SOLUTION: For the problem: R(x)= -x^5+5x^3-4x I got the y-int to be (0,0) by setting x=0 And then I got the x-int to be(0,0),(2,0),(-2,0),(1,0),(-1,0) by setting the entire problem to

Question 950146: For the problem: R(x)= -x^5+5x^3-4x
I got the y-int to be (0,0) by setting x=0
And then I got the x-int to be(0,0),(2,0),(-2,0),(1,0),(-1,0) by setting the entire problem to 0.
like -x(x^4-5x^2+4)
(x^2-4)(x^2-1)=0
But I checked my graphing calculator and it states that only 1,0 is present and (-0.8,0). Why isn't -1,-2,2 present and where does -0.8 come from?

THANKS!

Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!

, the second polynomial factor being in quadratic form.

Roots or Zeros are :
0, -2, 2, -1, 1.

NO such root of -0.8.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
For the problem: R(x)= -x^5+5x^3-4x
I got the y-int to be (0,0) by setting x=0
And then I got the x-int to be(0,0),(2,0),(-2,0),(1,0),(-1,0) by setting the entire problem to 0.
like -x(x^4-5x^2+4)
(x^2-4)(x^2-1)=0
But I checked my graphing calculator and it states that only 1,0 is present and (-0.8,0). Why isn't -1,-2,2 present and where does -0.8 come from?
------
You have made some error with your graphing calculator.
Find the intersection of R(x) = y = 0 and you will
see the x-intercepts.
Cheers,
Stan H.
==========

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