"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.

- anonymous

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- anonymous

ok,
x^3 = -64
ok?

- anonymous

ok?

- anonymous

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 \[x=\frac{ -(-64)\sqrt{(-64)^{2}-4(x)(4096)} }{ 2a }\]

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- anonymous

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

- anonymous

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

- anonymous

(-4)^3 = -64 :)

- anonymous

got it?

- anonymous

@wealdooo

- anonymous

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

- anonymous

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

- anonymous

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- anonymous

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).

- anonymous

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

- DebbieG

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.

- DebbieG

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.

- anonymous

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

- anonymous

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

- DebbieG

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

- anonymous

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

- anonymous

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

- anonymous

-64, sorry

- anonymous

Also, remember to take the angle mod 2pi

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

All other solutions are congruent to these mod 2pi

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Yeah, it should be

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

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- anonymous

That would probably be the easiest way to do it.

- DebbieG

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

- anonymous

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.

- DebbieG

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

- anonymous

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

- DebbieG

I did it above, although I see I had an exponent in the wrong place.
\(\Large x ^{3}+4 ^{3}=(x+4)(x ^{2}-4x+4^2)=(x+4)(x^2-4x+16)\)

- DebbieG

b=4. NOT b=64.

- anonymous

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

- anonymous

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

- anonymous

oops, 80x

- DebbieG

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

- anonymous

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

- anonymous

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

- DebbieG

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

- DebbieG

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

- DebbieG

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.

- anonymous

so is a = 1?

- DebbieG

exactly :)

- anonymous

so is a = 1?

- anonymous

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

- DebbieG

under the radical.... 16 - 64 = ?
Then you'll be able to simplify the radical part.

- DebbieG

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

- anonymous

Oops. -48 not 80

- DebbieG

right now you can simplify that radical....

- anonymous

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

- anonymous

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

- anonymous

or would it be 2?

- anonymous

2i

- DebbieG

I don't know what you mean, "with 3 or 4".... just simplify the expression you have.
Start with the radical. \(\sqrt{-48}\) = ??

- anonymous

got -6.93

- DebbieG

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.

- anonymous

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

- DebbieG

Factor what's under the radical sign:
\(\sqrt{-48}=\sqrt{-1\cdot 16 \cdot3}\)
The radical is the square root sign.

- DebbieG

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 \(\sqrt{-48}\)

- anonymous

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

- DebbieG

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?

- anonymous

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

- DebbieG

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.

- DebbieG

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.

- anonymous

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?

- DebbieG

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).

- anonymous

oh thank you!so we have 3 answers!

- DebbieG

over the set of complex numbers, yes. :)

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