Question

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

Answer 1

Did you try to factor the expression?

Answer 2

I have no idea how to do this problem at all

Answer 3

This might be a start to factor it:
$$-2x^3(x+2)-2x^2+3x+8$$

Answer 4

Can you characterize the roots you'll get?

Answer 5

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

Answer 6

Oh, that takes all the challenge out of it :-)

Answer 7

Do you know calculus?

Answer 8

no im in algebra

Answer 9

Okay, do you know about Descartes' Rule of SIgns?

Answer 10

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.

Answer 11

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 :(

Answer 12

anyone?

Answer 13

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 $$-2\times0^4 - -4\times0^3 - 0 + 8= 8$$.

Answer 14

so would it be a, b, c or d

Answer 15

Its not B.

Answer 16

If x=1 the expression turns +3

Answer 17

There is definetely a zero between 1 and 2. So its either A or D.

Answer 18

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

Answer 19

But that wouldn't fit any of the possible answers?

Answer 20

If you've chosen an answer, I can show you a graph of the function in the relevant area...

Answer 21

And so the answer must be A.

Answer 22

we had a zero between x=-2, x=-1, and another between x = 1, x=2

Answer 23

I didnt think of doing a graph online, that is deviously genius.

Answer 24

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:
\[\{\{x\to -0.709177-1.25956 i\},\{x\to -0.709177+1.25956 i\},\]\[\{x\to -1.70468\},\{x\to 1.12303\}\}\]

Answer 25

Though the symbolic version is quite striking :-)

Answer 26

Thanks for explaining the Decartes Rule of Signs, its been some time I saw that and didnt remember it.

Answer 27

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 :-)