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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 Divide x 2 +5x+6 by x+2 2) Divide x 3 – 1 by x – 1
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Long Division Write the dividend and divisor in descending powers of variable and insert placeholders with zeros for missing powers of the variable • 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 Quotient Divisor Remainder
Long Division of Polynomials f(x) = 6x3 – 19x2 + 16x – 4Divide 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 • Divide x4 – 10x2 -2x + 4 by x + 3 • 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 Test If 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 • f(x) = x3 + 3x2 – 8x + 3 2) f(x) = 2x 4 – 5x2 + 3x – 8
Descartes’ Rule of Signs Let 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 Rules Let 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