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Algebra II Chapter 5 Notes: Polynomials Name Day 24: Section P-2 Exponents and Scientific Notation; Section P-3 Radicals and Exponents Properties of Exponents

The Negative-Exponent Rule: (a) Evaluate: 2−3 (b) Evaluate: 2

1

3

The Zero-Exponent Rule: (a) Evaluate: 20 (b) Evaluate: −30 (c) Evaluate: (−4)0

The Product Rule: (a) Evaluate: 23•22 (b) Evaluate: x−4•x7 (c) Evaluate: x3y5•x4y−3

The Quotient Rule: (a) Evaluate: 9

7

3

3 (b) Evaluate:

4

7

x

x

(c) Evaluate: 4 5

2 2

x y

x y

If b is any real number other than 0 and n is a natural number, then:

If b is any real number other than 0, then:

If b is any real number or algebraic expression, and m and n are integers, then:

If b is a nonzero real number or algebraic expression, and m and n are integers, then:

Products Raised to Powers: (a) Evaluate: (−3x3y)2 (b) Evaluate: (−3x3y)3

If a and b are real numbers or algebraic expressions, and n is an integer, then:

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Scientific Notation Converting from Scientific to Decimal Notation Practice: In 1-4, write each number in decimal notation: 1. 53.8 10 2. 62.431 10

3. 33.012 10 4. 29.34 10

Converting from Decimal to Scientific Notation

Practice: In 1-4, write each number in scientific notation: 1. 2,103,000 2. −0.000213 3. 0.00000010143 4. 34 Operations with Scientific Notation Practice: In 1-2, perform the indicated operation and write your final answer in scientific notation:

Quotients Raised to Powers: (a) Evaluate: 4

2

3

(b) Evaluate: 4

2

x

(c) Evaluate: 5

2

z

If a and b are real numbers, b≠0, or algebraic expressions, and n is an integer, then:

A number is written in scientific notation when it is expressed in the form 10 ,na

Where the absolute value of a is greater than or equal to 1 and less than 10 1 10 ,a and

n is an integer.

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1: 5 48.0 10 2.0 10 2: 3

8

1.1 10

5.5 10

Classwork: The following problems must be completed in order to get full credit for your note packet. Evaluate: Put the answer in simplified form.

1.) 33 24 2.) 232 = 3.) 24 4 =

4.) 2

2 38 = 5.) 5

2

3

3= 6.)

23

5

=

7.) 2

2

7

= 8.) 3

2

2

2 = 9.) 3 4

4 4

3 3

=

10.) 504 555 = 11.)

221

3

12.)

23

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13.) Brittany drove her car from Encinitas to Tijuana on Saturday at a rate of 50 miles per hour. On her way back home she hit rush hour traffic that slowed her down to only 30 miles per hour. Find the time going and returning if the time returning was four hours longer than the time going.

HW24: Pg. 333#4-40 by 4's Pg. 335: #55-59, 65-68 all

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Day 25: 5.2 Evaluate and graph polynomial functions Polynomial function: 1 2 1

1 2 1 0( ) ...n n nn n nf x a x a x a x a x a

Where: 0na and the exponents are all _____ numbers, and the coefficients are all _____ numbers.

Leading coefficient: The coefficient of the term with the highest exponent. Degree: The degree is the value of the highest exponent.

Common Polynomial Functions Degree Type Standard Form Example

0 Constant 0( )f x a ( ) 14f x

1 Linear 1 0( )f x a x a ( ) 5 7f x x

2 Quadratic 2 12 1 0( )f x a x a x a 2( ) 2 9f x x x

3 Cubic 3 3 13 2 1 0( )f x a x a x a x a 3 2( ) 3f x x x x

4 Quartic 4 3 2 14 3 2 1 0( )f x a x a x a x a x a 4( ) 2 1f x x x

Decide whether the function is a polynomial function. If so write in standard form, state its degree, type, and leading coefficient.

1. 4 21( ) 3

4h x x x 2. 2( ) 7 3g x x x

3. 2 1( ) 5 3f x x x x 4. 5( ) 2 0.6xk x x x Evaluate by direct substitution: Plug in value given for x into function wherever x is. Use direct substitution to evaluate 4 3( ) 2 5 4 8f x x x x when x=3.

4 3(3) 2(3) 5(3) 4(3) 8f = Try: #1: 4 3 2( ) 2 3 7; 2f x x x x x #2: 3 2( ) 5 6 1; 4g x x x x x

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Synthetic Substitution: Evaluate (2)f using synthetic substitution to evaluate: 5 3( ) 6 2 3 1f x x x x 1. Arrange the polynomial in descending powers, 1. 6x5 + 0x4 – 2x3 + 0x2 – 3x + 1 with a 0 coefficient for any missing term. 2. Since you want to evaluate (2)f in a box. 2. 2 6 0 −2 0 −3 1 Write the coefficients of the dividend. 3. Write the leading coefficient of the dividend 3. 2 6 0 −2 0 −3 1 on the bottom row. Bring down 6 6 4. Multiply c (in this case 2) times the value just 4. 2 6 0 −2 0 −3 1 written on the bottom row. Write the product 12 in the next column in the second row. 6 5. Add the values in this new column, writing 5. 2 6 0 −2 0 −3 1 the sum in the bottom row. 12 Add 12 6. Repeat this process of multiplications and 6. 2 6 0 −2 0 −3 1 additions until all columns are filled in. 12 24 Add 6 12 22 7. The final number is the answer to evaluating the polynomial. So (2)f = 171. 2 6 0 −2 0 −3 1 12 24 44 88 170 6 12 22 44 85 171 Practice: Synthetic substitution: How to evaluate a function more easily than direct substitution Use synthetic substitution to evaluate 4 3( ) 2 5 4 8f x x x x when x=3. Step 1: Write coefficients in descending order using zeros for terms missing. 3 2 -5 0 -4 8 Step 2: Bring down first coefficient. Multiply that by x-value. Write the product under the second coefficient. Repeat until done. Step 3: Final sum is value of f(x) when x = 3.

Multiply by 2: 2 • 6 = 12

Multiply by 2: 2 • 12= 24

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Try these: #1: 4 3 2( ) 2 3 7; 2f x x x x x #2: 3 2( ) 5 6 1; 4g x x x x x Graphs of Polynomials are Smooth and Continuous Graphs Polynomial functions of degree 2 or higher have graphs that are smooth and continuous. By smooth, we mean that the graphs contain only rounded curves with no sharp corners. By continuous, we mean that the graphs have no breaks and can be drawn without lifting your pencil from the rectangular coordinate system. End Behavior of Graphs of Polynomial Functions

Graphs of Polynomials Functions Not Graphs of Polynomials Functions Smooth rounded corner

Smooth rounded corner

Smooth rounded corner

Smooth rounded corner

Smooth rounded corner

Discontinuous: a break in the graph

Sharp Corner

Sharp Corner

Odd-degree polynomial functions have graphs with opposite behavior at each end. Even-degree polynomial functions have graphs with the same behavior at each end.

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The Leading Coefficient Test The coefficient on the term with the largest exponent on the variable is called the leading coefficient.

If the largest exponent is an odd number:

If the leading coefficient is positive, the graph falls to the left and rises to the right.

If the leading coefficient is negative, the graph rises to the left and falls to the right.

Falls Left

Rises Right

Rises Left

Falls Right

The Leading Coefficient Test (continued) The coefficient on the term with the largest exponent on the variable is called the leading coefficient.

If the largest exponent is an even number:

If the leading coefficient is positive, the graph rises to the left and rises to the right.

If the leading coefficient is negative, the graph falls to the left and falls to the right.

Odd Degree; positive leading coefficient

Odd Degree; negative leading coefficient

Rises Right Rises

Left

Falls Right

Falls Left

Even Degree; positive leading coefficient

Even Degree; negative leading coefficient

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Practice: In #1-4, use the leading coefficient test to determine the end behavior of the graph of each function: 1. 2 35 2 3f x x x x 2. 4 3 2( ) 2 3 5 20f x x x x x

3. f(x) = 7x7 4. f(x) = 5x5 −3x6 Graphing polynomial functions: Make a table of values and plot the points. Connect the points into a smooth curve and check end behaviors. Graph 3 2( ) 3 3f x x x x x -3 -2 -1 0 1 2 3 y Graph 4 3 2( ) 4 4f x x x x x -3 -2 -1 0 1 2 3 y

HW25: Pg. 333: #6-30 x4’s Pg. 341#4-58 by 4's

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Day 26: 5-3 Add, Subtract,and multiply polynomials and 5-5 Long Division of Polynomials Add polynomials: collect like terms; put answer in descending order. Ex: Add 3 22 5 3 9x x x and 3 26 11x x = (horizontal) Or 3 22 5 3 9x x x (vertical) + 3 26 0 11x x x

Ex: Subtract 3 23 2 7x x x from 3 28 5 1x x x Horizontally: ( 3 28 5 1x x x ) - ( 3 23 2 7x x x ) = Always distribute the negative first then add. Vertically: 3 28 5 1x x x

– ( 3 23 2 7x x x )

Ex: Multiply 22 3 6y y and 2y Box method keeps you better organized. Combine like terms in diagonals.

Ex: Multiply 3 binomials: 5, 1, 3x x x Multiply first two by foiling. Then multiply that product by third binomial using box. (x – 5)(x+1)= ( ) (x + 3) = For 1-3, find the product: 1.) 3 2( 3)( 2 5)x x x 2.) 2 3 ( 1)x x x

3.) 32 1x

10

For #4-5, find the sum or difference:

4.) 3 3 22 5 7 4 3 11x x x x 5.) 2 3 24 5 9 8 5 11 9x x x x x

5.5 Polynomial long division Long Division of Polynomials In #6-7, divide using long division. State the quotient, q(x) and the remainder, r(x).

1. (12x2 + x – 4) ÷ (3x – 2) 2. 5 4 3 2

3

2 8 2

2 1

x x x x

x

1. Arrange the terms of both the dividend and the divisor in descending powers of the variable. This is called standard form and include “0” coefficient for any missing terms.

2. Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient.

3. Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up.

4. Subtract the product for the dividend. 5. Bring down the next term in the original dividend and write it next to the remainder to form a

new dividend. 6. Use this new expression as the dividend and repeat this process until the remainder can no

longer be divided. This will occur when the degree of the remainder (the highest exponent on a variable in the remainder) is less than the degree of the divisor.

Helpful acronym for the steps in long division is DMSB

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3. (2a3 + a – 3) (a + 2) 4. (6x4 + 5x2 – 3x3 + 2x – 6) (3x2 – 2) Day 27: 5-4 Factor and solve polynomial equations Factoring Polynomials Greatest Common Factor (GCF) Practice: In 1-3, Factor 1. 10x3 – 4x2 2. 3x(x – 7) + 3(x – 7) 3. 5x2 + 10x + 20 Factor by Grouping Practice: In 1-2, Factor 1. x3 + 5x2 – 2x – 10 2. 3x3 – 2x2 – 9x + 6

HW26: Pg. 333: #17-31 odd Pg. 349: #4-44 by 4's Pg. 366: #4-10(even)

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Factoring Trinomials To factor a trinomial of the form ax2 + bx + c, use the X-Box method Practice: In 1-4, Factor. 1. x2 + 13x + 40 2. y2 – 5y – 14 3. 6x2 + 19x – 7 4. 2x2 – 7xy + 3y2 Factoring The Difference of Two Squares Practice: In 1-4, Factor. 1. x2 – 81 2. 36x2 – 25 3. 5y2 + 125 4. 81x4 – 16 Factoring The Sum and Difference of Two Cumbes Practice: In 1-4, Factor. 1. x3 + 27 2. 16u5 – 250u2 3. y3 + 64 4. 125 – k3

If A and B are real numbers, variables, or algebraic expressions, then: A2 – B2 = (A + B)(A – B).

In words: The difference of the squares of two terms factors as the product of the sum and a difference of those terms.

1. Factoring the Sum of Two Cubes: A3 + B3 = (A + B)(A2 – AB + B2) Example: x3 + 8 = (x)3 + (2)3 = (x + 2)(x2 – 2x + 4)

2. Factoring the Difference of Two Cubes: A3 – B3 = (A – B)(A2 + AB + B2) Example: 125x3 – 1 = (5x)3 – (1)3 = (5x – 1)(25x2 + 5x + 1)

Same Sign Different Sign

Same Sign Different Sign

Helpful Hint: S.O.A.P. which means “Same, Opposite, and Always positive”

S. O. A.P.

S. O. A.P.

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To solve --- set each factor = 0 and solve for the variable. Find the all the solutions of: 1. 3 24 40 36 0x x x 2. 2 22 24 14x x 3. 3 8 0x 4. 3 27 9 63x x x 5. 6 4 24 20 24x x x

A Strategy for Factoring a Polynomial 1. If there is a common factor, factor out the GCF. 2. Determine the number of terms the polynomial and try factoring as follows:

If there are two terms, can the binomial be factored by one of the following special forms? a. Difference of Two Squares b. Sum of Two Cubes c. Difference of Two Cubes

If there are three terms, is the trinomial a prefect square trinomial? If so factor it by one of the following special forms:

a. A2 + 2AB + B2 = (A + B)2 b. A2 – 2AB + B2 = (A – B)2

If there are four or more terms, try factoring by grouping. 3. Check to see if any factors with more than one term in the factored polynomial can be factored further. If so, factor completely.

HW27: Pg. 356#4-48 by 4's Pg. 407#1-13 odd

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Day 28: 5-5 Synthetic Division Synthetic Division: Use only when divisor is a linear expression. To divide a polynomial by x – c Example

5 32 6 2 3 1x x x x

1. Arrange the polynomial in descending powers, 1. 6x5 + 0x4 – 2x3 + 0x2 – 3x + 1 with a 0 coefficient for any missing term. 2. Write c for the divisor, x – c. To the right, 2. 2 6 0 −2 0 −3 1 Write the coefficients of the dividend. 3. Write the leading coefficient of the dividend 3. 2 6 0 −2 0 −3 1 on the bottom row. Bring down 6 6 4. Multiply c (in this case 2) times the value just 4. 2 6 0 −2 0 −3 1 written on the bottom row. Write the product 12 in the next column in the second row. 6 5. Add the values in this new column, writing 5. 2 6 0 −2 0 −3 1 the sum in the bottom row. 12 Add 12 6. Repeat this process of multiplications and 6. 2 6 0 −2 0 −3 1 additions until all columns are filled in. 12 24 Add 6 12 22 2 6 0 −2 0 −3 1 12 24 44 88 170 6 12 22 44 85 171 7. Use the numbers in the last row to write the Quotient, plus the remainder above the divisor. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in this row is the remainder.

5 32 6 2 3 1x x x x

Multiply by 2: 2 • 6 = 12

Multiply by 2: 2 • 12= 24

4 3 2 1716 12 22 44 85

2x x x x

x

Written Form 6 12 22 44 85 171

The last row of the synthetic division

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Practice: In 1-2, divide using synthetic division.

1. (x2 – 5x – 5x3 + x4) ÷ (5 + x) 2. 7 5 310 12

2

x x x

x

3. 5 33 18 15 1x x x x 4. 5 43 16 48

2

x z z

z

(FYI: Synthetic Division is the same process as synthetic substitution.) Remainder Theorem: If a polynomial is divided by a factor (x-k) then the remainder(r) is = f(k). Use synthetic division: Divide 3 2( ) 2 8 5f x x x x by 3x k = -3 -3 2 1 -8 5 2 Notice that ( 3) 16f according to the remainder theorem.

3 2( 3) 2( 3) ( 3) 8( 3) 5f Try: divide 3 2( ) 2 9 14 5f x x x x by 3x

( 3)f

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Factor theorem: If f(k) = 0 then (x-k) is a factor of f(x). When you use synthetic division you will have __ _ remainder if (x-k) is a factor of f(x). Ex#1 Given one _______ of a polynomial, find the other factors. Hint: Use synthetic division to factor the polynomial until you can factor without synthetic division. Factor 3 2( ) 3 4 28 16f x x x x completely given that 2x is a factor. -2 3 -4 -28 -16 3 Fill in variables in descending order starting with one degree lower: 23 10 8 0x x Now factor and get all zeros of function. Try: Factor 3 2( ) 2 11 3 36f x x x x completely given that 3x is a factor. Ex #2 Given one ______ of a polynomial function, find the other zeros. Hint: A zero is a solution of the polynomial. Use synthetic division with this number to break down polynomial until you can factor. Set each factor equal to zero to solve for zeros. One zero of 3 2( ) 2 23 60f x x x x is 3x . What is another zero? 3 1 -2 -23 60 1 Try: One zero of 3 2( ) 16 16f x x x x is 4. What is another zero?

HW28: Pg. 366#12-36(even) Pg. 377#52-56(even) Pg. 407: #2-16 even Quiz on Chapter 5.1-5.5 tomorrow

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Day 29: Chapter 5 Quiz Sections 1-5 Day 30: Finding Rational Zeros Rational Zero Theorem: To find all possible zeros of a function: ___all factors of constant term____ all factors of leading coefficient List possible zeros of f using rational zero theorem. 1. 3 2( ) 2 11 12f x x x x 2. 4 3 2( ) 4 3 9 10f x x x x x To find the zeros using the rational zero theorem: Find all possible zeros and then test them one at a time using synthetic division. If there is no remainder then it is a solution. Find all real zeros of 3 2( ) 8 11 20f x x x x

HW29: Pg. 359: 66-72 all Pg. 368: 49-60 all

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Try: Find all the real zeros of 3 2( ) 4 15 18f x x x x Try: Find all the zeros of 4 3 2( ) 10 11 42 7 12f x x x x x Using a graph to shorten the list of possible zeros: Try: Use the graph to shorten the list of possible rational zeros of the function. Then findl all zeros of the function: 3 2( ) 6 25 16 15f x x x x

HW30: Pg. 366#3-9(odd) Pg. 374#12-36 by 4's

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Day 31: Review Day 32: 5-7 Apply the fundamental theorem of Algebra The Fundamental Theorem of Algebra: If f(x) is a polynomial of degree n where n>0, then the equation f(x)=0 has at least one solution that is a complex number. If f(x) is a polynomial of degree n where n>0 , then it has n solutions including repeats. How many solutions does this equation have? 1. 3 25 4 20 0x x x 2. 4 3 28 18 27 0x x x Find all the zeros of a polynomial function.

5 4 3 2( ) 4 4 10 13 14f x x x x x x 1. Find all possible zeros. Test using synthetic division. 2. When you can no longer factor use quadratic formula to get imaginary solutions. Complex conjugates theorem: When you find an imaginary solution to the function there is also a complex conjugate solution to the function. Irrational conjugates theorem: When you find a irrational solution to a function there is also a irrational conjugate solution to the function If the function has a solution of 3 5i then it also has a solution of___________.

If a function has a solution of 2 3 5 then it also has a solution of __________. Use zeros to write a polynomial function

1. write the factored form of the solutions 2. Regroup when have a radical of imaginary number to get rid of radical or

imaginary part 3. Multiply and expand the binomial 4. Combine like terms

HW31: Pg 377: 1-12 all Pg 392: 51, 55-62 all

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3. Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and zeros of: 2 and 2 5i . 4. Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and zeros of: -3, 4, and 5i. 5. Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1,

and zeros of: 2, 3 , and – 3

HW32: Pg. 383#4-40 by 4's Pg 1014 #1-17 odd

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Day 33: 5-8 Analyze graphs of polynomial functions and 5-9 Writing Equations of Polynomial Functions Zero: k is a solution to a function Factor: (x – k) is a factor of the function Solution: k is a solution of the function x-intercept: if k is a real number, k is an intercept of the graph of the polynomial function. The graph passes through k on the x-axis. If the function is in factored form – you can use x-intercepts to graph the function

1. Graph the function 21( ) ( 3)( 2)

6f x x x

Plot the intercepts Plot some points between intercepts Determine end behavior and draw

ends of graph correctly 2. Try: Graph the function

2( ) ( 2)( 1)f x x x Turning Points: Local maximum: the y-coordinate of a turning point where it is higher than all nearby points Local minimum: the y-coordinate of a turning point where it is lower than all nearby points The graph of every polynomial function of degree n has at most n -1 turning points. If a polynomial function has n distinct real zeros, then its graph has exactly n – 1 turning points.

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Graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. 1. f(x) = −x4 + 16x2 b. f(x) = x2(x – 1)3(x + 2)

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5-9 Write polynomial functions Write a cubic function: ( ) ( )( )( )f x a x p x q x r

Use three given x-intercepts to write the function in factored form. Find the value of a by substituting the coordinates of the fourth point

1. Read x-intercepts off graph: 2. Write function as factors:

3. Solve for a 4. Re-write function in factored form with “a” value.

Try These, write the polynomial function in standard form of the function graphed below. 1. The function passes through (1, −6) 2. The function passes through (−1, −4) Day 34 Chapter 5 Review Day

HW34: Chapter 5 Review Worksheet Chapter 5 Test Next Class Print Ch 6 Note Packet for Wednesday

HW 33 Pg. 390: #4-48 x 4 Pg. 397: 4 – 10 even

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Algebra 2 HW#34 Name: Period:

Chapter 5 Review Sheet

Complete each of the following problems on a separate sheet of paper. Show all of your work! NO WORK = NO CREDIT! For questions 1-3, simplify each expression.

1. 2 1 5

3 1

2 2

3 10

x x y

xy y

2.

22 5

32

x y z

x

3.

2 43 2 4 7

4 2 32 8

x y x y

x x y

For questions 4-11, write each expression in standard form.

4. 3 2 3 2(2 3 5 2) ( 3 6 2)x x x x x 5. 2 3 2(4 5 1) ( 3 4)x x x x x

6. 3 4 3 2(2 6 4) (3 2 4 1)x x x x x 7. (3 4)(8 1)x x

8. 2( 1)( 2 1)x x x 9. 3 3 2( 2 1)( 5)x x x x

10. 3 4 3 2(2 )( 3 2 1)x x x x x 11. 2(2 3 )x y For 12-18, factor each expression.

12. 327 8x 13. 3125 64y 14. 3 23 5 15x x x

15. 3 23 12 2 8x x x 16. 416 81x 17. 4 25 6x x

18. 4 210 21x x For 19-20, find the real-number solutions of each equation.

19. 3 212 3 4x x x 20. 5 4 23 8 24x x x x For 21-22, divide using polynomial long division.

21. 3 2 2( 3 2 6) ( 3 1)x x x x x 22. 3 2(4 2 6 1) (2 3)x x x x For question 23-24, given one zero of the polynomial function, find the other zeros.

23. 3 2( ) 8 5 14; 2f x x x x 24. 3 2 1( ) 12 8 13 3;

2f x x x x

For questions 25-26, find all real zeros of the function.

25. 3 2( ) 6 49 20 2f x x x x 26. 4 3 2( ) 4 14 20 3f x x x x x For questions 27-28, write a polynomial function of least degree that has real coefficients, the given zeros, and a leading coefficient of 1. 27. 2, , i i 28. 2, 4 i