If x=1 is the root of 5x^3-4x^2+7x-8=0, what are the other roots of equation? (use the remainder theorem)

If x = 1 is the root of polynomial 5x^3-4x^2+7x-8, then
the polynomial is divided by x - 1.

We'll apply reminder
theorem:

5x^3-4x^2+7x-8 = (x-1)(ax^2 + bx +
c)

We'll remove the
brackets:

5x^3-4x^2+7x-8 = ax^3 + bx^2 + xc - ax^2 - bx -
c

We'll combine like terms from the right
side:

5x^3-4x^2+7x-8 = ax^3 + x^2(b - a) + x(c - b) -
c

Comparing, we'll get:

a =
5

b - a = -4

b = a -
4

b = 5 - 4

b =
1

c - b = 7

c = b +
7

c = 1 + 7

c =
8

We'll get the quotient:

5x^2
+ x + 8

We'll put 5x^3-4x^2+7x-8 =
0

5x^3-4x^2+7x-8 = 0 <=> (x-1)(5x^2 + x + 8)
= 0

5x^2 + x + 8 = 0

We'll
apply quadratic formula:

x2 = [-1+sqrt(1 -
160)]/10

x2 = (-1 +
isqrt159)/10

x3 = (-1 -
isqrt159)/10

The other 2 roots of the
polynomial are: { (-1 - isqrt159)/10 ; (-1 +
isqrt159)/10}.