volume polynomial word problem?

if the square is 18 by 18 and you're going to be cutting squares out of the corners then you can imagine the three sides of your new box that will be made. And I'm assuming that you meant 432 cubic cm or that the square is 18 inches on each side

If each square is x by x then x will be the height of the new box and 18 - x - x = 18 - 2x will be the length of the sides (2x b/c you are cutting off a length of x from each corner so each side loses x inches twice)

So the volume of this new box is

x(18 - x)(18 - x) = x^3 - 36x^2 + 324x = 432 => x^3 - 36x^2 + 324x - 432 = 0

If you graph y = x^3 - 36x^2 + 324x - 432 and see where it intersects the x-axis you will have your answer. The key though is that x can't be so large that the corners you cut off will end up cutting off more tin then you actually have.

There will be three zeros, but only one of them will actually work for your problem. Hope this helps

3 and 5 ft longer. the only ingredient which you ought to rigidity approximately in this issue is factoring the hot volume equation. it quite is the type you do it: ingredient an x from each and each term. x remains representing the dimensions of an component of the unique cube. something it quite is accomplished to it extremely is going to be further or subtracted from this x. for that reason, the subject says that Hector further to the size. x(x^2+8x+15) Now, when you consider that Hector in easy terms enlarged 2 dimensions (the backside of the container), once you ingredient the quadratic in parentheses, you will get the extra advantageous lengths of the two factors. it quite is assuming which you will set the above = 0. x^2+8x+15 (x+3)(x+5) So including the x that we pulled out interior the commencing up, x=0.. This is clever - in easy terms 2 factors have been altered, and that they have got been altered via including 3 and 5 ft to them. The 0.33 area replaced into no longer altered, so its result's 0 ft of adjusting.