Factoring: z^4 - 81?

Question

^ means to the power of

Answer 1

Let z^2 = h
Therefore, z^4 -81
= h^2 - 81
Oh, this is familiar
here's a factorizing rule you should be familiar with:
a^2 - b^2 = (a+b)(a-b)
therefore, statement can be factorized to (h+9)(h-9)
h = z^2
therefore, statement is (z^2 + 9)(z^2 - 9)
again, simply the (z^2 - 9) term using the rule to give (z^2 + 9)(z+9)(z-9)
This answer would now be sufficient for a basic maths class.

If however you are doing comlpex numbers, then we can factorize further using rule.
The answer then becoems (z-3i)(z+3i)(z+9)(z-9), where i=(-1)^0.5

Overall steps:
z^4 - 81
= (z^2)^2 - 9^2
= (z^2 + 9)(z^2 - 9)
= (z^2 + 9)(z+9)(z-9)
if doing comlpex numbers....
= (z+3i)(z-3i)(z+9)(z-9)

Answer 2

3

Answer 3

z^4 - 81
= (z^2 + 9) (z^2 - 9)
= (z^2 + 9) (z + 3) (z - 3)

Answer 4

this is a difference of two squares, as 81 can be rewritten as 9^2
and z^4 can be rewritten as (z^2)^2

(z^2)^2 - 9^2
(z^2 - 9)(Z^2 + 9)

THIS IS NOT THE ANS AS THERE IS NOW ANOTHER DIFFERENCE OF TWO SQUARES AS 9 CAN BE REWRITTEN AS 3^2

U MUST REMEMBER THAT IT IS A DIFFERENCE THEREFORE
Z^2 +9 IS COMPLETELY FACTORIZED

(Z^2 - 3^2)(Z^2 + 9)
(Z - 3)(Z + 3 )(Z^2 +9)

Answer 5

Looks like a sum and diffence of squares (both are 4th power, their square roots are 2d power), so we factor to
(z^2-9)(z^2+9),
We can do one more "round" on the first factor to get (z+3)(z-3)(z^2+9)

Answer 6

z^4-81
(z^2+9)(z^2-9)
(z^2+9)(z+3)(z-3)

z^2+9 has no real factors.
complex factors are (z+3i)(z-3i)

Answer 7

(z ^2 + 9) (z^2 - 9)
(z^2 + 9) (z + 3) (z -3)
z equals positive 3 or negative 3 to make the equation equal zero.

Hope this helps!

Answer 8

(z-3)^4 or (z-3)(z-3)(z-3)(z-3)