Question

"Solving a Polynomial Equation"
x^3+64=0

So far, I have:
(x)^3+64^(3)
(x+64)(x^2-64x+64^2)
(x+64)(x^2-64x+4096)
x=-64 (?)

According to the textbook example, I'm supposed to use the Quadratic formula next, which I don't understand how.

Answer 1

ok,
x^3 = -64
ok?

Answer 2

ok?

Answer 3

I don't understand at all.

I know to use the "difference" of cubes" formula, \[a ^{3}+b ^{3}=(a+b)(a ^{2}-ab+b ^{2})\] to simplify.
Which I did. And then simplify. And then that makes (x+64) a factor and -64 a root (I think?).

But then, you have to use the Quadratic formula. And then I got to \[x=\frac{ -(-64)\sqrt{(-64)^{2}-4(x)(4096)} }{ 2a }\]

Answer 4

And then \[x= \frac{ 65\sqrt{-12288} }{ 2x }\] And then I don't know. Probably got that wrong. Hate that formula.

Answer 5

OH!It isn't necessary! x = -4 :)

Answer 6

(-4)^3 = -64 :)

Answer 7

got it?

Answer 8

@wealdooo

Answer 9

Oh, I know what you mean but I did that the first time I did the assignment and the instructor marked it incorrect.

Answer 10

did he say why?
It's quite true! :) i'm sure :)

Answer 11

Well, see, I think it's cause there's a big difference between an equation and a polynomial equation. So if it was solving an equation, it'd be correct. But they want us to solve it as a polynomial (Algebra 2, by the way).

Answer 12

I'm sure!but I will ask...
@DebbieG ...

Answer 13

Weird, I'm not familiar with that distinction, but ooookkkk.... let's run with that.

so you have:

\(\Large x ^{3}+4 ^{3}=(x+4)(x ^{2}-4x+4) ^{2})=(x+4)(x^2-4x+16)\)

Clearly x=-4 is a solution from the first factor.

The trinomial factor only has complex solutions... you could give those too, if you are looking for all real or complex solutions; but if only solving over the reals, then x=-4 is it.

Answer 14

Hmm, actually, I guess I see how THAT explains the distinction. If you just apply the odd root property, you ONLY get the real solutions, not the complex ones.

Answer 15

-4 isn't the only cube root of -64

Answer 16

If you do it the other way, and take all of the cube roots of 64, you will get all of the answers.

Answer 17

Actually, -4 is the only cube root of -64

and 4 is the only cube root of 64.

Answer 18

\[ x^3 = 64e^{\pi i}\]
\[ re^{ i 3\theta} = 64 e^{\pi i} \]

Answer 19

Solve this complex equation to get all of the roots of 64

Answer 20

-64, sorry

Answer 21

Also, remember to take the angle mod 2pi

Answer 22

also, I made a typo, it should be:
\[r^3e^{i3\theta} = 64e^{i\pi}\]

Answer 23

It is obvious that:
\[r = 4\]
All we have left to do is find theta

Answer 24

\[3\theta = \pi + 2\pi n : n \in Z\]

Answer 25

This yields the solutions:
\[\theta = \pi, \frac{\pi}{3}, \frac{-\pi}{3}\]

Answer 26

All other solutions are congruent to these mod 2pi

Answer 27

Therefore, our solutions are:
\[4e^{i\pi}, 4e^{i\pi / 3}, 4e^{-i \pi / 3}\]

Answer 28

Rewriting these, we get:
\[-4, 2-2i\sqrt3 , 2+ 2i\sqrt 3\]

Answer 29

Which are the solutions we would get by using the sum of cubes formula.

Answer 30

I'm sorry, but that's even more confusing than my textbook. :/

Answer 31

Yeah, it should be

Answer 32

This is much more advanced than algebra 2, and you don't need to know this

Answer 33

I was just showing, that, if you take complex roots as well, either way works

Answer 34

There is, in fact, no distinction between a "polynomial equation" and a "regular equation"

Answer 35

Okay, well this is what my textbook is telling me to do (attatchment) But with x^3+64=0

Answer 36

That would probably be the easiest way to do it.

Answer 37

That's just what I said above. :) lol

Answer 38

That's my problem! I don't understand the Quadratic formula, that's why I made this post. I'm just asking how to do the quadratic part. I got to \[(x+64)(x ^{2}-64x+4096)\]
And I set up the quadratic part as \[x= \frac{-(-64)\sqrt{(-64)^{2}-4(x)(4096)} }{ 2x }\] and then got \[x= \frac{ 64\sqrt{-12288} }{ 2x }\] If that's even a tad bit right, I don't understand where to go from there. I do online schooling so it's really important that I understand this.

Answer 39

ahhhhh, no no no. You have done the factoring wrong. No wonder you are confused!

Answer 40

You did the quadratic formula right, but you didn't factor the polynomial correctly.

Answer 41

I did it above, although I see I had an exponent in the wrong place.

\(\Large x ^{3}+4 ^{3}=(x+4)(x ^{2}-4x+4^2)=(x+4)(x^2-4x+16)\)

Answer 42

b=4. NOT b=64.

Answer 43

Oh oh oh, okay! I see now. I see what I did wrong there

Answer 44

So the quadratic changes to \[x= \frac{ -(-4)\pm \sqrt{(4)^{2}-4(x)(16)} }{ 2x }\] ? \[x=\frac{ 4\pm \sqrt{80} }{ 2x }\]
?

Answer 45

oops, 80x

Answer 46

??? the quadratic formula expression shouldn't have an x in it. The QF is a formula that SOLVES for x.

Answer 47

I have no clue then.
I just know, according to the textbook explanation, to use (x^2-4x+16) part of the equation for the quadratic formula

Answer 48

If a polynomial is of the form:
\[ax^2 + bx + c \]
its roots are given by:
\[\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 49

Yes, but you are using the QF wrong.

If you have \(ax^2+bx+c=0\) then you compute the solutions with the QF as:

\(\Large x= \dfrac{ -b\pm \sqrt{b^{2}-4ac} }{ 2a }\)

Answer 50

You put in "x" where you need to put in "a".

Answer 51

You aren't as far off as you think.... don't be discouraged. Just need to fine-tune that QF part. :)

You have \(\Large x^2-4x+16=0\)

So a=?
You have b=-4 and c=16 correct.

Answer 52

so is a = 1?

Answer 53

exactly :)

Answer 54

so is a = 1?

Answer 55

\[x=\frac{ -(-4)\pm \sqrt{(4)^{2}-4(1)(16)} }{ 2(1) }\]
\[x=\frac{ 4\pm \sqrt{80} }{ 2 }\]

Answer 56

under the radical.... 16 - 64 = ?

Then you'll be able to simplify the radical part.

Answer 57

(remember, we are expecting these to be complex solutions)

Answer 58

Oops. -48 not 80

Answer 59

right now you can simplify that radical....

Answer 60

In the textbook, the next step uses i
Example: \[\frac{ 3\pm \sqrt{-27} }{ 18 }\]
\[=\frac{ 3\pm3i \sqrt{3} }{ 18 }\]
\[=\frac{ 1\pm i \sqrt{3} }{ 6 }\]

Answer 61

So do I do that. And if so, with 3 or 4?

Answer 62

or would it be 2?

Answer 63

2i

Answer 64

I don't know what you mean, "with 3 or 4".... just simplify the expression you have.

Start with the radical. \(\sqrt{-48}\) = ??

Answer 65

got -6.93

Answer 66

hm. weeeeird. It's kinda like you used a calculator or something. ;)

And entered \(\sqrt{48}\), took the result, and slapped a negative sign on that.

Uh.uh. no no no. :)

SIMPLIFY the RADICAL.

Answer 67

I don't know what a radical is.....

Answer 68

Factor what's under the radical sign:

\(\sqrt{-48}=\sqrt{-1\cdot 16 \cdot3}\)

The radical is the square root sign.

Answer 69

Now anything under the radical that is a perfect square, the square root comes outside, right?

e.g.
\(\sqrt{18}=\sqrt{2\cdot 9}=\sqrt{2}\sqrt{9}=3\sqrt{2}\)
and
\(\sqrt{-18}=\sqrt{-1\cdot 2\cdot 9}=\sqrt{-1}\sqrt{2}\sqrt{9}=3i\sqrt{2}\)

Now you do that, for \(\sqrt{-48}\)

Answer 70

I got \[\frac{ 2\pm i \sqrt{7} }{ 1 }\] as root

Answer 71

You are not going one step at a time. That's not correct, but I need you to stay with me, one step at a time, so see where your error is. I think you are having trouble simplifying the square root.

You have:
\(\sqrt{-48}=\sqrt{-1\cdot 16 \cdot3}\)

How does that simplify?

Answer 72

I did do one step at a time. I did it by the books steps though.

Answer 73

I don't know what you mean, "by the book's steps". I'm sure the book is not telling you anything different than I am; it's just that the example in the book has different numbers in it.

Can you show me, step by step, how you simplified

\(\sqrt{-48}\) ?

Because there SHOULD be an i outside, as you have, but not a 7 inside.

Answer 74

Here is what "the book did":

\(\Large \sqrt{-27} =\sqrt{-1\cdot9\cdot3} =\sqrt{-1}\cdot \sqrt{9}\cdot \sqrt{3} =i\cdot 3\cdot \sqrt{3} =3i\sqrt{3} \)

That's how they simplified the radical part from the quad. formula. Then simplified the whole larger expression by reducing.

But first, you MUST understand how to simplify that radical part. Your PROCESS should follow the book's process exactly, but your numbers are different because what's under your radical sign is different.

Answer 75

OH, @DebbieG , @wealdooo , @MrMoose , I'm confused too :D
so what is the answer?we should use quadratic formula or we should say that -4 is the answer?

Answer 76

If you are only seeking solutions over the real numbers, then the odd root property will suffice and x=-4 is the only solution.

But there are also 2 complex solutions, the roots of the quadratic part obtained from factoring the sum of cubes:

\(\Large x ^{3}+4 ^{3}=(x+4)\color{red}{(x^2-4x+16)}\)

You don't get those complex solutions if you only apply the odd root property, you only get them if you factor and apply the zero factor property (or, use @mrmoose's approach, but that is beyond the scope of an Algebra 2 class; factoring and using the QF, is not).

Answer 77

oh thank you!so we have 3 answers!

Answer 78

over the set of complex numbers, yes. :)