Question

algebra ii honors mp2 day 2 point value board
you may complete as many problems as needed to reach your goal! your goal is to earn 10 points.

1. a) given ( f(x) = x^4 - 2x^3 + 9x^2 - 6x - 10 ) and ( g(x) = x - 2 ). what is the remainder of the quotient ( \frac{f(x)}{g(x)} )? is ( x - 2 ) a factor? explain.

b) find ( k ) such that ( k^2x^2 + (-2k + 2)x - 1 ) is divisible by ( x - 1 ).
2 points (a.apr.b.2)

1. consider these functions:

( h(n) = \frac{1}{3}n - 5 )
( f(x) = h(n)x^2 + 4h(n)x ), ( n
eq 15 )
a) determine the zeros of ( f(x) ) when ( n = 6 ).
b) determine the values of ( n ) for which ( f(x) ) has a maximum value. show your work.
2 points (a.rei.a.1)

1. consider the polynomial function:

( f(x) = -(x + 2)(x + 1)(x - 1)(x - 2) )
describe the end behavior and number of maximums the function has.
1 point (f.if.b.4)

1. which of the following functions are even?

( f(x) = 6x^3 - 8x )
( g(x) = 14x^4 + 3x )
(image of two graphs here)

1. what is remainder of the quotient: ( (2x^3 - 5x^2 + 3x - 4) div (x - 2) )

1 point (f.bf.b.3)

1. list the linear factors of the following graph:

(image of a graph here)
2 points (a.apr.d.6)

1. find the zeros of the following function:

( f(x) = (x^2 - 5x + 6)(x - 4) )

1. which is a factor of ( 72x^2 - 200x^3 )?

a. ( 3x - 5 )
b. ( 2x + 3 )
c. ( 4x - 5 )
d. ( 5x + 3 )
1 point (a.apr.b.3)

1. a tennis ball was 2 feet off the ground when a tennis player hit it so that the ball traveled up in the air before coming back to the ground. the height of the tennis ball is described by the graph shown. numbers along the x - axis represent the time, in seconds, after the ball was hit, and the numbers along the y - axis represent the height, in feet, of the ball at time ( x ).

use the graph to estimate the average rate of change of the height of the ball for the first 0.25 second after being hit.
1 point (a.sse.a.2)
1 point (f.if.b.6)

Explanation:

To solve these Algebra II problems, we'll address a few of them step by step. Let's start with problem 1a:

Problem 1a:

Question: Given \( f(x) = x^4 - 2x^3 + 9x^2 - 6x - 10 \) and \( g(x) = x - 2 \). What is the remainder of the quotient \( \frac{f(x)}{g(x)} \)? Is \( x - 2 \) a factor? Explain.

Step 1: Use the Remainder Theorem

The Remainder Theorem states that the remainder of \( \frac{f(x)}{x - a} \) is \( f(a) \). Here, \( a = 2 \) (since \( g(x) = x - 2 \), so \( a = 2 \)).

Step 2: Evaluate \( f(2) \)

Substitute \( x = 2 \) into \( f(x) \):
\[

$$\begin{align*} f(2) &= (2)^4 - 2(2)^3 + 9(2)^2 - 6(2) - 10 \\ &= 16 - 2(8) + 9(4) - 12 - 10 \\ &= 16 - 16 + 36 - 12 - 10 \\ &= (16 - 16) + (36 - 12 - 10) \\ &= 0 + 14 \\ &= 14 \end{align*}$$

\]

Step 3: Determine if \( x - 2 \) is a factor

A polynomial \( x - a \) is a factor of \( f(x) \) if and only if \( f(a) = 0 \). Since \( f(2) = 14
eq 0 \), \( x - 2 \) is not a factor of \( f(x) \).

Problem 3:

Question: Find the zeros of the function \( f(x) = (x^2 - 5x + 6)(x - 4) \).

Step 1: Factor the quadratic

First, factor \( x^2 - 5x + 6 \). We need two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3:
\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]

Step 2: Find the zeros

Now, the function becomes:
\[
f(x) = (x - 2)(x - 3)(x - 4)
\]
To find the zeros, set \( f(x) = 0 \):
\[
(x - 2)(x - 3)(x - 4) = 0
\]
Using the Zero Product Property, we get:
\[
x - 2 = 0 \implies x = 2 \\
x - 3 = 0 \implies x = 3 \\
x - 4 = 0 \implies x = 4
\]

Problem 5:

Question: What is the remainder of the quotient \( (2x^3 - 5x^2 + 3x - 4) \div (x - 2) \)?

Step 1: Use the Remainder Theorem

Again, the Remainder Theorem tells us the remainder is \( f(2) \) where \( f(x) = 2x^3 - 5x^2 + 3x - 4 \).

Step 2: Evaluate \( f(2) \)

Substitute \( x = 2 \):
\[

$$\begin{align*} f(2) &= 2(2)^3 - 5(2)^2 + 3(2) - 4 \\ &= 2(8) - 5(4) + 6 - 4 \\ &= 16 - 20 + 6 - 4 \\ &= (16 - 20) + (6 - 4) \\ &= -4 + 2 \\ &= -2 \end{align*}$$

\]

Final Answers:

1. Problem 1a:

• Remainder: \( \boldsymbol{14} \)
• \( x - 2 \) is not a factor (since \( f(2)

eq 0 \)).

1. Problem 3:

• Zeros: \( \boldsymbol{x = 2, \, x = 3, \, x = 4} \)

1. Problem 5:

• Remainder: \( \boldsymbol{-2} \)

Answer:

To solve these Algebra II problems, we'll address a few of them step by step. Let's start with problem 1a:

Problem 1a:

Question: Given \( f(x) = x^4 - 2x^3 + 9x^2 - 6x - 10 \) and \( g(x) = x - 2 \). What is the remainder of the quotient \( \frac{f(x)}{g(x)} \)? Is \( x - 2 \) a factor? Explain.

Step 1: Use the Remainder Theorem

The Remainder Theorem states that the remainder of \( \frac{f(x)}{x - a} \) is \( f(a) \). Here, \( a = 2 \) (since \( g(x) = x - 2 \), so \( a = 2 \)).

Step 2: Evaluate \( f(2) \)

Substitute \( x = 2 \) into \( f(x) \):
\[

$$\begin{align*} f(2) &= (2)^4 - 2(2)^3 + 9(2)^2 - 6(2) - 10 \\ &= 16 - 2(8) + 9(4) - 12 - 10 \\ &= 16 - 16 + 36 - 12 - 10 \\ &= (16 - 16) + (36 - 12 - 10) \\ &= 0 + 14 \\ &= 14 \end{align*}$$

\]

Step 3: Determine if \( x - 2 \) is a factor

A polynomial \( x - a \) is a factor of \( f(x) \) if and only if \( f(a) = 0 \). Since \( f(2) = 14
eq 0 \), \( x - 2 \) is not a factor of \( f(x) \).

Problem 3:

Question: Find the zeros of the function \( f(x) = (x^2 - 5x + 6)(x - 4) \).

Step 1: Factor the quadratic

First, factor \( x^2 - 5x + 6 \). We need two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3:
\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]

Step 2: Find the zeros

Now, the function becomes:
\[
f(x) = (x - 2)(x - 3)(x - 4)
\]
To find the zeros, set \( f(x) = 0 \):
\[
(x - 2)(x - 3)(x - 4) = 0
\]
Using the Zero Product Property, we get:
\[
x - 2 = 0 \implies x = 2 \\
x - 3 = 0 \implies x = 3 \\
x - 4 = 0 \implies x = 4
\]

Problem 5:

Question: What is the remainder of the quotient \( (2x^3 - 5x^2 + 3x - 4) \div (x - 2) \)?

Step 1: Use the Remainder Theorem

Again, the Remainder Theorem tells us the remainder is \( f(2) \) where \( f(x) = 2x^3 - 5x^2 + 3x - 4 \).

Step 2: Evaluate \( f(2) \)

Substitute \( x = 2 \):
\[

$$\begin{align*} f(2) &= 2(2)^3 - 5(2)^2 + 3(2) - 4 \\ &= 2(8) - 5(4) + 6 - 4 \\ &= 16 - 20 + 6 - 4 \\ &= (16 - 20) + (6 - 4) \\ &= -4 + 2 \\ &= -2 \end{align*}$$

\]

Final Answers:

1. Problem 1a:

• Remainder: \( \boldsymbol{14} \)
• \( x - 2 \) is not a factor (since \( f(2)

eq 0 \)).

1. Problem 3:

• Zeros: \( \boldsymbol{x = 2, \, x = 3, \, x = 4} \)

1. Problem 5:

• Remainder: \( \boldsymbol{-2} \)