Please help me with steps

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.

hreryryrranswer:

step-by-step explanation:

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)