# 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)

justaguide | Certified Educator

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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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giorgiana1976 | Student

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

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}.

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