Question

Find all the zeroes of the equation.

x^4 - 6x^2 -7x -6 = 0

Answer 1

@Will.H

Answer 2

ew will

Answer 3

Do you need the endless steps and walk you through it or do you want me to solve it and provide you with a graph?

Answer 4

The steps c:

Answer 5

She wants understanding over a free answer, respect

Answer 6

Okay.
1st by the fundamental theorem of Algebra we know there will be 4 solutions to this since the highest degree is 4.

Now we will have to make predictions of some kind using P/q where P in our case is 6 and q is 1
Find their factors.
Note: I am doing the steps but since am using my phone it might not appear that am typing

Answer 7

Anyways
Factors of 6: +-1, +-2,+-3,+-6
Factors 1: +-1

Factors P/Q: +-1, +-2,+-3,+-6

Luckily we don't have much potentials.

Now we need to find the positive possible solutions.
x^4 - 6x^2 -7x -6 = 0
The sign changed 1 time in here

Therefore there can be only 1 positive solution.

Let's see how many negative solutions there will be
But 1st you'll have to change every x into -x in the expression and then simplify...

Answer 8

The negative possible solutions:

x^4 + 0x^3 -6x^2 -7x -6 = 0
Change every x into -x
(-x)^4 + 0(-x)^3 -6(-x)^2 -7(-x) -6 = 0
Simplify
X^4 -0x^3 -6x^2 + 7x -6 = 0
Possible negative solutions are: 2 or 0

Now complex possible solutions can be predicateD by counting how many solutions left if we anticipated the negative and positive solutions

Positive: 1
Negative: 2 or zero
Complex:1 or 4

Answer 9

Now we know there will be positive solution so let's try and find it from the possibilities that we made earlier.
The positive integer that would set the expression equal to zero will be a solution of the polynomial. And in return that would help us find the deritiave equation.

x^4 + 0x^3 -6x^2 -7x -6 = 0
Possible solutions: 1,2,3,6

After using my calculator I found the positive solution would be 3

So out of the 4 solutions one of them will be (3,0)

Now use that to find the deritiave equation

3| 1. 0. -6. -7. -6
=====3==9===9==6====
1. 3. 3. 2. 0

X^3 + 3x^2 + 3x + 2
Now we will do try to anticipate another solution which will be negative this time (perhaps) based on the analysis we made

Answer 10

X^3 + 3x^2 + 3x + 2
Possible negative solutions: -1,-2,-3,-6
Use calculator to find which of them would be a solution
Turned out it would be -2

-2| 1. 3. 3. 2
=====-2==-2==-2===
1 1. 1. 0

X^2 + x + 1
Now we can factor this out using groups factoring.
But remember out of the 4 solutions we found out that 3 and -2 are solutions. So only 2 more left.

For x^2 + x + 1
Try factor by groups.

Answer 11

So for x^2 + x+ 1 we couldn't factor it by groups. That means there won't be real solutions. They will be (imaginary)
We can still find them using
The famous quadratic formula
Which I will type below but am sure it will not be clear because again am using my phone and can't use the equation button.

Famous quadratic formula
(-B +- sqrt (b^2 -4ac))/2a

Substitute
-1 +- sqrt(1^2 -4(1)(1))/2(1)

The 1st solution would be
(-1 + sqrt (-3))/2
Simplify
(-1 + i sqrt 3)/2 that's the 1st imaginary solution

The second
(-1 - sqrt (-3))/2
Simplify
(-1 - i sqrt 3)/2 that's the second imaginary solution.

Consequently
We have the 4 complete solutions of
x^4 - 6x^2 -7x -6 = 0
As
(3,0) (-2,0) and the imaginary 2 solutions above (complex solutions)

Hope that helps ;)

Answer 12

Questions?

Answer 13

Whoa, how did you type all that ^ ??

Answer 14

It takes all that to solve XD

Answer 15

Okay sorry, I was afk for a bit. I think I got it, but thank you for taking the time to help me! :) @Will.H

Answer 16

You welcome :)