ...
$\displaystyle x^3 - 8x^2 + 17x - 10 = 20$
$\displaystyle x^3 - 8x^2 + 17x - 30 = 0$

i then found a factor of (x-6), and after putting that through senthetic division i was left with $\displaystyle x^2 - 2x + 5$, but this dosent really factor, so i think i've done something wrong...help please
• Dec 16th 2009, 11:30 AM
earboth
Quote:

Originally Posted by extraordinarymachine
hi i need help with this question, i got some of it...

The volume of a cardboard box is $\displaystyle x^3 - 8x^2 + 17x + 10$ Factor this polynomial.

i got (x-1)(x-5)(x-2)<<< = x^3-8x^2+17x-10

if these factors represent the length, width, and height, find the demensions of a box that has a volume of $\displaystyle 20cm^3$

i got ...
$\displaystyle x^3 - 8x^2 + 17x + 10 = 20$
$\displaystyle x^3 - 8x^2 + 17x\bold{\color{red} - 10} = 0$

i then found a factor of (x-6), and after putting that through senthetic division i was left with $\displaystyle x^2 - 2x + 5$, but this dosent really factor, so i think i've done something wrong...help please

...
• Dec 16th 2009, 12:01 PM
extraordinarymachine
Quote:

Originally Posted by earboth
.The volume of a cardboard box is http://www.mathhelpforum.com/math-he...ec2a950d-1.gif Factor this polynomial.

i got (x-1)(x-5)(x-2)<<< = x^3-8x^2+17x-10

if these factors represent the length, width, and height, find the demensions of a box that has a volume of http://www.mathhelpforum.com/math-he...be79b41f-1.gif

i got ...
http://www.mathhelpforum.com/math-he...7a441c43-1.gif
http://www.mathhelpforum.com/math-he...f475dc27-1.gif..

i'm not sure what you did here, or what your trying to say.
the question was actually $\displaystyle x^3 - 8x^2 + 17x - 10$
• Dec 16th 2009, 12:09 PM
earboth
Quote:

Originally Posted by extraordinarymachine
i'm not sure what you did here, or what your trying to say.
the question was actually $\displaystyle x^3 - 8x^2 + 17x - 10$

I noticed that you have corrected the typo in your first post.

Now all your calculations are OK.

You can't factor $\displaystyle x^2-2x+5$ because this term is greater zero for all $\displaystyle x \in \mathbb{R}$.
• Dec 16th 2009, 12:15 PM
extraordinarymachine
and here i thought i was wrong lol
• Dec 16th 2009, 10:12 PM
earboth
Quote:

Originally Posted by extraordinarymachine
and here i thought i was wrong lol

To finish the original question:

You certainly have noticed that you get only one solution: x = 6.

Thus the demensions of the box are 5 * 1 * 4 which yields indeed 20.