anonymous
  • anonymous
for the given function determine consecutive values of x between which each real zero is located f(x)=-2x^4-4x^3-2x^+3x+8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Did you try to factor the expression?
anonymous
  • anonymous
I have no idea how to do this problem at all
anonymous
  • anonymous
This might be a start to factor it: $$-2x^3(x+2)-2x^2+3x+8$$

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More answers

whpalmer4
  • whpalmer4
Can you characterize the roots you'll get?
anonymous
  • anonymous
The possible answers are: A. there is a zero between x=1, x=2,x=-1,x=-2 B.zero between x=2, x=3,x=1,x=0,x=-2,x=-3 C.zero between x=-1,x=-2 D. zero between x=1 and x=2
whpalmer4
  • whpalmer4
Oh, that takes all the challenge out of it :-)
whpalmer4
  • whpalmer4
Do you know calculus?
anonymous
  • anonymous
no im in algebra
whpalmer4
  • whpalmer4
Okay, do you know about Descartes' Rule of SIgns?
anonymous
  • anonymous
Just start to plug in those values into the equation, in the order: -2, -1, 0, 1, 2, 3 If one value is positive and the following is negative you have a zero between them.
anonymous
  • anonymous
I don't know how to do that? I'm only in algebra one and I need this done in ten minutes to turn in :(
anonymous
  • anonymous
anyone?
anonymous
  • anonymous
Calculate $$-2x^4-4x^3-2x^2+3x+8$$ When x= -2, then x=-1 and the other values. Example: if x=0 the answer is $$-2\times0^4 - -4\times0^3 - 0 + 8= 8$$.
anonymous
  • anonymous
so would it be a, b, c or d
anonymous
  • anonymous
Its not B.
anonymous
  • anonymous
If x=1 the expression turns +3
anonymous
  • anonymous
There is definetely a zero between 1 and 2. So its either A or D.
whpalmer4
  • whpalmer4
With the equation written in descending order of powers of x, count the sign changes. That number, possibly minus a multiple of 2, is the number of positive roots. By my calculation, the answer is 1 positive root. Next, rewrite the equation substituting (-x) wherever you see x, and simplifying. For this equation, that gives -2x^4+4x^3-2x^2-3x+8. Count the sign changes again. This number, possibly reduced by a multiple of 2, gives the number of negative real roots. Here I make it to be 3 or 1. You'll always have as many total roots as the highest power exponent in the polynomial, so here there will be 4. Any not accounted for above are complex, and come in conjugate pairs. I think your best bet here is to do as @DanielM_113 suggests. I'll do a few to help: \[-2(-2)^4-4(-2)^3-2(-2)^2+3(-2)+8=-32-32-8-6+8 = -70\]\[-2(-1)^4-4(-1)^3-2(-1)^2+3(-1)+8=-2+4-2-3+8=5\]So there must be a zero between x = -2 and x = -1 You know that the value is positive at x = 0, so no root between x = -1 and x = 0. On to x=1: \[-2(1)^4-4(1)^3-2(1)^2+3(1)+8 = -2-4-2+3+8=3\]still positive, so no root between x=0 and x=1 \[-2(2)^4-4(2)^3-2(2)^2+3(2)+8 = -32-32-8+5+8 = -57\]so we have a zero between x=1 and x=2 \[-2(3)^4-4(3)^3-2(3)^2+3(3)+8 = -162-108-36+9+8 < 0\]so no zero between x=2 and x = 3
anonymous
  • anonymous
But that wouldn't fit any of the possible answers?
whpalmer4
  • whpalmer4
If you've chosen an answer, I can show you a graph of the function in the relevant area...
anonymous
  • anonymous
And so the answer must be A.
whpalmer4
  • whpalmer4
we had a zero between x=-2, x=-1, and another between x = 1, x=2
whpalmer4
  • whpalmer4
1 Attachment
anonymous
  • anonymous
I didnt think of doing a graph online, that is deviously genius.
whpalmer4
  • whpalmer4
As I was saying about Descartes' Rule of Signs (which it sounds like you probably haven't had yet), we had 1 positive root, and either 1 or 3 negative roots. Turns out to be 1 negative root. The other two are complex numbers. The full set of solutions is: \[\{\{x\to -0.709177-1.25956 i\},\{x\to -0.709177+1.25956 i\},\]\[\{x\to -1.70468\},\{x\to 1.12303\}\}\]
whpalmer4
  • whpalmer4
Though the symbolic version is quite striking :-)
anonymous
  • anonymous
Thanks for explaining the Decartes Rule of Signs, its been some time I saw that and didnt remember it.
whpalmer4
  • whpalmer4
I didn't know about it either until someone asked a question about it on OpenStudy. I had to learn quickly in order to be able to answer :-)

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