<h3>
Answer: The zeroes of the equation are x=1, 3/2, 5i, -5i.</h3>
Step-by-step explanation:
Given equation ![2x^4-5x^3+53x^2-125x+75=0](https://tex.z-dn.net/?f=2x%5E4-5x%5E3%2B53x%5E2-125x%2B75%3D0)
Applying rational roots theorem.
The constant term is 75 and leading coefficient is 2.
Factors of 75 are 1,3,5,15. and factors of 2 are 1, and 2.
Therefore, possible rational roots would be ±1,3,5,15,1/2, 3/2, 5/2 and 15/2.
Let us check first x=1 if it is a root or not.
Plugging x=1 in given equation, we get
would give us 0.
Therefore, first root would be x=1 so the first factor would be x-1.
Dividing given polynomial using syntactic division
________________________
1 | 2 -5 53 -125 75
2 -3 +50 -75
_______________________
2 -3 +50 -75 0
So the other factored polynomial, we get
![2x^3-3x^2+50x-75](https://tex.z-dn.net/?f=2x%5E3-3x%5E2%2B50x-75)
Factor it by grouping
![(2x^3-3x^2)+(50x-75)](https://tex.z-dn.net/?f=%282x%5E3-3x%5E2%29%2B%2850x-75%29)
![x^2(2x-3) +25(2x-3)](https://tex.z-dn.net/?f=x%5E2%282x-3%29%20%2B25%282x-3%29)
![(2x-3)(x^2+25).](https://tex.z-dn.net/?f=%282x-3%29%28x%5E2%2B25%29.)
Setting each of the factors equal to 0, we get
2x-3=0
2x=3
x= 3/2.
![x^2+25 =0.](https://tex.z-dn.net/?f=x%5E2%2B25%20%3D0.)
![x^2 = -25.](https://tex.z-dn.net/?f=x%5E2%20%3D%20-25.)
Taking square root on both sides, we get
x = ±5i
<h3>Therefore, the zeroes of the equation are x=1, 3/2, 5i, -5i.</h3>