To find the zeroes of the equation, we can use the Rational Root Theorem and synthetic division to test different possible rational zeroes.
The Rational Root Theorem states that if a polynomial has a rational zero, then it can be expressed in the form p/q, where p is a factor of the constant term (75 in this case) and q is a factor of the leading coefficient (2 in this case).
The possible rational zeroes to test are:
±1, ±3/2, ±5, ±15, ±25, ±75
We can start by trying x = 1:
2(1)^4 - 5(1)^3 + 53(1)^2 - 125(1) + 75 = 2 - 5 + 53 - 125 + 75 = 0
Therefore, x = 1 is a zero of the equation.
Using synthetic division, we can divide the equation by (x - 1):
2 -5 53 -125 75
-------------------------------------
1 | 2 - 3 50 - 75 0
The resulting quotient is 2x^3 - 3x^2 + 50x - 75.
We can now repeat the process with the cubic equation. Using the Rational Root Theorem, the possible rational zeroes are:
±1, ±3/2, ±5, ±15, ±25, ±75
Trying x = 1:
2(1)^3 - 3(1)^2 + 50(1) - 75 = 2 - 3 + 50 - 75 = -26 ≠ 0
Trying x = -1:
2(-1)^3 - 3(-1)^2 + 50(-1) - 75 = -2 - 3 - 50 - 75 = -130 ≠ 0
Trying x = 3/2:
2(3/2)^3 - 3(3/2)^2 + 50(3/2) - 75 = 27/4 - 27/4 + 75/2 - 75 = 135/4 - 75/2 - 75 = -175/4 ≠ 0
Trying x = -3/2:
2(-3/2)^3 - 3(-3/2)^2 + 50(-3/2) - 75 = -27/4 - 27/4 - 75/2 - 75 = -135/4 - 75/2 - 75 = -325/4 ≠ 0
Trying x = 5:
2(5)^3 - 3(5)^2 + 50(5) - 75 = 250 - 75 + 250 - 75 = 350 ≠ 0
Trying x = -5:
2(-5)^3 - 3(-5)^2 + 50(-5) - 75 = -250 - 75 - 250 - 75 = -650 ≠ 0
None of the rational zeroes tested resulted in a zero value for the cubic equation. Therefore, we can conclude that the cubic equation has no rational zeroes.
The remaining zeroes can be calculated using numerical methods or a graphing calculator. The solutions are approximately -1.326409, 0.532698 ± 4.066604i.
Therefore, the correct answer is:
Option 1. -1, -3/2, ± 5i
Find all the zeroes of the equation.
2x^4 - 5x^3 + 53x^2 - 125x + 75 = 0
1. -1, -3/2, +_ 5i
2. 1, 3/2, +_ 5i
3. 1, 3/2, +_ 5
4. -1, -3/2, +_ 5
1 answer