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)

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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 =...

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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

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