determine the number of each type of root of the equation
x^3-4x^2+x+26=0
how many imaginary roots?
a) 0
b)1
c)2
d)3
how many real roots?
a)0
b)1
c)2
d)3
show all work
To determine the number of each type of root of the equation x^3 - 4x^2 + x + 26 = 0, we can use Descartes' Rule of Signs and the Rational Root Theorem.
1. Descartes' Rule of Signs:
The number of positive roots of the polynomial is given by counting the sign changes in the coefficients when written in standard form. In this case, there is only one sign change from -4x^2 to + x, so there is 1 positive root.
The number of negative roots of the polynomial is given by counting the sign changes in the coefficients when written in standard form. In this case, there is no sign change, so there are 0 negative roots.
2. Rational Root Theorem:
The Rational Root Theorem states that any rational roots of the polynomial equation are of the form p/q, where p is a factor of the constant term (26) and q is a factor of the leading coefficient (1).
The factors of 26 are ±1, ±2, ±13, ±26.
The factors of 1 are ±1.
Plugging these factors into the polynomial equation, we find that 2 is a root. Therefore, (x-2) is a factor of the polynomial.
Using polynomial long division or synthetic division, we can divide x^3 - 4x^2 + x + 26 by (x-2) to obtain x^2 - 2x - 13. This quadratic equation can now be solved by factoring or using the quadratic formula.
The solutions to the quadratic equation are x = 3 and x = -1.
Therefore, the types of roots are:
- 1 positive real root (x = 2)
- 2 real roots (x = 3, x = -1)
- 0 imaginary roots
Therefore, the answers are:
How many imaginary roots?
a) 0
How many real roots?
c) 2