Question

How many real solutions does the equation x^7+14x^5+16x^3+30x-560=0 have?

Answer 1

What kind of approach are you looking for? A solution by hand, or by calculator?

Answer 2

by hand of course ! and i want to know how to solve this fat...shortcut maybe

Answer 3

*fast

Answer 4

x^7+14x^5+16x^3+30x-560=0 this is your problem yeah? Okay, first take 560 away on both sides, now you have x^7+14^5+16^3+30x = 560 yeah? Okay, now that you have simplified and solved the problem a little now divide both sides by 30. So you now have x^7+14x^5+16x^3+x = 18.6 repeating yeah? Since both sides share a common factor of 2, then divide both sides by 2. Now you have x^7+7x^5+8x^3+x = 9.3. You can't simplify this anymore.

Answer 5

if we divide both sides by 30 X^7 will become x^7/30 and so on for all the terms u forgot to write that...

Answer 6

The most significant coefficient is 1, not 30....
@StudiousHojea

Answer 7

As for the number of real solutions, I just graphed the equation, there are none.

Answer 8

A polynomial equation of nth degree has n solutions, no?

Answer 9

yes @aleroth but all need not be real right?

Answer 10

If I recall correctly, it has n real solutions

Answer 11

So the answer should be 7

Answer 12

nope, it has 7 possible, but none are real

Answer 13

Actually, that is not correct....for ex:even qudratic eq can have non-real roots...that is why we have...discriminant...remember @aleroth D=b^2-4ac...

Answer 14

The graph show that it never crosses the x-axis, therefore no real solutions.

Answer 15

are u sure @roadjester ?

Answer 16

Oops, copied in the equation wrong, but I see 1 real solution

Answer 17

Are you supposed to use a graph drawer for this?

Answer 18

I want to know how to solve algebraically..not through graph...

Answer 19

Well, what class is this for? So I know what I can and can't use.

Answer 20

11th

Answer 21

I mean, is this for Algebra 1, Algebra 2, Pre-Calculus, etc

Answer 22

Okay, my bad, what I meant was divide both sides by 14 so that the problem will be like this: x^7+14x^5+16x^3+30x = 40
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14 Do you see where this going? The 14 cancels out the other 14 so you are left with: x^7+x^5+16x^3+30x = 40. Then you divide both sides by 10 so that the 30 becomes 3 and the 40 becomes 4 like this. So this is how the equation will look x^7+x^5+16x^3+3x = 4. Then you divide both sides by 4 so that the 16 in the equation will be reduced to 4 so this is how the equation looks now:
x^7+X^5+4x^3+3x = 1. Then divide both sides by x. This is how the equation looks now:
x^7+x^5+4x^3+3 = 1/x. I got this because I cancelled out the x in the equation but not the ones with the exponents.

Answer 23

Does anyone know Descarter's rule of signs?

Answer 24

@StudiousHojea I have no clue what you're trying to do since you're not being consistent in your division of terms. Your x^7 term isn't divided. O.o

Answer 25

That's the thing, we can't do anything about the variables with the exponents.

Answer 26

ok thanks all...i found the answer

Answer 27

And as for descartes rule of signs, i remember learning bout it when I was in high school. Is that what you're supposed to use?

Answer 28

Oh, that's great!

Answer 29

Ok, sweet! Good luck in your studies!

Answer 30

thank u

Answer 31

Sorry roadjester, I like math but people tell me I am not very good at it so I was just telling her the stuff I learned in Algebra 1.