Note: This page has been translated by MathWorks. Click here to see

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

This example shows several different methods to calculate the roots of a polynomial.

The `roots`

function calculates
the roots of a single-variable polynomial represented by a vector
of coefficients.

For example, create a vector to represent the polynomial $${x}^{2}-x-6$$, then calculate the roots.

p = [1 -1 -6]; r = roots(p)

r = 3 -2

By convention, MATLAB^{®} returns the roots in a column vector.

The `poly`

function converts
the roots back to polynomial coefficients. When operating on vectors, `poly`

and `roots`

are
inverse functions, such that `poly(roots(p))`

returns `p`

(up
to roundoff error, ordering, and scaling).

p2 = poly(r)

p2 = 1 -1 -6

When operating on a matrix, the `poly`

function
computes the characteristic polynomial of the matrix. The roots of
the characteristic polynomial are the eigenvalues of the matrix. Therefore, `roots(poly(A))`

and `eig(A)`

return
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 $$\theta $$ that solve the equation

$$3{\mathrm{cos}}^{2}(\theta )-\mathrm{sin}(\theta )+3=0.$$

Use the fact that $${\mathrm{cos}}^{2}(\theta )=1-{\mathrm{sin}}^{2}(\theta )$$ to express the equation entirely in terms of sine functions:

$$-3{\mathrm{sin}}^{2}(\theta )-\mathrm{sin}(\theta )+6=0.$$

Use the substitution $$x=\mathrm{sin}(\theta )$$ to express the equation as a simple polynomial equation:

$$-3{x}^{2}-x+6=0.$$

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 $$\theta ={\mathrm{sin}}^{-1}(x)$$. 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 $$\theta $$ that solve the original equation (within roundoff error).

f = @(Z) 3*cos(Z).^2 - sin(Z) + 3; f(theta)

ans =2×1 complex10^{-14}× -0.0888 + 0.0647i 0.2665 + 0.0399i

Use the `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 $$3{x}^{7}+4{x}^{6}+2{x}^{5}+4{x}^{4}+{x}^{3}+5{x}^{2}$$.

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 $$[-2,1]$$.

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 `-1.5`

. Use `fzero`

to calculate and plot the root that is near `-1.5`

.

Z = fzero(p, -1.5)

Z = -1.6056

`plot(Z,p(Z),'r*')`

If you have Symbolic Math
Toolbox™, then there are additional
options for evaluating polynomials symbolically. One way is to use
the `solve`

function.

```
syms x
s = solve(x^2-x-6)
```

s = -2 3

Another way is to use the `factor`

function
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.