SOLUTION: I am trying to solve this problem but am having some issues. Find all real and imaginary zeros for each polynomial function f(x)=4x^4-1 Here is what I have started but I

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I am trying to solve this problem but am having some issues. Find all real and imaginary zeros for each polynomial function f(x)=4x^4-1 Here is what I have started but I       Log On


   

Question 88393: I am trying to solve this problem but am having some issues.
Find all real and imaginary zeros for each polynomial function
f(x)=4x^4-1
Here is what I have started but I am unsure if this is right or how to finish it.
f(x)=4x^4-1
4x^4-1=0
+1=+1
4x^4=1
-------
4 4
x^4=-1/4
Thank you for your assistance.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
f%28x%29+=+4x%5E4-1 Do you recognise this as the difference of two squares?
f%28x%29+=+%282x%5E2%29%5E2-1%5E2
The difference of two squares can be factored thus:
A%5E2-B%5E2+=+%28A%2BB%29%28A-B%29 Applying this to your problem, you get:
f%28x%29+=+%282x%5E2%2B1%29%282x%5E2-1%29 Now set this equal to zero.
%282x%5E2%2B1%29%282x%5E2-1%29+=+0 Apply the zero products principle.
2x%5E2%2B1+=+0 or 2x%5E2-1+=+0 Solve each of these quadratic equations.
2x%5E2%2B1+=+0 Subtract 1 from both sides.
2x%5E2+=+-1 Divide both sides by 2.
x%5E2+=+-1%2F2 Finally, take the square root of both sides.
x+=+sqrt%28-1%2F2%29 or x+=+-sqrt%28-1%2F2%29
...and for the other equation:
2x%5E2-1+=+0 Add 1 to both sides.
2x%5E2+=+1 Divide both sides by 2.
x%5E2+=+1%2F2 Take the square root of both sides.
x+=+sqrt%281%2F2%29 or x+=+-sqrt%281%2F2%29
So you have two real but irrational roots:
x+=+sqrt%281%2F2%29
x+=+-sqrt%281%2F2%29
...and two imaginary roots:
x+=+sqrt%28-1%2F2%29 which can also be written as: x+=+sqrt%281%2F2%29i
x+=+-sqrt%28-1%2F2%29 which can also be written as: x+=+-sqrt%281%2F2%29i