x = (3 − √5) / 2, (3 + √5) / 2, and 1
Step-by-step explanation:
Using rational root theorem, ±1 are possible roots. By trial and error, we find that +1 is a root. Therefore, x − 1 is a factor. Knowing this, we can use grouping or long division. Using grouping:
x⁴ − 5x³ + 8x² − 5x + 1 = 0
x⁴ − 5x³ + 4x² + 4x² − 5x + 1 = 0
x² (x² − 5x + 4) + 4x² − 5x + 1 = 0
x² (x − 4) (x − 1) + (4x − 1) (x − 1) = 0
(x² (x − 4) + 4x − 1) (x − 1) = 0
(x³ − 4x² + 4x − 1) (x − 1) = 0
Repeating the process, possible rational roots of x³ − 4x² + 4x − 1 are ±1. We already know -1 isn't a root. By trial and error, we find that +1 is again a root. Using grouping to factor:
(x³ − 4x² + 3x + x − 1) (x − 1) = 0
(x (x² − 4x + 3) + x − 1) (x − 1) = 0
(x (x − 3) (x − 1) + x − 1) (x − 1) = 0
(x (x − 3) + 1) (x − 1) (x − 1) = 0
(x² − 3x + 1) (x − 1)² = 0
To find the roots of x² − 3x + 1, we use the quadratic formula.
x = [ -(-3) ± √((-3)² − 4(1)(1)) ] / 2(1)
x = [ 3 ± √(9 − 4) ] / 2
x = (3 ± √5) / 2
So the roots are x = (3 − √5) / 2, (3 + √5) / 2, and 1.
