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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 Divide x 2 +5x+6 by x+2 2) Divide x 3 – 1 by x – 1

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2.3 Real Zeros of Polynomial Functions

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  1. 2.3 Real Zeros of Polynomial Functions

  2. 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

  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

  4. Long Division of Polynomials f(x) = 6x3 – 19x2 + 16x – 4Divide by 2x2– 5x + 2

  5. Remainder Divide x2 + 3x + 5 by x + 1

  6. 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

  7. 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

  8. 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)

  9. 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)

  10. 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

  11. Examples!! f(x) = 2x3 + x2 – 5x + 2 (x + 2) and (x – 1)

  12. 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

  13. Finding the possible real zeros • f(x) = x3 + 3x2 – 8x + 3 2) f(x) = 2x 4 – 5x2 + 3x – 8

  14. 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.

  15. Using Descartes’ Rule of Signs Describe the possible real zeros of f(x) = 3x3 – 5x2 + 6x - 4

  16. 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

  17. Finding the Zeros of a Polynomial Function f(x) = 6x3 – 4x2 + 3x - 2

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