Using the remainder theorem, as x = 1 is a root of the expression 5x^3-4x^2+7x-8=0, the expression is divisible by ( x - 1).
(ax^2 + bx + c)(x - 1) = 5x^3-4x^2+7x-8
=> ax^3 + bx^2 + cx - ax^2 - bx - c = 5x^3-4x^2+7x-8
=> a =...
See This Answer Now
Start your subscription to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
Using the remainder theorem, as x = 1 is a root of the expression 5x^3-4x^2+7x-8=0, the expression is divisible by ( x - 1).
(ax^2 + bx + c)(x - 1) = 5x^3-4x^2+7x-8
=> ax^3 + bx^2 + cx - ax^2 - bx - c = 5x^3-4x^2+7x-8
=> a = 5
=> b - a = -4 => b = -4 + 5 = 1
=> c - b = 7 => c = 8
5x^2 + x + 8 = 0
=> x1 = [-1 + sqrt (1 - 160)]/10
=> x1 = -1/10 + i*sqrt 159 /10
x2 = -1/10 - i*sqrt 159 / 10
Therefore the other roots of the expression are -1/10 + i*sqrt 159 /10 and -1/10 - i*sqrt 159 / 10