anonymous
  • anonymous
Find the zeros of the parabola f(x) = x^3 - 4x^2 - 25x + 100 algebraically.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
x^3 is not a parabola ...
amistre64
  • amistre64
let x = u-4/3 perhaps for a subsitution .... unless your good at factoring
amistre64
  • amistre64
might need to bea u+4/3

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wolf1728
  • wolf1728
|dw:1378843200116:dw| Graph would look like this and as amistre64 said it's not a parabola. Then again, I don't think you need a graph for solving the equation.
amistre64
  • amistre64
x^3 - 4x^2 - 25x + 100 (x^3 - 4x^2) - (25x - 100) seems factorable to me
anonymous
  • anonymous
sorry, it's a polynomial, a cubic function to be more accurate. It has three 'zero' values. I know that the zeros are at -5, 4, and 5 because I looked at a graphing calculator; I have no Idea how to get there though. (x^3 - 4x^2) - (25x - 100) x^2 (x - 4) - 25 (x - 4) <- pretty sure I'm not doing this right At least I'm winding up with 4 as one of my zeros... even if it's two of them... How do I proceed?
anonymous
  • anonymous
since you know that (x-4) is one of the solution, try long division.
anonymous
  • anonymous
or, try this: \[x^2 (x-4) - 25x + 100\]
anonymous
  • anonymous
with long division I wound up with x^2 - 25, no remainder
anonymous
  • anonymous
\[x^2 ( x - 4 ) + 25 ( x + 4 )\] okay, so there's two of the values.
anonymous
  • anonymous
which gives you (x+5)(x-5) since it's a perfect square...so your solutions are (x+5), (x-5), and (x-4)
anonymous
  • anonymous
from the long division...
anonymous
  • anonymous
so if the last value is a perfect square, you can just simplify it down? where are you getting the x in (x-5) from?
amistre64
  • amistre64
you did fine factoring, but you need to undistribute (factor again really) the (x-4) \[x^2 ( x - 4 ) - 25 ( x - 4 )\] \[(x-4)~(x^2 - 25)\] from here we should be bale to see that when x=4, then x-4 goes to zero and when x = 5 or -5, then x^2-25 goes to zero

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