one solution of x^3-5x^2+5x-1=0 is 1. find the other two solutions.
so either long division or synthetic division, whichever you learned or know, by (x-1).
x^3-5x^2+5x-1 = (x-1)(x^2 - 4x + 1)
then x^3-5x^2+5x-1 = 0
(x-1)(x^2 - 4x + 1) = 0
x = 1 or x = 2 ± √3
thank you so much Reiny, i got an exam and am sorry i gave you a hard time with this. thank you again, i really appreciate it.
sorry my sis last used this site under the name zena... the reply was from me not her but i forgot to put my name instead of hers. sorry
To find the other two solutions of the equation x^3 - 5x^2 + 5x - 1 = 0, we can utilize polynomial long division or synthetic division to divide the polynomial by (x - 1), which is the known root. But first, let's confirm that x = 1 is a solution.
Substitute x = 1 into the equation:
(1)^3 - 5(1)^2 + 5(1) - 1 = 1 - 5 + 5 - 1 = 0.
The result is indeed 0, so x = 1 is a solution.
Now, let's divide the polynomial by (x - 1) to find the other two solutions:
Performing polynomial long division or synthetic division with the polynomial x^3 - 5x^2 + 5x - 1 divided by (x - 1), we get:
x^2 - 4x + 1
_______________________
x - 1 | x^3 - 5x^2 + 5x - 1
By dividing, we obtain a quotient of x^2 - 4x + 1 and no remainder.
Now, to find the remaining solutions, we need to solve the equation x^2 - 4x + 1 = 0.
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -4, and c = 1.
Plugging these values into the quadratic formula, we have:
x = (-(-4) ± √((-4)^2 - 4(1)(1))) / (2(1))
x = (4 ± √(16 - 4)) / 2
x = (4 ± √12) / 2
x = (4 ± 2√3) / 2
x = 2 ± √3
Therefore, the other two solutions are x = 2 + √3 and x = 2 - √3.