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Section 4.2 Polynomial Functions of Higher Degree

Polynomial Function P(x) Graph

P(x) = a degree 0 Horizontal line through (0,a)

P(x) = ax +b (degree 1) line with y­intercept (0,b) and slope a

P(x) = ax2+bx+c (degree 2) parabola with vertex (­b/2a, P(­b/2a) )

Polynomials degree 3 or higher

P(x) = anxn + an­1xn­1 + . . . + a1x + a0

The leading term anxn is the dominating term

Far­right and far­left behavior of the graph of a polynomial can be determined by looking at the leading coefficient an.

n is even n is odd

an > 0 Up to left and up to right Down to left and up to right

an < 0 Down to left and down to right Up to left and down to right

GraphP(x) = 2x4 + 3x + 1 R(x) = 3x3 ­ 2x + 1

S(x) = ­4x6 + 2x2 ­ x + 1 T(x) = ­2x5 + 1

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turning points

Turning Points

The graph of a polynomial function of degree n has at most n­1 turning points

Relative Minimum and Relative Maximum

If there is an open interval I containing c on whichf(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.

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Use the calculator to find the relative max and min of P(x) = x4 ­ 6x2 ­ 2

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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.

(b,P(b))

P

(a,P(a))

ba c

P(a)

P(b)

P(c)=0

Use the Zero Location Theorem to verify that P has a zero between a and b.

P(x) = 2x3 + 3x2 ­ 23x ­ 42; a=3, b=4

3 2 3 ­23 ­42

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Even and Odd Powers of (x ­ c) TheoremIf c is a real number and the polynomial function P(x) has (x­c) as a factor exactly k times, then the graph of P will

• if k is even, intersect but not cross the x­axis at (c,0)

• if k is odd, cross the x­axis at (c,0)

Graph: P(x) = (x+1)3(x­2)2 : Crosses at ­1, intersects at 2

P(x) = (x ­1)3(x ­ 3)(x +1)4 :

Application: Advertising Expenses

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The rational zero theorem helps us find zeros when we have large numbers that are hard to factor

Rational Zero Theorem If P(x) = anxn + …+ a1x + a0, has integer

coefficients (an≠ 0) and p/q is a rational zero of P, then• p is a factor of the constant term a0 and• q is a factor of the leading term coefficient an.

Example : Use the Rational Zero Theorem to list all possible zeros of

P(x) = 3x3 + 11x2 ­ 6x ­ 8

p: ±1, ±2, ±4, ±8q: ±1, ±3

possible rational zeros are ±1, ±2, ±4, ±8, ±1/3, ±2/3, ±4/3, ±8/3

other real zero is not rational

an

ana0

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Rational Numbers can be represented as p/q where p, q are integers.

Some real numbers that are not : π, e, √2

Example: x2 + 2x ­ 2 = 0

p: ±1, ±2q: ±1

p/q = ±1, ±2

Solve x2 + 2x ­ 2 = 0x2 + 2x = + 2

+ 1 +1(x + 1)2 = 3

x = ­1 ±√3

Use the Rational Zero Theorem to list the possible rational zeros.

P(x) = x3 ­ 19x ­30

p = ±1,±2, ±3, ±5, ±6, ±10, ±15, ±30q = ±1

p/q = ±1,±2, ±3, ±5, ±6, ±10, ±15, ±30

Graph x = 5 is a zero

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Upper­ and Lower­Bound theoremLet 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) = 3x3 + 11x2 ­ 6x ­ 8

3 11 ­6 ­8­6 3 11 ­6 ­8­5

Not uniquelower bound

Upper­ and Lower­Bound theoremLet 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) = 3x3 + 11x2 ­ 6x ­ 8

3 11 ­6 ­83 3 11 ­6 ­82

Upperbound

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If P (x) = anxn + an−1xn−1 + . . . + a2x2 + a1x + a0 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.

Descartes' Rule of Signs

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Determine the possible number of positive and negative real zeros of P(x) = x3 + 2x2 + 5x + 4.

EXAMPLE: Using Descartes’ Rule of Signs

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) = 2x4 − 5x2 − 12.

1. P(x) sign changes (pos zeros):

2. P(­x) sign changes (neg. zeros) :

Factor to solve :

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Find the Real Zeros of P(x) = 3x3 + 11x2 ­ 6x ­ 8

1. Degree is 3. At most 3 zeros.

2. Descartes Rule of Signs: 1 positive real zero

P(­x) = ­ 3x3 + 11x2 + 6x ­ 82 or 0 neg. real zeros

3. Rational Zero Theorem : p factors of 8 = ±1, ±2, ±4, ±8q 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) = 3x3 ­ x2 ­ 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:

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P(x) = ­0.03x3 + 0.02x2 + 1.5x ­ 2.1

Advertising Expenses A company manufactures computers. the company estimates that the profit from computer sales is

where P is profit in millions of dollars and x is the amount, in hundred­thousands 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 AlgebraIf P(x) is a polynomial function of degree n≥1 with complex coefficients, then P(x) has at least one complex zero.

Linear Factor TheoremIf P(x) is a polynomial function of degree n≥1 with leading coefficient an≠ 0.

P(x) = anxn + …+ a1x + a0then P(x) has exactly n linear factors

P(x) = an(x ­ c1)(x ­ c2) .... (x ­ cn)where c1, c2, ..., cn are complex numbers.

The Number of Zeros TheoremIf 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.

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Find the Zeros and linear factors of a Polynomial Function

P(x) = x3 ­ 3x2 + 7x ­ 5

p: ±1, ±5q: ±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 TheoremIf 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.

Find all the zeros of P(x) = 2x3 ­ 5x2 + 6x ­ 2, given that (1 + i) is a zero.

1 + i 2 ­5 6 ­2

remember i2 = ­1

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Find all the zeros of P(x) = 8x4 ­ 2x3 +199x2 ­ 50x ­ 25, given that ­5i is a zero.

­5i 8 ­2 199 ­50 ­25

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­40i

­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) (x2 + 4)

Zeros : 2 + 3i, 2 ­ 3i, ­5, 2