WCCUSD Algebra II Benchmark 2 Study Guide...WCCUSD Algebra II Benchmark 2 Study Guide Page 7 of 8...

WCCUSD Algebra II Benchmark 2 Study Guide

of 8 MCC@WCCUSD (WCCUSD) 11/12/15

1 Identifying Zeros from Graphs and Equations

A-APR.3

1´ You try: Which of the following functions share at least one real zero with the graphed function below? A) f (x) = x2 −3x −10 B) f (x) = x2 −16 C) f (x) = x3 − 4x D) f (x) = x4 + 4x2 − 5

Which of the following functions share at least one real zero with the graphed function below?

A)

B)

C)

D)

Solution: Recognize from the graph that all real zeros are found at the x-intercepts: , ,

and .

Factor each equation to compare zeros with the graphed function:

A, C and D have real zeros in common with the graphed function.

A)

B)

C)

D)



2 The Remainder Theorem and Polynomial Division

A-APR.2

2´ You try:

The Remainder Theorem states: For a polynomial and a number a, the remainder on division by is , so if and only if is a

factor of . In other words, if you get zero when you substitute an input value, a, into a function, then is a factor of that function. The converse of that statement is also true.

Example 1: Given a function , if , then is a factor of the function because would make equal zero. If , then what is a factor of that function?

Solution 1: is a factor of the function because would make equal zero.

Example 2: If is a factor of a polynomial , then which of these statements is true?

A) B) C) D)

Solution 2: C is the correct answer because when , the value of is zero. So, .

Example 3: Is a factor of the polynomial ?

Solution 3: If a factor, then should make the polynomial equal zero. Evaluate for to find out:

Since , then is not a factor of the polynomial. (You could also use polynomial or synthetic division…see #3 on next page.)

A. When evaluating a polynomial function, , Tommy found that

. He therefore thinks that is a factor of the polynomial. Is

he correct? Why or why not?

B. Maria thinks that if is a factor of a polynomial, then . Is she correct? Why or why not?

C. Is a factor of the polynomial ?



3 Polynomial Division

A-APR.6

3´ You try: A. Divide 2x3 + x2 −13x + 6 by x − 2 using

long division.

B. Divide 2x3 + x2 −13x + 6 by x − 2 using

synthetic division.

Example 1: Dividing Using Long Division Divide by . Solution: Notice that the linear term in the dividend is missing. Rewrite the numerator, inserting . This allows like terms to line up during the long division process.

So, .

Example 2: Dividing Using Synthetic Division Divide by . Solution: When dividing by , use a divisor that will make equal zero. This would be .

We get the same result:

.



3 cont

Polynomial Division Continued

A-APR.6

4 Solving Radical Equations

Solve x −1 = x − 7 .

Check for extraneous solutions:

A-REI.2

3´ C. Simplify 4x2 −16x2 +15x − 9

4´ You try:

Solve 3− x = x −3 .

Solution:

--Square each side. --Multiply out the right side.

--Rewrite in Standard Form.

--Factor if possible. If not, use the Quadratic Formula or Complete the Square.

Therefore, is the solution and is an extraneous solution.

Example 3: Simplifying by Factoring Simplify:



5 Solving Rational Equations

Solve 3x + 2

−1x −1

=7

x2 + x − 2

A-REI.2

5´ You try:

Solve 8x2 − 2x

−2x=

4x − 2

Solution: The trinomial can be factored to . This also happens to be the LCM.

To clear the fractions, multiply each term by the LCM.

The value of is a solution because it does not make any denominators in the original equation equal zero.



6 Transformations of Radical Functions

F-IF.7b

6´ You try:

Sketch the following functions if f (x) = x : A. f (x − 4) B. f (x)−3 C. f (x + 5)− 2 D. Name the function:

End of Study Guide

Transformation rules affect radical functions similarly to quadratic transformations. Below is the parent function .

Examples: A vertical transformation occurs when a value is added to the output.

A horizontal transformation occurs when a value is added to the input.

Both Vertical and Horizontal:



You Try Solutions: 1´ The zeros for the graphed function are -4, -1,

0, 2 and 3.

A) The zeros are 5 and -2. No match. B) The zeros are 4 and -4. Match. C) The zeros are 2, 0 and -2. Match. D) The zeros are 1, -1 and ± 5 . Match.

2´ A. Tommy is incorrect because x = −3will

not make the factor (x −3) equal zero. He should have said that (x +3) is the factor if f (−3) = 0 .

B. Maria is correct because x = −2will make the factor (x + 2) equal zero. Therefore, f (−2) = 0 .

C. If is a factor of f (x) , then

f (−3) = 0 . So we will evaluate for f (−3) .

f (x) = x3 + 2x2 − 4x −3f (−3) = (−3)3 + 2(−3)2 − 4(−3)−3f (−3) = −27+18+12−3f (−3) = 0

∴(x + 2) is a factor of f (x).

3´ A. Divide 2x3 + x2 −13x + 6 by x − 2 using long division.

B. Divide 2x3 + x2 −13x + 6 by x − 2 using synthetic division.

(x +3)

So B, C and D all share at least one real zero with the graphed function.



3´ C. Simplify:

4x2 −16x2 +15x − 9

=(2x −1)(2x +1)3(2x2 + 5x −3)

=(2x −1)(2x +1)3(2x −1)(x +3)

=2x +1

3(x +3) or 2x +1

3x + 9

4´ Solve

3− x = x −3

3− x( )2= x −3( )2

3− x = x2 − 6x + 9 0 = x2 − 5x + 6 0 = (x −3)(x − 2)

∴x = 3 and x = 2

.

Check for extraneous roots:

5´ Solve

6´ A. f (x − 4) B. f (x)−3

C. f (x + 5)− 2 D. f (x +10)+ 5 or f (x) = x +10 + 5

However, would make at least one of the denominators zero and therefore is not a solution to the equation. There is no solution.

8x2 − 2x

−2x=

4x − 2

8x2 − 2x

−2x=

4x − 2

8x(x − 2)

−2x=

4x − 2

8x(x − 2)

x(x − 2)[ ]− 2xx(x − 2)[ ] = 4

x − 2x(x − 2)[ ]

8− 2(x − 2) = 4x 8− 2x + 4 = 4x 12 = 6x 2 = x

Therefore, is the solution and is an extraneous solution.

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