SOLUTION: Find all zeroes of each equation 1. x^5-3x^4-15x^3-45x^2-16x+48=0 2. x^3-x^2-5x-5=0 3. 7x^2-144=-x^4 Please help. I've tried figuring them out for hours and I cannot ge

Algebra ->  Test -> SOLUTION: Find all zeroes of each equation 1. x^5-3x^4-15x^3-45x^2-16x+48=0 2. x^3-x^2-5x-5=0 3. 7x^2-144=-x^4 Please help. I've tried figuring them out for hours and I cannot ge      Log On


   

Question 671904: Find all zeroes of each equation
1. x^5-3x^4-15x^3-45x^2-16x+48=0
2. x^3-x^2-5x-5=0
3. 7x^2-144=-x^4
Please help. I've tried figuring them out for hours and I cannot get it.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
1. You need to use a calculator since it's impossible to this by hand.
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2.

Are you sure it's not

-x^3-x^2-5x-5=0

or

x^3-x^2-5x+5=0

or is it something else entirely?
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3.

7x%5E2-144=-x%5E4

x%5E4+%2B+7x%5E2-144+=+0

%28x%5E2%29%5E2+%2B+7x%5E2-144+=+0

z%5E2+%2B+7z-144+=+0 Let z=x%5E2

%28z%2B16%29%28z-9%29+=+0

z%2B16+=+0 or z-9+=+0

z=-16 or z=9

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If z=9, then...

z=9

x%5E2=9

x=sqrt%289%29 or x=-sqrt%289%29

x=3 or x=-3

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If z=-16, then...

z=-16

x%5E2=-16

but it's impossible for you to multiply a real number by itself to get a negative number.

So there are NO real numbered solutions in this equation here.

We can find the complex solutions to get

x%5E2=-16

x=sqrt%28-16%29 or x=-sqrt%28-16%29

x=4i or x=-4i


So the only real numbered solutions are x=3 or x=-3

If you want to throw in the complex solutions, then you would add x=4i or x=-4i, but it will depend on the teacher/book.