Solve: 0 = x3 + 5x2 + 2x – 8
Solve: 0 = x3 – x2 – 11x – 10
3 answers
use your graphing calculator or plot it on a graph
Use the Rational Roots Theorem.
Suppose
a x^n + b x^(n-1) + .... e = 0
If x is a rational number of the form p/q with p and q integers that don't have factors in common (i.e. the fraction p/q is simplified as much as possible), then p must divide e and q must divide a.
In case of the equation:
x^3 + 5x^2 + 2x – 8 = 0
this means that x = integer that divides 8. x = 1, x= -2 and x = -4 are solutions.
Suppose
a x^n + b x^(n-1) + .... e = 0
If x is a rational number of the form p/q with p and q integers that don't have factors in common (i.e. the fraction p/q is simplified as much as possible), then p must divide e and q must divide a.
In case of the equation:
x^3 + 5x^2 + 2x – 8 = 0
this means that x = integer that divides 8. x = 1, x= -2 and x = -4 are solutions.
x^3 – x^2 – 11x – 10 = 0
Assuming again that the roots are rational numbers, it follows from the Rational Roots Theorem that the roots are integers that divide the number 10.
So, the only possible rational roots are:
x = -10, x = -5, x = -2, x = -1, x = 1, x = 2, x = 5, x = 10
Only x = -2 satisfies the equation.
Factoring the polynomial gives:
x^3 – x^2 – 11x – 10 =
(x+2)(x^2 -3x - 5)
The zeroes of the quadratic factor are the other zeroes (you already know that they cannot be rational).
Assuming again that the roots are rational numbers, it follows from the Rational Roots Theorem that the roots are integers that divide the number 10.
So, the only possible rational roots are:
x = -10, x = -5, x = -2, x = -1, x = 1, x = 2, x = 5, x = 10
Only x = -2 satisfies the equation.
Factoring the polynomial gives:
x^3 – x^2 – 11x – 10 =
(x+2)(x^2 -3x - 5)
The zeroes of the quadratic factor are the other zeroes (you already know that they cannot be rational).