Solve for x: x3+72=32
6
5
-6
-5
Given, x3+72=32
Transposing 72 to RHS, we get
⇒ x3=32−72
⇒ x3=−42 ⇒ x3=−2
Multiplying both sides by 3, we get
x=−6
The maximum number of zeroes the polynomial x7+x5+x3+x+1 can have is _______.
Solve:- 4x−2−2x+1=0