This example shows several different methods to calculate the roots of a polynomial.
roots function calculates
the roots of a single-variable polynomial represented by a vector
For example, create a vector to represent the polynomial , then calculate the roots.
p = [1 -1 -6]; r = roots(p)
r = 3 -2
By convention, MATLAB® returns the roots in a column vector.
poly function converts
the roots back to polynomial coefficients. When operating on vectors,
inverse functions, such that
to roundoff error, ordering, and scaling).
p2 = poly(r)
p2 = 1 -1 -6
When operating on a matrix, the
computes the characteristic polynomial of the matrix. The roots of
the characteristic polynomial are the eigenvalues of the matrix. Therefore,
the same answer (up to roundoff error, ordering, and scaling).
You can solve polynomial equations involving trigonometric functions by simplifying the equation using a substitution. The resulting polynomial of one variable no longer contains any trigonometric functions.
For example, find the values of that solve the equation
Use the fact that to express the equation entirely in terms of sine functions:
Use the substitution to express the equation as a simple polynomial equation:
Create a vector to represent the polynomial.
p = [-3 -1 6];
Find the roots of the polynomial.
r = roots(p)
r = 2×1 -1.5907 1.2573
To undo the substitution, use . The
asin function calculates the inverse sine.
theta = asin(r)
theta = 2×1 complex -1.5708 + 1.0395i 1.5708 - 0.7028i
Verify that the elements in
theta are the values of that solve the original equation (within roundoff error).
f = @(Z) 3*cos(Z).^2 - sin(Z) + 3; f(theta)
ans = 2×1 complex 10-14 × -0.0888 + 0.0647i 0.2665 + 0.0399i
fzero function to find the roots of a polynomial in a specific interval. Among other uses, this method is suitable if you plot the polynomial and want to know the value of a particular root.
For example, create a function handle to represent the polynomial .
p = @(x) 3*x.^7 + 4*x.^6 + 2*x.^5 + 4*x.^4 + x.^3 + 5*x.^2;
Plot the function over the interval .
x = -2:0.1:1; plot(x,p(x)) ylim([-100 50]) grid on hold on
From the plot, the polynomial has a trivial root at
0 and another near
fzero to calculate and plot the root that is near
Z = fzero(p, -1.5)
Z = -1.6056
If you have Symbolic Math
Toolbox™, then there are additional
options for evaluating polynomials symbolically. One way is to use
syms x s = solve(x^2-x-6)
s = -2 3
Another way is to use the
to factor the polynomial terms.
F = factor(x^2-x-6)
F = [ x + 2, x - 3]
See Solve Algebraic Equation (Symbolic Math Toolbox) for more information.