Section 4.2 Polynomial Functions of Higher Degree Polynomial Function P(x) P(x) = a degree 0 P(x) = ax +b (degree 1) Graph Horizontal line through (0,a) line with y intercept (0,b) and slope a P(x) = ax 2 +bx+c (degree 2) parabola with vertex ( b/2a, P( b/2a) ) Polynomials degree 3 or higher P(x) = a n x n + a n 1 x n 1 +... + a 1 x + a 0 The leading term a n x n is the dominating term Far right and far left behavior of the graph of a polynomial can be determined by looking at the leading coefficient a n. n is even n is odd a n > 0 Up to left and up to right Down to left and up to right a n < 0 Down to left and down to right Up to left and down to right Graph P(x) = 2x 4 + 3x + 1 R(x) = 3x 3 2x + 1 S(x) = 4x 6 + 2x 2 x + 1 T(x) = 2x 5 + 1 1
Turning Points The graph of a polynomial function of degree n has at most n 1 turning points turning points Relative Minimum and Relative Maximum If there is an open interval I containing c on which f(c) f(x) for all x in I, then f(c) is a relative minimum of f. f(c) f(x) for all x in I, then f(c) is a relative maximum of f. 2
Use the calculator to find the relative max and min of P(x) = x 4 6x 2 2 3
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The Zero Location Theorem Let P(x) be a polynomial and let a and b be two distinct real numbers. If P(a) and P(b) have opposite signs, then there is at least one real number c between a and b such that P(c) = 0. P P(b) (b,p(b)) P(c)=0 a c b P(a) (a,p(a)) Use the Zero Location Theorem to verify that P has a zero between a and b. P(x) = 2x 3 + 3x 2 23x 42; a=3, b=4 3 2 3 23 42 5
Even and Odd Powers of (x c) Theorem If c is a real number and the polynomial function P(x) has (x c) as a factor exactly k times, then the graph of P will if k is even, intersect but not cross the x axis at (c,0) if k is odd, cross the x axis at (c,0) Graph: P(x) = (x+1) 3 (x 2) 2 : Crosses at 1, intersects at 2 P(x) = (x 1) 3 (x 3)(x +1) 4 : Application: Advertising Expenses 6
The rational zero theorem helps us find zeros when we have large numbers that are hard to factor Rational Zero Theorem If P(x) = a n x n + + a 1 x + a 0, has integer coefficients (a n 0) and p / q is a rational zero of P, then p is a factor of the constant term a 0 and q is a factor of the leading term coefficient a n. Example : Use the Rational Zero Theorem to list all possible zeros of P(x) = 3x 3 + 11x 2 6x 8 p: ±1, ±2, ±4, ±8 q: ±1, ±3 possible rational zeros are ±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3 a n a n a 0 other real zero is not rational 7
Rational Numbers can be represented as p/q where p, q are integers. Some real numbers that are not : π, e, 2 Example: x 2 + 2x 2 = 0 p: ±1, ±2 q: ±1 p/q = ±1, ±2 Solve x 2 + 2x 2 = 0 x 2 + 2x = + 2 + 1 +1 (x + 1) 2 = 3 x = 1 ± 3 Use the Rational Zero Theorem to list the possible rational zeros. P(x) = x 3 19x 30 p = ±1,±2, ±3, ±5, ±6, ±10, ±15, ±30 q = ±1 p/q = ±1,±2, ±3, ±5, ±6, ±10, ±15, ±30 Graph x = 5 is a zero 8
Upper and Lower Bound theorem Let P(x) be a polynomial function with real coefficients and a positive leading coefficient. Use synthetic division to divide P(x) by x b, where b is a nonzero real number: Upper bound If b > 0 and all the numbers in the bottom row of the synthetic division of P by x b are either positive or zero, then b is an upper bound for the real zeros of P. Lower bound If b < 0 and all the numbers in the bottom row of the synthetic division of P by x b alternate sign (zero either pos. or neg.), then b is a lower bound for the real zeros of P. Example: P(x) = 3x 3 + 11x 2 6x 8 lower bound Not unique 6 3 11 6 8 5 3 11 6 8 Upper and Lower Bound theorem Let P(x) be a polynomial function with real coefficients and a positive leading coefficient. Use synthetic division to divide P(x) by x b, where b is a nonzero real number: Upper bound If b > 0 and all the numbers in the bottom row of the synthetic division of P by x b are either positive or zero, then b is an upper bound for the real zeros of P. Lower bound If b < 0 and all the numbers in the bottom row of the synthetic division of P by x b alternate sign (zero either pos. or neg.), then b is a lower bound for the real zeros of P. Example: P(x) = 3x 3 + 11x 2 6x 8 Upperbound 3 3 11 6 8 2 3 11 6 8 9
Descartes' Rule of Signs If P (x) = a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 be a polynomial with real coefficients. 1. The number of positive real zeros of P(x) is either equal to the number of sign changes of P (x) or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of P(x) is either equal to the number of sign changes of P ( x) or is less than that number by an even integer. If P ( x) has only one variation in sign, then f has exactly one negative real zero. 10
EXAMPLE: Using Descartes Rule of Signs Determine the possible number of positive and negative real zeros of P(x) = x 3 + 2x 2 + 5x + 4. 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for P (x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. There are three sign changes, so there are either three or one negative real zero EXAMPLE 2: Using Descartes Rule of Signs Determine the possible number of positive and negative real zeros of P(x) = 2x 4 5x 2 12. 1. P(x) sign changes (pos zeros): 2. P( x) sign changes (neg. zeros) : Factor to solve : 11
Find the Real Zeros of P(x) = 3x 3 + 11x 2 6x 8 1. Degree is 3. At most 3 zeros. 2. Descartes Rule of Signs: 1 positive real zero P( x) = 3x 3 + 11x 2 + 6x 8 2 or 0 neg. real zeros 3. Rational Zero Theorem : p factors of 8 = ±1, ±2, ±4, ±8 q factors of 3 = ±1, ±3 p/q = ±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3 4. Synthetic Division Reduce polynomial: Find the Real Zeros of P(x) = 3x 3 x 2 6x + 2 1. Degree is 3. At most 3 zeros. 2. Descartes Rule of Signs: positive real zero P( x) = neg. real zeros 3. Rational Zero Theorem : p factors of 2 = q factors of 3 = p/q = 4. Synthetic Division Reduce polynomial: 12
Advertising Expenses A company manufactures computers. the company estimates that the profit from computer sales is P(x) = 0.03x 3 + 0.02x 2 + 1.5x 2.1 where P is profit in millions of dollars and x is the amount, in hundred thousands of dollars, spent on advertising. Determine the minimum amount, rounded to the nearest thousand dollars, the company needs to spend on advertising if it is to receive a profit of $2,000,000. Section 4.4 The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra If P(x) is a polynomial function of degree n 1 with complex coefficients, then P(x) has at least one complex zero. Linear Factor Theorem If P(x) is a polynomial function of degree n 1 with leading coefficient a n 0. P(x) = a n x n + + a 1 x + a 0 then P(x) has exactly n linear factors P(x) = a n (x c 1 )(x c 2 )... (x c n ) where c 1, c 2,..., c n are complex numbers. The Number of Zeros Theorem If P(x) is a polynomial function of degree n 1 with complex coefficients, then P(x) has exactly n complex zeros, provided each zero is counted according to its multiplicity. 13
Find the Zeros and linear factors of a Polynomial Function P(x) = x 3 3x 2 + 7x 5 p: ±1, ±5 q: ±1 p/q: ±1, ±5 Graph: Try x = 1. 1 1 3 7 5 P(x) = (x 1)(x 1 2i)(x 1 + 2i) The Conjugate Pair Theorem If a + bi is a complex zero of a polynomial function with real coefficients, then the conjugate a bi is also a complex zero of the polynomial function. remember i 2 = 1 Find all the zeros of P(x) = 2x 3 5x 2 + 6x 2, given that (1 + i) is a zero. 1 + i 2 5 6 2 14
Find all the zeros of P(x) = 8x 4 2x 3 +199x 2 50x 25, given that 5i is a zero. 5i 8 2 199 50 25 40i 8 2 40i 5i, 5i, 1/4, 1/2 Find a polynomial function P(x) with real coefficients that has the indicated zeros Zeros : 3, 2i, 2i P(x) = (x 3)(x 2i)(x + 2i) Multiply = (x 3) (x 2 + 4) Zeros : 2 + 3i, 2 3i, 5, 2 15