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Plowing Through Sec. 2.4b with Two New Topics: Synthetic Division Rational Zeros Theorem Synthetic Division Synthetic Division is a shortcut method for the division of a polynomial by a linear divisor, x – k. Notes: This technique works only when dividing by a linear polynomial… It is essentially a “collapsed” version of the long division we practiced last class… Synthetic Division – Examples: 2 x 3x 5 x 12 3 Evaluate the quotient: Zero of divisor: 2 x 3 Coefficients of dividend: 3 2 –3 6 2 3 –5 –12 9 12 4 0 Remainder Quotient 2 x 3x 5 x 12 3 2 x 3 2 x 3x 4 2 Synthetic Division – Examples: 4 2 Divide x 2 x 3 x 3 by x 2 and write a summary statement in fraction form. –2 1 0 –2 –2 4 3 –3 –4 2 1 –2 –1 –1 2 x 2 x 3x 3 1 3 2 x 2x 2x 1 x2 x2 4 2 Verify Graphically? Rational Zeros Theorem Real zeros of polynomial functions are either rational zeros or irrational zeros. Examples: f x 4 x 9 2x 3 2x 3 2 The function has rational zeros –3/2 and 3/2 f x x 2 x 2 x 2 2 The function has irrational zeros – 2 and 2 Rational Zeros Theorem Suppose f is a polynomial function of degree n > 1 of the form f x an x an1 x n n 1 a0 with every coefficient an integer and a0 0. If x = p /q is a rational zero of f, where p and q have no common integer factors other than 1, then • p is an integer factor of the constant coefficient • q is an integer factor of the leading coefficient a0 , and an . RZT – Examples: Find the rational zeros of f x x 3x 1 3 2 The leading and constant coefficients are both 1!!! The only possible rational zeros are 1 and –1…check them out: f 1 1 3 1 1 1 0 3 2 f 1 1 3 1 1 3 0 3 2 So f has no rational zeros!!! (verify graphically?) RZT – Examples: 3 2 Find the rational zeros of f x 3x 4 x 5x 2 Potential Rational Zeros: Factors of –2 Factors of 3 1, 2 1, 3 1 2 1, 2, , 3 3 Graph the function to narrow the search… Good candidates: 1, – 2, possibly –1/3 or –2/3 Begin checking these zeros, using synthetic division… RZT – Examples: 3 2 Find the rational zeros of f x 3x 4 x 5x 2 1 3 3 4 –5 –2 3 7 2 7 2 0 Because the remainder is zero, x – 1 is a factor of f(x)!!! f x 3x 7 x 2 x 1 2 Now, factor the remaining quadratic… f x 3x 1 x 2 x 1 The rational zeros are 1, –1/3, and –2 RZT – Examples: Find the polynomial function with leading coefficient 2 that has degree 3, with –1, 3, and –5 as zeros. First, write the polynomial in factored form: f x 2 x 1 x 3 x 5 Then expand into standard form: f x 2 x 2 x 3 x 5 2 2 x 5 x 2 x 10 x 3x 15 3 2 2 2 x 6 x 26 x 30 3 2 RZT – Examples: Using only algebraic methods, find the cubic function with the given table of values. Check with a calculator graph. x –2 –1 1 5 f(x) 0 24 0 0 (x + 2), (x – 1), and (x – 5) must be factors… f x k x 2 x 1 x 5 f 1 24 : k 1 2 1 1 1 5 24 k 2 But we also have f x 2 x 2 x 1 x 5 2 x x 2 x 5 2 x3 8 x 2 14 x 20 2 A New Use for Synthetic Division in Sec. 2.4: Upper and Lower Bounds What are they??? A number k is an upper bound for the real zeros of f if f (x) is never zero when x is greater than k. A number k is a lower bound for the real zeros of f if f (x) is never zero when x is less than k. What are they??? Let’s see them graphically: y f x c d c is a lower bound and d is an upper bound for the real zeros of f Upper and Lower Bound Tests for Real Zeros Let f be a polynomial function of degree n > 1 with a positive leading coefficient. Suppose f (x) is divided by x – k using synthetic division. • If k > 0 and every number in the last line is nonnegative (positive or zero), then k is an upper bound for the real zeros of f. • If k < 0 and the numbers in the last line are alternately nonnegative and nonpositive, then k is a lower bound for the real zeros of f. Cool Practice Problems!!! Prove that all of the real zeros of the given function must lie in the interval [–2, 5]. f x 2x 7 x 8x 14x 8 4 3 2 The function has a positive leading coefficient, so we employ our new test with –2 and 5: 5 2 –7 –8 14 10 15 35 245 3 7 49 253 2 8 This last line is all positive!!! 5 is an upper bound –2 2 –7 –4 –8 14 8 22 –28 28 2 –11 14 –14 36 This last line has alternating signs!!! –2 is a lower bound Let’s check these results graphically… Cool Practice Problems!!! Find all of the real zeros of the given function. f x 2x 7 x 8x 14x 8 4 3 2 From the last example, we know that all of the rational zeros must lie on the interval [–2, 5]. Next, use the Rational Zero Theorem…potential rational zeros: Factors of 8 Factors of 2 1 1, 2, 4, 8 1, 2, 4, 8, 2 1, 2 Look at the graph to find likely candidates: Let’s try 4 and –1/2 Cool Practice Problems!!! Find all of the real zeros of the given function. f x 2x 7 x 8x 14x 8 4 4 2 –7 8 –8 14 8 4 –16 –8 2 –4 1 –2 3 2 0 f x x 4 2x x 4x 2 3 –1/2 2 1 –1 –4 0 –2 2 2 0 –4 0 2 1 2 f x x 4 x 2x 4 2 Cool Practice Problems!!! Find all of the real zeros of the given function. f x 2x 7 x 8x 14x 8 4 3 2 1 2 f x x 4 x 2x 4 2 1 2 f x 2 x 4 x x 2 2 1 f x 2 x 4 x x 2 x 2 2 The zeros of f are the rational numbers 4 and –1/2 and the irrational numbers are – 2 and 2 Cool Practice Problems!!! Prove that all of the real zeros of the given function lie in the 5 2 interval [0, 1], and find them. f x 10 x 3x x 6 Check our potential bounds: 0 10 0 0 0 0 –3 0 1 0 –6 0 10 0 0 –3 1 –6 0 is a lower bound!!! 1 10 0 0 –3 10 10 10 10 10 10 7 1 7 –6 8 8 2 1 is an upper bound!!! Cool Practice Problems!!! Prove that all of the real zeros of the given function lie in the 5 2 interval [0, 1], and find them. f x 10 x 3x x 6 Possible rational zeros: 1 3 1 2 3 6 1 3 1, 2, 3, 6, , , , , , , , 2 2 5 5 5 5 10 10 Check the graph (with 0 < x < 1) to select likely candidates… The function has no rational zeros on the interval!!! Are there any zeros??? Lone Real Zero: 0.951