The distance, d, in kilometers traveled by a place after t hours can be represented by d(t) = -4t^3 + 40t^2 + 500t, where 0 <= t <= 10. How long does the place take to fly 4088km.

I can't seem to get the answer. This is my approach.

1) Sub in distance

d(t) = -4t^3 + 40t^2 + 500t
4088 = -4t^3 + 40t^2 + 500t

2) Make equation equal to 0 and factor greatest common factor.

0 = -4t^3 + 40t^2 + 500t - 4088
0 = -4t (t^2 -10t -125 + 1022)
0 = -4t (t^2 -10t + 897)

3) No clue, the discriminate of the quadratic equation gives you a negative root which you obviously can't do.


The answer is a real root of 7 hours. Any help would be appreciated.

0 = -4t^3 + 40t^2 + 500t - 4088
0 = -4t (t^2 -10t -125 + 1022)
0 = -4t (t^2 -10t + 897)

This part is mistaken. You can't factor -4t from 4088.

oh man. ty

-4 (-7+t) (-146-3 t+t^2)

since I'm too lazy to factor myself, I plugged it in wolfram and got this.

You can solve it by hand by realizing that if you have a root of 7, you can probably factor (x-7) term from the polynomial. THen you use polynomial division.

Edit: Please people, when you have tedious algebra to do plug it in Wolfram! Saves you time! End of PSA

This post was edited by zackill4 on Sep 12 2013 07:54pm

oh we are learning it different. we simply plug in factors of the constant 1022 into the equation until we get a value of 0. in this case plugging in 7 worked therefore our factor is (x - 7).

I have another question I was stuck at for a while.
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The passenger section of a train has width (2x-7), length (2x+3), height (x-2), with all dimensions in meters. Solve a polynomial equation to determine the dimensions of the section if the volume is 117m cubed.

This is my approach.

1) Sub in knowns.

V = lwh
117 = (2x-7) (2x+3) (x-2)

2) Expand and set equation equal to 0

117 = (2x-7) (2x+3) (x-2)
117 = (4x^2 -8x - 21) (x-2)
117 = (4x^3 -8x^2 - 21x - 8x^2 - 16x +42)
117 = (4x^3 - 16x^2 - 5x + 42)
0 = 4x^3 - 16x^2 - 5x + 42 - 117
0 = 4x^3 - 16x^2 - 5x -75

3) Now I have no clue what to do.

Yes, well you can guess the zeroes like you said. You just need to guess one and then you can factor.
No matter what "approach" you take, it'd be very difficult to avoid polynomial division. I don't think you're method is any different. The guessing was just to find the first 0.

Anyway, just plug this into Wolfram alpha again to get the answer lol... I don't get what's so hard about that.


Answer is
(-5+x) (15+4 x+4 x^2)


Here's the link
http://www.wolframalpha.com/

Edit: You plug in that last expression you derived into wolfram, or any expression set to 0. It will factor it for you.

This post was edited by zackill4 on Sep 12 2013 08:39pm

if you don't want to use wolfram or similar, approximate it manually using the last formula- or have a closer look at 117 = (2x-7) (2x+3) (x-2)
for the right to be positive x needs to be 4 or larger, but 4 won't work because (4-2)=2 is even and 117=3.3.13 is odd
go one up to 5 and you get 117=(10-7)(10+3)(5-2)=3.13.3 - hurray we got it

postscript:





This post was edited by brmv on Sep 12 2013 08:42pm

I was under the impression he wanted to ensure he got all the zeroes. Of course you can just guess one zero...

Under normal circumstances, that would not be sufficient to solve the problem. If these are all constructed so there's only one zero then it's a way of doing it

This post was edited by zackill4 on Sep 12 2013 08:45pm

it's not guessing but applying certain principles
but why do it the easy way if you can make it more complicated?

Well, as I said, it works for very simple cases but it won't work for all cases. He was saying that the approach I was using was not "relevant" to what he's learning, but it really is. That's all i'm saying. For most cases where the problem isn't written out nicely you would have to use polynomial division. That is the general way of solving these problems.

This post was edited by zackill4 on Sep 12 2013 08:51pm