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Chapter 2 Review

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Chapter 2 Review

Polynomial Functions

First, some corrections on 2 slides……

CORRECTIONS on SLIDE 1

- P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0
- Evaluate by
- direct substitution or
- Remainder Theorem: When P(x) is divided by (x – a), the remainder is P(a). a.k.a Synthetic Substitution

- Zeros and Factors
- a is a zero of P(x), if P(a) = 0 [A root refers to the soln to an eqn]
- (x – a) is a factor of P(x), if P(x) divided by (x – a) leaves a zero remainder. [Factor Theorem, (x – a) is factor of P(x),iffP(a) = 0]

EXERCISES: Give real zeros of each function.

- P(x) = (x – 7)2(2x + 1)
- f(x) = x3 + 2x2 + x, Also evaluate f(-2) by direction & synthetic substitution
- g(x) = 6x2 – 7x – 20
- h(x) = x3 – 5x2 + x – 5
- k(x) = 4x4 – 17x2 + 18

CORRECTIONS

on Last Slide

L2.1–2.2: 1. 7, –1/2; 2. 0, –1,f(–2) = –2; 3. 5/2, –4/3; 4. 5; 5.

L2.3a: 1. 2. 3.

L2.3b: 1. y = (x – 1)(x – 3)2 2. y = 2(x – 1)(x – 3)2(x – 4)

3. y = ½(x – 2)3 4. y = –(x – 1)3(x – 3)

L2.4: 1. The minimum product is –9 and the numbers are –3 and 3.

2. a) The base of the box is a square with length = width = 10 – 2x.

The height is x. Volume, V(x) = x(10 – 2x)2

b) x(10 – 2x)2 > 0 so 0 < x < 5

L2.6: 1. , x = -5; 2. ; 3. x = -2,

L2.7: 1. x3 – 10x2 + 33x – 34 = 0

2. other roots: , –2 3. other roots: , 0, –6

L2.4

L2.6

??

NOW LETS START THE PRESENTATION:

Chapter 2 Review

Polynomial Functions

- P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0
- Evaluate by
- direct substitution or
- Remainder Theorem: When P(x) is divided by (x – a), the remainder is P(a). a.k.a Synthetic Substitution

- Zeros and Factors
- a is a zero of P(x), if P(a) = 0 [A root refers to the soln to an eqn]
- (x – a) is a factor of P(x), if P(x) divided by (x – a) leaves a zero remainder. [Factor Theorem, (x – a) is factor of P(x),iffP(a) = 0]

EXERCISES: Give real zeros of each function.

- P(x) = (x – 7)2(2x + 1)
- f(x) = x3 + 2x2 + x, Also evaluate f(-2) by direct & synthetic substitution
- g(x) = 6x2 – 7x – 20
- h(x) = x3 – 5x2 + x – 5
- k(x) = 4x4 – 17x2 + 18

- Leading coefficient test: if an > 0 RHS , if an < 0 RHS
- If degree of P(x) is even, both ends go in the same direction
odd, the ends go in opposite directions

- Multiplicity of zeros: (x – a)n is a factor of P(x)
- If n is even, graph is tangent to x-axis at a (touches but doesn’t cross)
- If n is odd, graph [flattens and ]crosses the x-axis at a.

- To graph: identify zeros and create a sign graph to determine intervals where graph is above and below the x-axis. Verify by examining degree and sign of an.

EXERCISES:

- Sketch graphs.
- y = –x2(x + 4)(x – 4)
- y = x(x + 3)2
- y = x3(2 – x)

3.

1.

2.

(2, 1)

(2, -4)

4.

- Write an equation for each graph

x

- Quadratic Functions – max or min occurs at vertex
Value is

- Cubic Functions – have “local” max’s and min’s.
Odd degree functions are unbounded, so have min or max points only within an interval. Use a graphing calculator to approximate max’s or min’s.

- EXERCISES:
- Two numbers have a difference of 6. Find their minimum possible product.
- An open box is to be formed by cutting squares from a 10cm square sheet of metal and then folding up the sides as shown.
- Write a function, V(x), to describe the volume of
- the resulting box.
- Find the domain of V(x).

x

10cm

- Use Quadratic Techniques (Factor, CTS, Quad Formula)
- Recast higher degree polynomials to Quadratic Form
- Rational Root Theorem – potential rational roots are of the form , where p = factors of a0 and q = factors of an.
Test roots using synthetic division and find subsequent roots using depressed polynomial.

EXERCISES: Solve.

- x3 + 5x2 – 4x – 20 = 0
- 2x4 – x2 – 3 = 0
- 3x4 + 13x3 + 15x2 – 4 = 0

- Fundamental Theorem of Algebra: a polynomial of degree n has exactly n roots in the complex number system.
- Complex and irrational roots occur in conjugate pairs.
- Odd degree polynomials have at least one real root and their graphs are unbounded.
- anxn + an-1xn-1 + … + a2x2 + a1x + a0 = 0
Sum of Roots: Product of Roots:

- Quadratics: sum = –b/a, product = c/a

EXERCISES:

- Find a cubic equation with integral coefficients that has 2 and 4 + i as roots.
- A cubic equation has no quadratic term. What are its roots if one of them is .
- A quartic (4th degree) equation has no cubic term and no constant term. What are its roots if one of them is .

L2.1–2.2: 1. 7, –1/2; 2. 0, –1,f(–2) = –2; 3. 5/2, –4/3; 4. 5; 5.

L2.3a: 1. 2. 3.

L2.3b: 1. y = (x – 1)(x – 3)2 2. y = 2(x – 1)(x – 3)2(x – 4)

3. y = ½(x – 2)3 4. y = –(x – 1)3(x – 3)

L2.4: 1. The minimum product is –9 and the numbers are –3 and 3.

2. a) The base of the box is a square with length = width = 10 – 2x.

The height is x. Volume, V(x) = x(10 – 2x)2

b) x(10 – 2x)2 > 0 so 0 < x < 5

L2.6: 1. , x = -5; 2. ; 3. x = -2,

L2.7: 1. x3 – 10x2 + 33x – 34 = 0

2. other roots: , –2 3. other roots: , 0, –6