2.3 real zeros of polynomial functions

2.3 Real Zeros of Polynomial Functions

Long Division

Write the dividend and divisor in descending powers of variable and insert placeholders with zeros for missing powers of the variable

1) Divide x2 +5x+6 by x+2

2) Divide x3 – 1 by x – 1

3)Divide 2x4 + 4x3 – 5x2 + 3x – 2 by x2 + 2x - 3

The Division Algorithm

If f(x) and d(x) are polynomials such that d(x) ≠ 0, and the degree of d(x) is less than or equal to the degree of f(x), there exist unique polynomials q(x) and r(x) such that

f(x) = d(x)q(x) + r(x)

Where r(x) = 0 or the degree of r(x) is less than the degree of d(x).

Dividend

Divisor

Quotient

Remainder

Long Division of Polynomials

f(x) = 6x3 – 19x2 + 16x – 4 Divide by 2x2 – 5x + 2

Remainder

Divide x2 + 3x + 5 by x + 1

Synthetic Division

Only can be used for linear divisors: (x – k) Divide by the zero!!! Not the factor

1) Divide x4 – 10x2 -2x + 4 by x + 3

2) Divide -2x3 + 3x2 + 5x – 1 by x + 2

The Remainder Theorem

If a polynomial f(x) is divided by x – k, the remainder is

r = f(k)

Use the remainder theorem to evaluate the following function at x = -2

f(x) = 3x3 + 8x2 + 5x – 7

Factor Theorem

• A polynomial f(x) has a factor (x – k) if and only if f(k) = 0

• You can use synthetic division to test whether a polynomial has a factor (x – k) – If the last number in your answer line is a zero,

then (x – k) is a factor of f(x)

Repeated Division

Show that (x – 2) and (x + 3) are factors of f(x) = 2x4 + 7x3 – 4x2 – 27x – 18

Then find the remaining factors of f(x)

Using the Remainder in Synthetic Division

In summary, the remainder r, obtained in the synthetic division of f(x) by (x – k), provides the following information.

1) The remainder r gives the value of f at x = k. That is, r = f(k)

2) If r = 0, (x – k) is a factor of f(x)

3) if r = 0, (k, 0) is an x-intercept of the graph f

Examples!!

f(x) = 2x3 + x2 – 5x + 2 (x + 2) and (x – 1)

Rational Zero TestIf the polynomial

f(x) = anxn + an-1xn-1 + …a2x2 + a1x + a0

Has integer coefficients, every rational zero of f has the form

Rational Zero = p/q

Where p and q have not common factor other than 1, p is a factor of the constant term a0, and q is a factor of the leading coefficient an

Finding the possible real zeros

1) f(x) = x3 + 3x2 – 8x + 3

2) f(x) = 2x 4 – 5x2 + 3x – 8

Descartes’ Rule of SignsLet f(x) = anxn + an-1xn-1 + …a2x2 + a1x + a0 be a polynomial with real coefficients and a0 ≠ 0

1) The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by an even integer

2) The number of negative real zeros of f is either equal to the number of variations in sign of f(-x) or less than that number by an even integer.

Using Descartes’ Rule of Signs

Describe the possible real zeros of f(x) = 3x3 – 5x2 + 6x - 4

Upper and Lower Bound RulesLet f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x – c, using synthetic division.

1. if c > 0 and each number in the last row is either positive or zero, c is an upper bound for the real zeros of f

2. If c < 0 and the numbers in the last row are alternately positive and negative (zeros count as either), c is a lower bound for the real zeros of f

Finding the Zeros of a Polynomial Function

f(x) = 6x3 – 4x2 + 3x - 2