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)
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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}.