Question

What are the real or imaginary solutions of each polynomial equation? Show work.

x^4-13x^2+36=0

x^3-8=0

Answer 1

@Directrix

Answer 2

FActor this:

x^4-13x^2+36=0

Answer 3

I can work with real solutions, but when it comes to imaginary it really messes me up.

Answer 4

Ok, hold on a sec.

Answer 5

Factor as a trinomial like y^2 - 13y + 36 = 0 but your y will be x^2

Answer 6

So, (x^2-9) (x^2-4)?

Answer 7

Yes, and factor each of the binomials again. They are differences of squares.

Answer 8

Wait so then would the solutions be + or - 3 and + or - 2?

Answer 9

Work in steps. It is not time to get to an answer yet. You are supposed to factor each of these: (x^2-9) (x^2-4)

Answer 10

Factor. That is all at this point.

Answer 11

So...

x^2-9=0
x^2=9
Square each side
x=square root of 9
x=+ or - 3

And the other one

x^2-4=0
x^2=4
Square each side
x=square root of 4
x=+ or - 2

Answer 12

Oh wait I did that wrong!

Answer 13

Hold on I think I can fix it.

Answer 14

No, FACTOR

x^4 -13x^2+36 =0
(x^2-9) (x^2-4) = 0
(x + 3)* (x - 3) * ( x + 2) * (x - 2) = 0

Do not lose sight of the entire problem.

Answer 15

(x^2-9) (x^2-4) = (x + 3)* (x - 3) * ( x + 2) * (x - 2)

That is factoring.

Answer 16

Yeah I just realized that! I'm doing this on paper with me as well so I just saw that. I totally forgot what you meant by factoring it. I'm so sorry. But yes that's what I got as well. And then I'd solve for 0 and then I'd get.. wait, it's still the same answer tho right?

Answer 17

I'm sorry lol that part just flew past my head.

Answer 18

It may be but here's the deal. All problems of this sort don't factor so easily and have integer roots. So, it is good to have a procedure.

There is no need to be sorry. You are learning.

I'm working on paper, too.

Answer 19

Alright, so it's important that I do that step as well. I'll remember that next time, thanks :)

Answer 20

Next Step:

(x + 3)* (x - 3) * ( x + 2) * (x - 2) = 0

Use the Zero Product Property to get the roots

Answer 21

x + 3 = 0 OR x-3 = 0 or x +3 = 0 or x -2 = 0

Answer 22

So set each one equal to 0 and solve correct?

x+3=0
x=-3

x-3=0
x=3

x+2=0
x=-2

x-2=0
x=2

So the solutions to this polynomial is x=-3, -2, 2, 3

Answer 23

Correct.

Answer 24

Thank you so much! :) But on the other one, after bringing the 8 to the other side and having x^3=8, I have no idea what to do then. Maybe a very vague idea but not a clear one.

Answer 25

What about this: x^3-8=0

Have you memorized the difference of two cubes factoring formula?

Answer 26

I think I have it written down here...

a^3-b^3=(a-b)(a^2+ab+b^2)

Answer 27

Here's a chart.

Answer 28

Wait, could I do x^3-2^3=0, and then factor then solve?

Answer 29

So apply that to x^3-8 = 0

Factor the x^3 - 8

Answer 30

Right. The second factor of x^3 - 8 is where you'll see some imaginaries (Complex roots)

Just factor and we'll worry about that later.

Answer 31

Alright, give me a sec :)

Answer 32

Wait, could I do x^3-2^3=0, and then factor then solve?

YES

Answer 33

Ok here's what I got:

x^3-2^3=(x-2)(x^2+2x+8)
=(x-2)(x+4)(x+2)

Answer 34

x^3 - 2^3 = (x-2)* (x^2 + 2x +4) = 0

Answer 35

Wait why would it be 4 not 8? The chart says y^3. So isn't 2^3 equal to 8?

Answer 36

Look at the factoring formula.

(x-2)* (x^2 + 2x +4) = x ^3 - 8

The 4 in the trinomial is multiplied by the -2 in the binomial to give -8.

Your factorizatin will give x^3 - 16 when multiplied back out.

Answer 37

Oh right! Because there's another -2 that's being multiplied by 4. Oh ok that makes sense.