Question

can someone help plz​

Answer 1

Answer:

4 possible real zeroes

solutions are

$x=\sqrt{3}\\x=-\sqrt{3}$

x = 1

x = -1

Step-by-step explanation:

4th order polynomial therefore there are 4 possible solutions.

factoring it:

(x^2-3)(x^2-1)

Solve for x

x^2 - 3 = 0

x^2 = 3 ; take the square root of both sides

$x=\sqrt{3}\\x=-\sqrt{3}$

x^2 - 1 = 0

x^2 = 1

x = 1

x = -1