Some graphing calculators may be used to perform algebraic operations on polynomials. For the TI-92 Graphing Calculator the  Algebra toolbar, in the Home screen, may be used to select the most commonly used algebraic operations.

Addition and subtraction of polynomials can done directly on the entry line

Example. Let f (x) = 3x⁴ - x³ +5x and g(x) = - x⁴ - 3x³ + x² - 7. Use a graphing calculator to find s(x) = f (x) + g(x).

Solution: We can enter the functions directly on the entry line. Use parentheses carefully.

3x  4  x  3  5x     x  4  3x  3  x  2  7    The answer is displayed: 2x⁴ - 4x³ + x² + 5x - 7

(a) s(x) = f (x) + g(x) and (b) d(x) = f (x) - g(x).

Multiplication of two polynomials is performed by selecting ‘3:Expand’ in the  Algebra toolbar of the Home screen.

Example. Let f (x) = 7x³ + 4x - 5 and g(x) = x - 2. Use a graphing calculator to find m(x) = f (x) g(x).

Solution: Be sure to use the correct number of parentheses.

Select and clear Home screen:    8

Display Algebra toolbar and select ‘expand’:  3

The functions to be multiplied must be enclosed by parentheses after ‘expand’. Note that the left parentheses is already displayed.

After expand( type:  7x  3  4x  5    x  2   

The answer is displayed: 7x⁴ - 14x³ + 4x² - 13x + 10

Factoring a polynomial to obtain its zeros is not always easy. It is useful to have some algebraic knowledge about zeros of a polynomial. We now state four properties. (Their proofs can be found in books on algebra.) Recall that the zeros of f (x) are the roots of the associated equation f (x) = 0.

1. An nth-degree equation has exactly n roots, where roots of multiplicity m are counted m times.
2. Complex roots always occur in conjugate pairs.
3. (a) There can be no more positive roots than the number of changes in sign in f (x). (If there are less, they are less by an even integer.)

(b) There can be no more negative roots than the number of changes in sign in f (-x). (If there are less, they are less by an even integer.)
4. A rational root can only be a factor of the constant term divided by a factor of the leading coefficient.

• 3x⁵ + 2x⁴ - x³ - x² + x + 1 = 0 has exactly 5 roots (degree 5).
• If f (x) = (x + 2)(x + 1)³ (x - 1)² , the factor (x + 2) is of multiplicity 1; the factor (x+ 1) is of multiplicity 3 (the exponent of the factor is 3); the factor (x - 1) is of multiplicity 2. Hence there are 1 + 3 + 2 = 6 zeros of f (x). They are: - 2, - 1, - 1, - 1, 1, 1.

Property 3 is known as Descartes' rule of signs. If the terms of a polynomial equation f (x) = 0 with real coefficients are written in standard form, a change of sign occurs when two consecutive terms differ in sign, those with zero coefficient being ignored. For example,

f (x) = 3x⁴ - 2x³ + x² + 5 = 0

f (x) = x⁵ + x⁴ - x³ - 2 = 0

Example. Given f (x) = x³ - 2x² + 4x + 1. Use Descartes' rule of signs (Property 3) to discuss the possibilities of the number of positive, negative, and complex zeros of f (x).

Solution: f (x) = x³ - 2x² + 4x + 1 has 2 changes of sign. Thus there may be 2 or no positive zeros (differ by an even integer.)

To obtain f (- x) replace each x by - x and simplify.

f (- x) = (- x)³ - 2(- x)² + 4(- x) + 1 or f (- x) = - x³ - 2x² - 4x + 1

and f (- x) has one change of sign. Thus there is one negative zero. (The difference of one and an even integer would be a negative number of zeros.)

There is a total of 3 zeros. The possibilities may be summarized in the following array:

The reason for decreasing the number of positive (or negative) zeros by an even integer is that complex zeros occur in pairs (Property 2.)

Problem. Use Descartes' rule of signs to discuss the possibilities of the number of positive, negative, and complex zeros of f (x) = x³ + 3x² - 5x + 1.

Example. Find the rational roots of x³ - 3x² + 3x - 2 = 0.

Solution: We make use of Property 4. Let the constant term be denoted by p and the leading coefficient by q. Here p = - 2 and q = 1. The factors for p are ± 1, ± 2, and for q are ± 1. The possible rational roots are the numbers  = ± 1 and ± 2. We test each of these numbers by substitution.
 

Problem. Find the rational roots of x³ + 4x² + 2x - 3 = 0.
 

To Table of Contents

To Section 8. Algebra Menu

To Section 11. Zeros of Polynomial Functions