# Math Help - Factor this beast.. Tricky Question

1. ## Factor this beast.. Tricky Question

In Canada, hundreds of thousands of Cubic metres of wood are harvested each year. The function

f(x)=1135x^4-8197x^3+15868x^2-2157x+176608 0<=x<=4

models the harvested volume from 1993 to 1997

Estimate the interval when less than 185000m^3 were harvested.

So I setup the equation being <185000

I brought the 185000 over and ended up with

1135x^4-8197x^3+15868x^2-2157x-8392<0

I can't find a factor using factor theorum, maybe i'm not trying hard enough but i've spent quite a bit on it. Could anyone assist?

2. Your Teacher surely will give you an integer solution. . . . .

Given: f(x)=1135x^4-8197x^3+15868x^2-2157x+176608 0<=x<=4

1135x^4-8197x^3+15868x^2-2157x-8392<0

why not substitute those values above? Please have a try on these . . . .

1) 1135x^4-8197x^3+15868x^2-2157x-8392 < 0, @ x = 0,

2) 1135x^4-8197x^3+15868x^2-2157x-8392 < 0, @ x = 1,

3) 1135x^4-8197x^3+15868x^2-2157x-8392 < 0, @ x = 2,

4) 1135x^4-8197x^3+15868x^2-2157x-8392 < 0, @ x = 3,

5) 1135x^4-8197x^3+15868x^2-2157x-8392 < 0, @ x = 4.

surely more than one of them will satisfy your given equation . . . . please try. Below is a plot of your inequality - 1135x^4-8197x^3+15868x^2-2157x-8392<0.

3. Hello Pankie

Welcome to Math Help Forum
Originally Posted by Pankie
In Canada, hundreds of thousands of Cubic metres of wood are harvested each year. The function

f(x)=1135x^4-8197x^3+15868x^2-2157x+176608 0<=x<=4

models the harvested volume from 1993 to 1997

Estimate the interval when less than 185000m^3 were harvested.

So I setup the equation being <185000

I brought the 185000 over and ended up with

1135x^4-8197x^3+15868x^2-2157x-8392<0

I can't find a factor using factor theorum, maybe i'm not trying hard enough but i've spent quite a bit on it. Could anyone assist?
Unless you've been told to look for factors, I think that the clue is the word 'Estimate', which probably indicates that a numerical/graphical method is required. I don't know what tools you are allowed to use, but if you have a calculator or spreadsheet that will produce a graph, you'll get something like the one I've attached.

This gives the interval (approximately) $[0, 1.2] \cup [2.8,3.8]$.