Algebra 2 Midterm (Units 1–3)

#	Answer	Lesson Origin
1A)		17, 18, 21
1B)	$\begin{matrix} Cubic & Square Root \\ Domain & \{x \| x \in ℝ, x \leq - 5\} & \{x \| x \in ℝ, - 4 < x \leq 5\} \\ Range & \{y \| y \in ℝ, y \leq 3\} & \{y \| y \in ℝ, - 6 \leq y < - 3\} \end{matrix}$	18
1C)	$y = \sqrt[3]{(x - 2)} - 6 when x \leq 3 y = {(x + 3)}^{2} - 4 when - 6 \leq x < - 3$	19
1D)	Sample: The inverse of the piecewise function will not also be a function because: there will be repeated domain values. the graph would not pass the vertical line test (VLT). when the inverse of the domain and range values are taken, it results in repeated domain values. And, the domain cannot repeat in a function.	20, 21
2A)	$y = 3 x^{2} - 12 x + 13$	1, 27
2B)	$x = 2 \pm \frac{i \sqrt{3}}{3} or \frac{6 \pm i \sqrt{3}}{3}$	24, 25
2C)	$y = 3 {(x - 2)}^{2} + 1$	24, 27
2D)	Sample: The equation is stretched by a factor of 3. The vertex is translated 2 spaces to the right and 1 space up from the origin.	17, 18
3)	A	1
4)	B	4
5)	D	3, 4, 7
6)	B	14
7)	C	10
8)	A	15, 16
9)	C	25
10)	C	26
11)	B	5
12)	D	18
13)	C	13
14)	A	17, 18, 19, 20
15)	A	27
16)	C	2
17)	D	17, 18, 22
18)	B	24, 30
19)	B	8, 9
20)	A	20
21)	D	11, 12, 13
22)	D	5, 6
23)	C	30
24)	B	9
25)	C	11
26)	B	1
27)	A	26, 28, 29
28)	A	3
29)	C	10
30)	C	11, 15, 23
31)	A	6
32)	$(x - 4) {(2 x + 5)}^{2}$	3, 4
33)	circle ellipse parabola	24, 27, 28, 29, 30
34)	$\pm 2 i 1 \pm \sqrt{5}$	15, 16, 23, 24, 25
35)	$(2, 3) (3, 0)$	1

Complete each part of the problem.

Graph: $f (x) = \{\begin{cases} {(x + 6)}^{3} + 2 & when x \leq - 5 \\ - \sqrt{(x + 4)} - 3 & when - 4 < x \leq 5 \end{cases}$

$f (x) = {(x + 6)}^{3} + 2 when x \leq 5 (h, k) = (- 6, 2); a = 1 f (x) = - \sqrt{(x + 4)} - 3 when - 4 < x \leq 5 (h, k) = (- 4, - 3); a = - 1$

Name the types of graphs and the domain and range from the piecewise function.

$\begin{matrix} Cubic & Square Root \\ Domain & {x | x \in ℝ, x \leq - 5} & {x | x \in ℝ, - 4 < x \leq 5} \\ Range & {y | y \in ℝ, y \leq 3} & {y | y \in ℝ, - 6 \leq y < - 3} \end{matrix}$

Algebraically determine the inverse to each expression in the piecewise function.

$y = {(x + 6)}^{3} + 2 x = {(y + 6)}^{3} + 2 \sqrt[3]{(x - 2)} = \sqrt[3]{{(y + 6)}^{3}} \sqrt[3]{(x - 2)} = y + 6 y = \sqrt[3]{(x - 2)} - 6 when x \leq 3$

$y = - \sqrt{(x + 4)} - 3 x = - \sqrt{(y + 4)} - 3 x + 3 = - \sqrt{(y + 4)} - 1 (x + 3) = - 1 (- \sqrt{(y + 4)}) {(- (x + 3))}^{2} = {(\sqrt{(y + 4)})}^{2} {(x + 3)}^{2} = y + 4 y = {(x + 3)}^{2} - 4 when - 6 \leq x < - 3$

Explain whether or not the inverse of $f (x)$ will also be a function.

Sample: The inverse of the piecewise function will not also be a function because:

there will be repeated domain values.
the graph would not pass the vertical line test (VLT).
when the inverse of the domain and range values are taken, it results in repeated domain values. And, the domain cannot repeat in a function.

Note

You should have one or more of the responses listed above. 

You may also sketch the inverse on the graph to help see the results.

Complete each part of the problem.

Write the quadratic equation in standard form containing the points $(0, 13), (- 1, 28), (2.5, 1.75) .$

$y = a x^{2} + b x + c (0, 13) 13 = a {(0)}^{2} + b (0) + c c = 13 (- 1, 28) 28 = a {(- 1)}^{2} + b (- 1) + c 28 = a - b + c 28 = a - b + (13) a - b = 15 (2.5, 1.75) 1.75 = a {(2.5)}^{2} + b (2.5) + c 1.75 = 6.25 a + 2.5 b + (13) 6.25 a + 2.5 b = - 11.25$

$(a - b = 15) (2.5) \Rightarrow 2.5 a + 2.5 b = 37.5 + 6.25 a - 2.5 b = - 11.25 8.75 a = 26.25 a = 3 a - b = 15 (3) - b = 15 - b = 12 b = - 12 y = 3 x^{2} - 12 x + 13$

Calculate the x-intercepts of the parabola using the equation from part A. Write the answer in simplified radical form.

$a = 3, b = - 12, c = 13 x = \frac{- (- 12) \pm \sqrt{{(- 12)}^{2} - 4 (3) (13)}}{2 (3)} x = \frac{12 \pm \sqrt{144 - 156}}{6} = \frac{12 \pm \sqrt{- 12}}{6} x = \frac{12 \pm 2 i \sqrt{3}}{6} x = 2 \pm \frac{i \sqrt{3}}{3} or \frac{6 \pm i \sqrt{3}}{3}$

Write the quadratic equation from part A in vertex form.

$3 x^{2} - 12 x + 13 = 0 3 x^{2} - 12 x = - 13 3 (x^{2} - 4 x + {(\frac{- 4}{2})}^{2}) = - 13 + 3 {(\frac{- 4}{2})}^{2} 3 {(x - 2)}^{2} = - 13 + 3 (4) 3 {(x - 2)}^{2} = - 13 + 12 3 {(x - 2)}^{2} = - 1 3 {(x - 2)}^{2} + 1 = 0 y = 3 {(x - 2)}^{2} + 1$

Describe the transformation of the quadratic equation from its parent equation.

Sample: The equation is stretched by a factor of 3. The vertex is translated 2 spaces to the right and 1 space up from the origin.

Multiple Choice

When a region is unbounded for linear programming, it means:

the region will not be completely enclosed and will continue infinitely in at least one direction.
the region does not exist.
the region will be enclosed in all directions.
nothing is known about the problem.

Note

B, C, and D do not correctly describe an unbounded region.

Determine the expression that, when set equal to ${(a x)}^{3} + {(b y)}^{3},$ would form a polynomial identity.

$(a x + b y) ({(a x)}^{2} - a b x y - {(b y)}^{2})$
$(a x + b y) ({(a x)}^{2} - a b x y + {(b y)}^{2})$
$(a x + b y) (a x^{2} - a b x y + b y^{2})$
$(a x - b y) ({(a x)}^{2} + 2 a b x y + {(b y)}^{2})$

This is the polynomial identity for the sum of cubes.

Note

The signs in the second expression are incorrect.
The coefficients a and b also need to be squared.
This option represents the difference of cubes. The formula for the sum of cubes is $(a^{3} + b^{3}) = (a + b) (a^{2} - a b + b^{2})$ .

State the restrictions on the denominator: $\frac{6 x^{2} + 7 x + 2}{2 x^{2} - 5 x - 12} \div \frac{3 x + 2}{3 x + 12}$

$x \neq - \frac{3}{2}, - \frac{2}{3}, 4$
$x \neq \pm 4, \frac{2}{3}, \frac{3}{2}$
$x \neq - \frac{3}{2}, \pm 4$
$x \neq - \frac{3}{2}, - \frac{2}{3}, \pm 4$

Note

A, C) These options ignore that both the numerator and denominator of the divisor are part of the restrictions.

These values have the opposite signs.

Solve: $\sqrt{4 x + 8} - 6 \leq 2$

Note

This option does not consider the restrictions to the solution.
This option only represents the values that make the radical greater than or equal to zero.
This option is the solution if the restriction inequality symbol is a less than or equal to sign.

$Restrictions \sqrt{4 x + 8} \geq 0 {(\sqrt{4 x + 8})}^{2} \geq {(0)}^{2} 4 x + 8 \geq 0 4 x \geq - 8 x \geq - 2 Solution \sqrt{4 x + 8} \leq 8 {(\sqrt{4 x + 8})}^{2} \leq {(8)}^{2} 4 x + 8 \leq 64 4 x \leq 56 x \leq 4$

Select the equation that best matches the graph.

$y = \frac{1}{x + 2} + 1$
$y = \frac{1}{x - 2} + 1$
$y = - \frac{1}{x + 2} + 1$
$y = - \frac{1}{x - 2} + 1$

$a = - 1, h = - 2, k = 1$
Vertical asymptote: x = –2
Horizontal asymptote: y = 1

Note

A, B) When the value of a is positive, the graph of a rational function is in quadrants 1 and 3 (using the asymptotes to determine quadrants).
B, D) With $(x - 2),$ the vertical asymptote is $x = 2 .$

Simplify: $\frac{- 2 i}{4 - 5 i}$

$\frac{10 - 8 i}{41}$
$\frac{10 + 8 i}{9}$
$\frac{- 2 i}{41}$
$\frac{10 - 8 i}{- 9}$

$\frac{- 2 i}{4 - 5 i} (\frac{4 + 5 i}{4 + 5 i}) \frac{- 8 i - 10 i^{2}}{16 - 25 i^{2}} \frac{- 8 i - 10 (- 1)}{16 - 25 (- 1)}$

Note

This option replaces $i^{2}$ with 1 rather than –1.
This option only multiplies the denominator by the conjugate.
This option has $+ 25 i^{2}$ in the denominator.

Determine the classification of the solutions to the quadratic equation: $- x^{2} + 5 x - 7 = 0$

Two rational solutions
One rational solution
Two complex solutions
No solution

$a = - 1, b = 5, c = - 7 b^{2} - 4 a c {(5)}^{2} - 4 (- 1) (- 7) 25 - 28 - 3$

Note

The discriminant is > 0 when there are two rational solutions.
The discriminant is = 0 when there is one rational solution.
When a quadratic has no real solutions, there are two complex solutions.

Calculate the distance between $(5, - 2)$ and $(- 7, - 4) .$

6.3
11.8
12.2
148

$d = \sqrt{{(5 - (- 7))}^{2} + {(- 2 - (- 4))}^{2}} d = 12.1655$

Note

Subtraction is not used in the formula.
The squared $x$ ‐ and y‐values are subtracted rather than added.
This option is the value before the square root is taken.

Divide $2 x^{3} - 3 x^{2} - 7 x + 2 by x^{2} - 3 x + 1 .$

$2 x - 9 + \frac{22 x - 6}{x^{2} - 3 x + 1}$
$2 x + 3$
$2 x + 3 - \frac{18 x + 6}{x^{2} - 3 x + 1}$
$x - 32$

$x^{2} - 3 x + 1 2 x + 3 2 x^{3} - 3 x^{2} - 7 x + 3 - (2 x^{3} - 6 x^{2} + 2 x) 3 x^{2} - 9 x + 3 - (3 x^{2} - 9 x + 3) 0$

Note

This option results when only the first term of each expression is subtracted.
The first step is completed correctly, but only the first terms are subtracted in the second step.
This option has a sign error for the constant.

Select the equation that best matches the graph.

$y = 2 \sqrt[3]{x + 6} - 4$
$y = 2 {(x + 6)}^{3} - 4$
$y = 2 \sqrt[3]{x - 6} - 4$
$y = 2 {(x - 6)}^{3} - 4$

Cubic equation, $a = 2, h = 6, k = - 4$

Note

A, C) The equation is not a cube root.
A, B) The value of h is not –6.

Solve: ${(x + 5)}^{\frac{2}{3}} - 7 = - 3$

3
± 3
–13, 3
61

${(x + 5)}^{\frac{2}{3}} = 4 {({(x + 5)}^{\frac{2}{3}})}^{\frac{3}{2}} = {(4)}^{\frac{3}{2}} \begin{matrix} \end{matrix} |x + 5| = 8 \begin{array}{lcl} Case 1 & Case 2 \\ x + 5 = 8 & - (x + 5) = 8 \\ x = 3 & x + 5 = - 8 \\ x = - 13 \end{array}$

Note

Case 2 is missing.
Five is added to –8 rather than subtracted.
The square root of both sides is not taken in Case 1.

Select the graph that represents the inverse of the function: $j (x) = \sqrt[3]{7 - x} - 3$

Note

B, C) Both options incorrectly have $k = - 7$ because the negative is not distributed correctly.

A parabola is the inverse of a square root, not a cube root.

$x = \sqrt[3]{7 - y} - 3 x + 3 = \sqrt[3]{7 - y} {(x + 3)}^{3} = {(\sqrt[3]{7 - y})}^{3} {(x + 3)}^{3} = 7 - y {(x + 3)}^{3} - 7 = - y y = - {(x + 3)}^{3} + 7 a = - 1, h = - 3, k = 7$

Select the parabola in the form $x = y^{2}$ , reflected over the y-axis and translated right eight spaces.

$x = - y^{2} + 8$
$x = - {(y + 8)}^{2}$
$x = - {(y - 8)}^{2}$
$x = - y^{2} - 8$

$x = a {(y - k)}^{2} + h a = - 1 k = 0 h = 8 x = - {(y)}^{2} + 8$

Note

This option shifts the graph down eight spaces.
This option shifts the graph up eight spaces.
This option shifts the graph left eight spaces.

Solve the system. Then calculate the product of x and the sum of y and z for the system:
$2 x - 3 y - 3 z = 22 2 x + y + z = 14$

16
8
–16
–2

$\begin{array}{rcl} \begin{matrix} (3) (2 x + y + z = 14) \Rightarrow 2 x - 3 y - 3 z = 22 + 6 x + 3 y + 3 z = 42 8 x = 64 x = 8 \end{matrix} \\ 2 (8) + y + z & = & 14 \\ 16 + y + z & = & 14 \\ y + z & = & - 2 \\ x (y + z) \\ 8 (- 2) & = & - 16 \end{array}$

Note

This option is the value of 2x.
This option is the value of x.
This option is the value of (y + z).

Determine the system that best represents the graph.

$y = |x + 5| - 3 y = - \sqrt{x + 4}$
$y < |x + 5| - 3 y \geq - \sqrt{x + 4}$
$y > |x + 5| - 3 y \geq \sqrt{- x + 4}$
$y > |x + 5| - 3 y \geq - \sqrt{x + 4}$

Note

The graph represents a system of inequalities.
The absolute value inequality symbol is incorrect.
The negative sign belongs outside of the radical for this graph.

Write the conic section equation
$x^{2} - 5 y^{2} - 10 y = 15$ in standard form.

$\frac{{(x + 1)}^{2}}{2} - \frac{y^{2}}{10} = 1$
$\frac{x^{2}}{10} - \frac{{(y + 1)}^{2}}{2} = 1$
$\frac{{(x + 1)}^{2}}{10} - \frac{y^{2}}{2} = 1$
$\frac{x^{2}}{10} + \frac{{(y + 1)}^{2}}{2} = 1$

$A = 1, C = - 5, hyperbola x^{2} - 5 (y^{2} + 2 y + {(\frac{2}{2})}^{2}) = 15 - 5 {(\frac{2}{2})}^{2} x^{2} - 5 {(y + 1)}^{2} = 15 - 5 \frac{x^{2}}{10} - \frac{5 {(y + 1)}^{2}}{10} = \frac{10}{10} \frac{x^{2}}{10} - \frac{{(y + 1)}^{2}}{2} = 1$

Note

A, C) The values of h, k, a, and b are substituted into the standard form incorrectly.

This option is an equation of an ellipse, and the given equation is a hyperbola.

Solve: $\frac{x - 4}{3 x} = \frac{x - 1}{x + 3} - 1$

–12
–12, 1
1
–1, 12

$LCD : 3 x (x + 3) (\frac{x + 3}{x + 3}) (\frac{x - 4}{3 x}) = (\frac{3 x}{3 x}) (\frac{x - 1}{x + 3}) - 1 (\frac{3 x (x + 3)}{3 x (x + 3)}) x^{2} - 4 x + 3 x - 12 = 3 x^{2} - 3 x - (3 x^{2} + 9 x) x^{2} - x - 12 = 3 x^{2} - 3 x - 3 x^{2} - 9 x x^{2} - x - 12 = - 12 x x^{2} + 11 x - 12 = 0 (x - 1) (x + 12) = 0 x = 1, - 12$

Note

A, C) These options are partially correct, but neither includes the entire solution.

This option is not factored correctly because the signs are switched.

Name the range for the inverse of the graph in interval notation.

$[1, \infty)$
$[2, \infty)$
$(1, \infty)$
$(2, \infty)$

Given point (1, 2)
Inverse (2, 1)

Note

This option is the range of the given graph, or the domain of the inverse.

C, D) Because the graph starts at a point, a left parenthesis is not correct notation.

The area of a rectangle is $29 - 7 \sqrt{17}$ square yards. The length of the rectangle is $4 - \sqrt{17}$ yards. Find the width of the rectangle.

$3 + 57 \sqrt{17}$
$3 + \sqrt{17}$
$- 29 + 7 \sqrt{17}$
$3 - \sqrt{17}$

Note

The negative symbol is dropped when multiplying by the conjugate.
The negative is not distributed to all terms.
The numerator is not multiplied by the conjugate.

$A = l w; w = \frac{A}{l} A = 29 - 7 \sqrt{17}, l = 4 - \sqrt{17} w = \frac{29 - 7 \sqrt{17}}{4 - \sqrt{17}} w = \frac{29 - 7 \sqrt{17}}{4 - \sqrt{17}} (\frac{4 + \sqrt{17}}{4 + \sqrt{17}}) w = \frac{116 + 29 \sqrt{17} - 28 \sqrt{17} - 7 (17)}{- 1} = \frac{116 + \sqrt{17} - 119}{- 1} w = - (- 3 + \sqrt{17}) w = 3 - \sqrt{17}$

Determine the value of $P (- 2)$ if $P (x) = x^{4} + 8 x^{2} + 23 x - 6 .$

P(–2) = –28
P(–2) = 80
P(–2) = 88
P(–2) = –4

Note

A, B) The third degree term $(0 x^{3})$ is missing when synthetic division is used.

B, C) Two is used in synthetic division rather than –2.

Determine the equation of the conic section from the given graph.

${(x - 2)}^{2} + {(y + 2)}^{2} = 10$
${(x + 2)}^{2} + {(y - 2)}^{2} = 10$
${(x - 2)}^{2} + {(y + 2)}^{2} = 25$
${(x + 2)}^{2} + {(y - 2)}^{2} = 25$

Center (2, –2)
r = 5
$r^{2} = 25$

Note

A, B) $r^{2} = 25$ , not 10

B, D) The values of h and k have the incorrect signs.

The sum of the reciprocals of two numbers is equal to $\frac{2}{3}$ . The difference between the two numbers is 4. Determine the solution if all values are whole numbers.

1, 6
2, 6
–3, 1
0, 4

Note

These options are the values of x when the equation is factored, but together they do not make the equation true.
This option is a solution, but –1 is not a natural number.
These options are the domain restrictions.

$\frac{1}{x} + \frac{1}{y} = \frac{2}{3} x - y = 4 x = y + 4 (2 x y) (\frac{1}{x} + \frac{1}{y} = \frac{2}{3}) \Rightarrow 3 y + 3 x = 2 x y 3 y + 3 (y + 4) = 2 y (y + 4) 3 y + 3 y + 12 = 2 y^{2} + 8 y 6 y + 12 = 2 y^{2} + 8 y 0 = 2 y^{2} + 2 y - 12 0 = 2 (y + 3) (y - 2) y = - 3, 2 y = 2 x = 6$

Simplify: ${(\frac{3}{4} x^{7} y^{10} z^{9})}^{\frac{1}{2}}$

$\frac{3}{8} x^{\frac{7}{3}} y^{5} z^{\frac{9}{2}}$
$| x^{3} y^{5} | z^{4} \sqrt{3 x z}$
$\frac{| x^{3} y^{5} | z^{4} \sqrt{3 x z}}{2}$
$\frac{x^{3} y^{5} z^{4} \sqrt{3 x z}}{2}$

$\frac{3^{\frac{1}{2}}}{4^{\frac{1}{2}}} x^{\frac{7}{2}} y^{\frac{10}{2}} z^{\frac{9}{2}} \frac{\sqrt{3}}{2} x^{3 \frac{1}{2}} y^{5} z^{4 \frac{1}{2}}$

Note

This option shows the coefficient multiplied by one-half.
The denominator is missing from the expression.
The absolute value bars are missing around the terms with odd exponents.

Use the objective function $f (x, y) = 2 y - 3 x$ to determine the maximum for the graphed system.

(0, 0)
(0, 6)
(3, 4)
(7, 0)

$f (x, y) = 2 y - 3 x f (0, 0) = 2 (0) - 3 (0) = 0 f (0, 6) = 2 (6) - 3 (0) = 12 f (3, 4) = 2 (4) - 3 (3) = - 1 f (7, 0) = 2 (0) - 3 (7) = - 21$

Note

A, C, D) These options are vertices of the system but do not result in the maximum.

Determine the midpoint between the centers of the conic sections: ${(x + 2.5)}^{2} + {(y - 9.8)}^{2} = 10 and \frac{{(x - 3.4)}^{2}}{6} + \frac{{(y - 7)}^{2}}{5} = 1$

(0.45, 8.4)
(2.95, 8.4)
(–0.45, –8.4)
(8, 7.5)

$Circle (- 2.5, 9.8) Ellipse (3.4, 7) Midpoint (\frac{- 2.5 + 3.4}{2}, \frac{9.8 + 7}{2}) = (0.45, 8.4)$

Note

The value 2.5 was used instead of –2.5.
The signs for the centers of both conics are reversed.
The values of a and b were used as center points.

Factor: $4 x^{3} - 32 y^{3}$

$4 (x - 2 y) (x^{2} + 2 x y + 4 y^{2})$
$- 4 (x + 2 y) (x^{2} + 2 x y + 4 y^{2})$
$4 (x - 2 y) (x^{2} - 2 x y + 4 y^{2})$
$4 (x + 2 y) (x^{2} - 2 x y + 4 y^{2})$

$4 (x^{3} - 8 y^{3}) 4 (x - 2 y) (x^{2} + 2 x y + 4 y^{2})$

Note

B, C) These options have an error in factoring the difference of cubes.

This option is the sum of cubes.

Select the rational function with a domain of all real numbers except 3 and a range of all real numbers except –2.

$f (x) = \frac{1}{x + 3} - 2$
$q (x) = - \frac{3}{x} + 2$
$m (x) = \frac{1}{x - 3} - 2$
$r (x) = \frac{1}{x - 3} + 2$

Domain: x ≠ 3
Range: y ≠ –2

Note

Domain: x ≠ –3, Range: y ≠ –2
Domain: x ≠ 0, Range: y ≠ 2
Domain: x ≠ 3, Range: y ≠ 2

Solve $x^{2} + \frac{7}{25} = 0$ under the set of complex numbers.

$\frac{\sqrt{7}}{5}$
$\frac{\pm \sqrt{7}}{5}$
$\frac{\pm i \sqrt{7}}{5}$
no solution

Note

A, B) These options are missing part of the solution.

A solution is possible because you are working with complex numbers.

$x^{2} = - \frac{7}{25} \sqrt{x^{2}} = \pm \sqrt{- \frac{7}{25}} x = \pm i \frac{\sqrt{7}}{\sqrt{25}} x = \frac{\pm i \sqrt{7}}{5}$

Calculate the value of n when $P (2) = - 1$ and $P (x) = x^{4} - 3 x^{2} + 4 x + n .$

–13
8
–5
11

Note

This option is the value of n for $P (- 1) = 2 .$
This option occurs when the coefficient 0 is not placed in the problem.
This option occurs when 12 and –1 are added.

$\begin{array}{rcl} n + 12 & = & - 1 \\ n & = & - 13 \end{array}$

Select all that apply.

Select the expressions that, when multiplied, will form a polynomial identity to $4 x^{2} (x + 1) - 5 (11 x + 20)$ .

$(x + 1)$
$(x - 4)$
${(2 x + 5)}^{2}$
$(11 x + 20)$

Note

The first and last options are expressions in the given problem, but the product does not result in a polynomial identity with the given expression.

$4 x^{2} (x + 1) - 5 (11 x + 20) 4 x^{3} + 4 x^{2} - 55 x - 100 (x - 4) {(2 x + 5)}^{2} (x - 4) (4 x^{2} + 20 x + 25) 4 x^{3} + 20 x^{2} + 25 x - 16 x^{2} - 80 x - 100 4 x^{3} + 4 x^{2} - 55 x - 100$

The process of completing the square can be used to write the equations for ____ in standard form.

circle
ellipse
parabola
rational

Note

A rational equation uses long division to write the equation in standard form.

Name all possible roots for the equation: $(4 x^{2} + 16) (x^{2} - 2 x - 4) = 0$ .

$\pm 2 i$
$\pm 4 i$
$1 \pm \sqrt{5}$
$1 \pm i \sqrt{3}$

Note

The second option did not factor out the GCF from the first set of parentheses.

The last option subtracted 4 and 16 inside of the radical rather than adding.

$\begin{matrix} 4 x^{2} + 16 = 0 4 (x^{2} + 4) = 0 x^{2} + 4 = 0 x^{2} = - 4 x = \pm 2 i & x = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (1) (- 4)}}{2 (1)} x = \frac{2 \pm \sqrt{20}}{2} x = \frac{2 \pm 2 \sqrt{5}}{2} x = 1 \pm \sqrt{5} \end{matrix}$

Select all vertices for the solution to the system of inequalities.

(2, 3)
(0, 9)
$(2, \frac{2}{3})$
(3, 0)

Note

The remaining ordered pairs are intersection points but not part of the region that represents the solution.