Question 1

Express $f ( x )$ in the form $a _ { n } x ^ { n } + a _ { n - 1 } x ^ { n - 1 } + \ldots + a _ { 1 } x + a _ { 0 }$ . Find a quadratic function that has a maximum value of 2 and that has - 2 and 4 as zeros.

A)

f ( x ) = - \frac { 4 } { 9 } x ^ { 2 } + \frac { 2 } { 9 } x - \frac { 16 } { 9 }

B)

f ( x ) = - \frac { 9 } { 2 } x ^ { 2 } - \frac { 16 } { 9 } x + \frac { 4 } { 9 }

C)

f ( x ) = - \frac { 2 } { 9 } x ^ { 2 } + \frac { 4 } { 9 } x + \frac { 16 } { 9 }

D)

f ( x ) = \frac { 2 } { 9 } x ^ { 2 } - \frac { 9 } { 4 } x - \frac { 16 } { 9 }

f ( x ) = - \frac { 2 } { 9 } x ^ { 2 } + \frac { 4 } { 9 } x + \frac { 16 } { 9 }

Accepted Answer

The quadratic function with zeros at -2 and 4 can be expressed as $f(x) = a(x + 2)(x - 4)$. To have a maximum value of 2, the vertex form is $f(x) = a(x - 1)^2 + 2$. Solving for $a$ using the zeros, we find $a = -\frac{2}{9}$. Thus, the function is $f(x) = -\frac{2}{9}x^2 + \frac{4}{9}x + \frac{16}{9}$.

Question 2

Determine the constants (denoted with capital letters) so that the equation is an identity. $\frac { 8 x + 3 } { ( x + 3 ) ^ { 2 } } = \frac { A } { x + 3 } + \frac { B } { ( x + 3 ) ^ { 2 } }$&#10;A) $\begin{array} { l } &#10;A = 7 \&#10;B = 21&#10;\end{array}$&#10;B) $\begin{array} { l } &#10;A = 7 \&#10;B = 22&#10;\end{array}$&#10;C) $\begin{array} { l } &#10;A = - 8 \&#10;B = - 21&#10;\end{array}$&#10;D) $\begin{array} { l } &#10;A = 8 \&#10;B = - 21&#10;\end{array}$&#10;E) $\begin{array} { l } &#10;A = - 8 \&#10;B = 21&#10;\end{array}$

Accepted Answer

Using partial fraction decomposition, we can write:
$$\frac{8x+3}{(x+3)^2}=\frac{A}{x+3}+\frac{B}{(x+3)^2}$$
Multiplying both sides by $(x+3)^2$ gives:
$$8x+3=A(x+3)+B$$
Expanding and grouping like terms, we get:
$$(A+8)x+3-3A=B$$
Since this equation must hold for all values of $x$, we can equate the coefficients of $x$ and the constants on both sides. This leads to the system of equations:
\begin{align*} A+8 &= 8 \ 3-3A &= -21 \end{align*}
Solving for $A$ and $B$, we get $A=8$ and $B=-21$, so the answer is $	extbf{(D)}$.

Question 3

Find a quadratic equation with rational coefficients, one of whose roots is the given number. Write your answer so that the coefficient of
$x ^ { 2 }$ is 1. $r _ { 1 } = 1 + \sqrt { 2 }$

A)

x ^ { 2 } - 3 x - 1 = 0

B)

x ^ { 2 } - 2 x - 8 = 0

C)

x ^ { 2 } - 2 x - 1 = 0

D)

x ^ { 2 } - 2 x - 9 = 0

E)

x ^ { 2 } - 2 x - 3 = 0

x ^ { 2 } - 2 x - 1 = 0

Accepted Answer

Since one of the roots is 1 + sqrt(2), the other root must be its conjugate, which is 1 - sqrt(2). The sum of the roots is 2, and the product of the roots is (1 + sqrt(2))(1 - sqrt(2)) = -1. Therefore, the quadratic equation with rational coefficients is:

(x - (1 + sqrt(2)))(x - (1 - sqrt(2))) = x^2 - 2x - 1 = 0

We can write this as x^2 - 2x - 1 = 0 by dividing both sides by the leading coefficient of 1. Hence, the best choice is C.

Question 4

List the distinct roots of the following equation. For both repeated and single roots, specify their multiplicity. Enter $\left( r _ { 1 } , m _ { 1 } \right) , \left( r _ { 2 } , m _ { 2 } \right)$ , ... where $r _ { 1 } , r _ { 2 }$ etc. are the roots of the polynomial and $m _ { 1 }$ is the multiplicity of $r _ { 1 } , m _ { 2 }$ is the multiplicity of $r _ { 2 }$ etc. $( x - 3 ) ( x - 2 ) ^ { 6 } ( x - 9 ) = 0$

A) (3, 1), (2, 4), (9, 1)
B) (3, 1), (2, 1), (9, 1)
C) (3, 1), (2, 5), (9, 1)
D) (3, 1), (2, 6), (9, 1)
E) (3, 4), (2, 4), (9, 4)

Accepted Answer

By the fundamental theorem of algebra, we know that a polynomial of degree n has exactly n roots (counting multiplicities). Therefore, we need to factor the given polynomial to determine its roots and their multiplicities.
We can distribute and simplify the given polynomial as follows: 
$$(x-3)(x-2)^6(x-9) = (x-3)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-2)(x-9)$$
From this, we can see that the roots and their multiplicities are:
$$\left(3,1\right), \left(2,6\right), \left(9,1\right)$$
Therefore, the answer is choice D.

Question 5

Use long division to find the quotient and the remainder. $\frac { 2 x ^ { 6 } + 2 } { x - 1 }$&#10;A) $\text { quotient: } 2 x ^ { 5 } - 2 x ^ { 4 } + 2 x ^ { 3 } - 2 x ^ { 2 } + 2 x - 2 \text {; remainder: } 0$&#10;B) $\text { quotient: } 2 x ^ { 5 } + 2 x ^ { 4 } + 2 x ^ { 3 } + 2 x ^ { 2 } + 2 x + 2 \text {; remainder: } 4$&#10;C) $\text { quotient: } 2 x ^ { 5 } + 2 x ^ { 3 } + 2 x + 2 ; \text { remainder: } 2$&#10;D) $\text { quotient: } 2 x ^ { 5 } + 2 x ^ { 4 } - 2 x ^ { 3 } + 2 x ^ { 2 } - 2 x + 2 \text {; remainder: } 4$&#10;E) $\text { quotient: } 2 x ^ { 5 } + 2 x ^ { 4 } + 2 x ^ { 3 } + 2 x ^ { 2 } + 2 x + 2 \text {; remainder: } 0$

Accepted Answer

The answer of Use long division to find the quotient...

Question 6

Find the rational roots and solve the equation. 4x³ + x² - 20x - 5 = 0 A) $\pm 5 , - \frac { 1 } { 4 }$ B) $- \frac { 1 } { \sqrt { 5 } } , \pm 4$ C) $\pm \sqrt { 5 } , - 4$ D) $\pm \sqrt { 5 } , - \frac { 1 } { 4 }$ E) $\pm \sqrt { 5 } , \frac { 1 } { 4 }$

Accepted Answer

By rational root theorem, the possible rational roots are $\pm 1, \pm 5, \pm \frac { 1 } { 2 }, \pm \frac { 1 } { 4 }$. Testing these values, we find that $x = - \frac { 1 } { 4 }$ is a root. Dividing $4x^3 + x^2 - 20x - 5$ by $4x + 1$, we get $4x^2 - 15x - 5$. Factoring this quadratic, we get $(4x + 5)(x - 1)$. Therefore, the other roots are $x = - \frac { 5 } { 4 }$ and $x = 1$.

Question 7

Use synthetic division to find the quotient and the remainder. $\frac { 4 x ^ { 4 } + 16 x ^ { 3 } + 24 x ^ { 2 } + 16 x + 4 } { x + 1 }$&#10;A) $\text { quotient: } 4 x ^ { 3 } + 20 x ^ { 2 } + 12 x + 4 \text {; remainder: } 0$&#10;B) $\text { quotient: } 4 x ^ { 3 } + 12 x ^ { 2 } + 12 x + 4 \text {; remainder: } 0$&#10;C) $\text { quotient: } 4 x ^ { 3 } + 4 x ^ { 2 } + 12 x - 12 \text {; remainder: } 3$&#10;D) $\text { quotient: } 4 x ^ { 3 } + 12 x ^ { 2 } + 12 x + 4 \text {; remainder: } 3$&#10;E) $\text { quotient: } 4 x ^ { 3 } - 12 x ^ { 2 } + 4 x + 12 \text {; remainder: } 0$

Accepted Answer

The synthetic division table will be as follows:

\begin{array}{c|ccccc} -1 & 4 & 16 & 24 & 16 & 4 \ \hline & & -4 & -12 & -12 & -4  \ & 4 & 12 & 12 & 4 & \ \end{array}

Therefore, the quotient is $4x^3 + 12x^2 + 12x + 4$, and the remainder is $0$. Option $(B)$ is the correct answer.

Question 8

Determine the constants (denoted by capital letters) so that the equation is an identity. $\frac { x - 17 } { x \left( x ^ { 2 } + 2 \right) } = \frac { A } { x } + \frac { B x + C } { x ^ { 2 } + 2 }$

A)

\begin{array} { l } A = 8 \frac { 1 } { 2 } \\B = 1 \\C = 8 \frac { 1 } { 2 }\end{array}

.
B)

\begin{array} { l } A = - 8 \frac { 1 } { 2 } \\B = 8 \frac { 1 } { 2 } \\C = 1\end{array}

C)

\begin{array} { l } A = 8 \frac { 1 } { 2 } \\B = 8 \frac { 1 } { 2 } \\C = - 1\end{array}

.
D)

\begin{array} { l } A = - 8 \frac { 1 } { 2 } \\B = 8 \frac { 1 } { 2 } \\C = 1\end{array}

.
E)

\begin{array} { l } A = - 8 \frac { 1 } { 2 } \\B = 8 \frac { 1 } { 2 } \\C = 2\end{array}

.

Accepted Answer

We can start by multiplying both sides by (x-2)(x+2) to get rid of the fractions:

$$9x-14=A(x+2) + B(x-2)$$

We want this equation to be true for all values of x, which means that the coefficients of x on both sides must match. Therefore,

$$9=A+B$$ 
and 
$$-14 = 2A-2B$$

Solving the system of equations, we get A=1 and B=8. Therefore, the answer is E.

Question 9

Use Descartes's rule of signs to obtain information regarding the roots of the equation. $x ^ { 9 } + 5 = 0$&#10;A) 1 negative root, 8 complex roots&#10;B) 9 complex roots&#10;C) 1 negative root, 9 complex roots&#10;D) 1 positive root, 1 negative real root, 7 complex roots&#10;E) 1 positive root, 8 complex roots

Accepted Answer

The equation $x^9 + 5 = 0$ can be rewritten as $x^9 = -5$, which indicates there is one real negative root (since raising a negative number to an odd power results in a negative number) and the rest of the roots (if any) would be complex. However, since it's a 9th degree polynomial, it must have 9 roots in total (counting multiplicities), so the remaining 8 roots are complex. There are no sign changes in the polynomial itself, indicating no positive roots.

Question 10

Find the rational roots and solve the equation. 3x³ - 19x² + 21x - 5 = 0 A) $\frac { 1 } { 3 } , 2,5$ B) $\pm \frac { 1 } { 3 } , \pm 1$ C) $\frac { 1 } { 3 } , 1,5$ D) $\frac { 2 } { 3 } , 1,5$ E) $- \frac { 1 } { 3 } , 2,5$

Accepted Answer

The answer of Find the rational roots and solve the...

Question 11

Use synthetic division to find the quotient and the remainder. $\frac { 5 x ^ { 2 } + 2 x - 1 } { x - 1 }$&#10;A) $\text { quotient: } - 6 x + 5 \text {; remainder: } - 7$&#10;B) $\text { quotient: } 5 x + 7 \text {; remainder: } 6$&#10;C) $\text { quotient: } 7 x + 6 \text {; remainder: } 7$&#10;D) $\text { quotient: } 7 x - 5 \text {; remainder: } 6$&#10;E) $\text { quotient: } 5 x + 7 \text {; remainder: } - 8$

Accepted Answer

The answer of Use synthetic division to find the quotient...

Question 12

Use a graph to determine whether the equation has at least one real root. $x ^ { 2 } - 3 x + 6.26 = 0$&#10;A) three real roots&#10;B) two real roots&#10;C) no real roots&#10;D) one real root&#10;E) It is not possible to make this determination by graphing the equation.

Accepted Answer

The answer of Use a graph to determine whether the...

Question 13

Use long division to find the quotient and the remainder. $\frac { 6 x ^ { 3 } - 3 x + 3 } { 2 x + 1 }$&#10;A) $\text { quotient: } 3 x ^ { 2 } - \frac { 3 } { 2 } x - \frac { 3 } { 4 } ; \text { remainder: } \frac { 15 } { 4 }$&#10;B) $\text { quotient: } 3 x ^ { 2 } - \frac { 3 } { 2 } x + \frac { 1 } { 4 } ; \text { remainder: } \frac { 15 } { 4 }$&#10;C) $\text { quotient: } 15 x ^ { 2 } + \frac { 3 } { 2 } x - \frac { 1 } { 4 } ; \text { remainder: } \frac { 3 } { 4 }$&#10;D) $\text { quotient: } 3 x ^ { 2 } - \frac { 3 } { 2 } x - \frac { 3 } { 4 } ; \text { remainder: } - \frac { 15 } { 4 }$&#10;E) $\text { quotient: } 3 x ^ { 2 } - \frac { 3 } { 2 } x - \frac { 3 } { 4 } ; \text { remainder: } \frac { 7 } { 4 }$

Accepted Answer

The answer of Use long division to find the quotient...

Question 14

Find the rational roots and solve the equation. 4x³ - 10x² - 25x + 4 = 0 A) $\frac { - 3 \pm \sqrt { 13 } } { 2 } , 4$ B) $\frac { - 3 \pm \sqrt { 15 } } { 4 } , 4$ C) $\frac { - 3 \pm \sqrt { 13 } } { 4 } , - 4$ D) $\frac { - 3 \pm \sqrt { 13 } } { 4 } , 4$ E) $\frac { 3 \pm \sqrt { 13 } } { 4 } , 4$

Accepted Answer

The answer of Find the rational roots and solve the...

Question 15

Determine which of the following real numbers is a root of the equation. $12 x - 8 = 136$&#10;A) $x = \frac { 8 } { 12 }$&#10;B) $x = 12$&#10;C) $x = 20$&#10;D) $x = 14$&#10;E) $x = 0$

Accepted Answer

The answer of Determine which of the following real numbers...

Question 16

Determine which of the following real numbers is a root of the equation. $x ^ { 2 } - 2 x - 3 = 0$&#10;A) $x = 1 + \sqrt { 4 }$&#10;B) $x = 1 - \sqrt { 4 }$&#10;C) $x = 1 - \sqrt { 5 }$&#10;D) $x = 0$&#10;E) $x = 1 - \sqrt { 6 }$

Accepted Answer

The answer of Determine which of the following real numbers...

Question 17

One root of the equation is given. Determine the remaining roots. $x ^ { 3 } - 9 x ^ { 2 } + 28 x - 40 = 0 ; x = 2 + 2 i$&#10;A) $x = 2 + 2 i , 4$&#10;B) $x = 2 - 2 i , - 5$&#10;C) $x = 2 + 2 i , - 5$&#10;D) $x = 2 - 2 i , 5$&#10;E) $x = 2 + 2 i , 9$

Accepted Answer

The answer of One root of the equation is given....

Question 18

The equation has exactly one positive root. Locate the root between successive hundredths. x³ - 3x² + 3x - 12 = 0 A) between 3.11 and 3.12 B) between 4.31 and 4.32 C) between 2.22 and 2.23 D) between 3.22 and 3.23 E) between 2.02 and 2.03

Accepted Answer

The answer of The equation has exactly one positive root....

Question 19

Use long division to find the quotient and the remainder. $\frac { x ^ { 3 } - x ^ { 2 } + 3 x - 6 } { x - 2 }$&#10;A) $\text { quotient: } x ^ { 2 } - x - 5 \text {; remainder: } 2$&#10;B) $\text { quotient: } x ^ { 2 } + x + 5 \text {; remainder: } 14$&#10;C) $\text { quotient: } x ^ { 2 } + x + 2 \text {; remainder: } - 5$&#10;D) $\text { quotient: } x ^ { 2 } + x + 5 \text {; remainder: } 4$&#10;E) $\text { quotient: } x ^ { 2 } + x - 5 \text {; remainder: } 6$

Accepted Answer

The answer of Use long division to find the quotient...

Question 20

Determine upper and lower bounds for the real roots of the equation. 5x⁴ - 10x - 12 = 0 A) 2 is an upper bound, - 1 is a lower bound B) 8 is an upper bound, - 1 is a lower bound C) 1 is an upper bound, - 2 is a lower bound D) 2 is an upper bound, - 4 is a lower bound E) 8 is an upper bound, - 4 is a lower bound

Accepted Answer

The answer of Determine upper and lower bounds for the...

Question 21

Determine the constants (denoted by capital letters) so that the equation is an identity. $\frac { x - 17 } { x \left( x ^ { 2 } + 2 \right) } = \frac { A } { x } + \frac { B x + C } { x ^ { 2 } + 2 }$

A)

\begin{array} { l } A = 8 \frac { 1 } { 2 } \\B = 1 \\C = 8 \frac { 1 } { 2 }\end{array}

.
B)

\begin{array} { l } A = - 8 \frac { 1 } { 2 } \\B = 8 \frac { 1 } { 2 } \\C = 1\end{array}

C)

\begin{array} { l } A = 8 \frac { 1 } { 2 } \\B = 8 \frac { 1 } { 2 } \\C = - 1\end{array}

.
D)

\begin{array} { l } A = - 8 \frac { 1 } { 2 } \\B = 8 \frac { 1 } { 2 } \\C = 1\end{array}

.
E)

\begin{array} { l } A = - 8 \frac { 1 } { 2 } \\B = 8 \frac { 1 } { 2 } \\C = 2\end{array}

.

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Accepted Answer

The answer of Determine the constants (denoted by capital letters)...

Question 22

Factor the denominator of the given rational expression, and then find the partial fraction decomposition. $\frac { 11 x - 3 \sqrt { 3 } } { x ^ { 2 } - 3 }$&#10;A) $\frac { 11 x } { x + \sqrt { 3 } } + \frac { - 3 \sqrt { 3 } } { x - \sqrt { 3 } }$&#10;B) $\frac { 4 } { x + \sqrt { 3 } } + \frac { 7 } { x - \sqrt { 3 } }$&#10;C) $\frac { 7 } { x + 3 } + \frac { 4 } { x - 3 }$&#10;D) $\frac { 7 } { x + \sqrt { 3 } } + \frac { - 4 } { x - \sqrt { 3 } }$&#10;E) $\frac { 7 } { x + \sqrt { 3 } } + \frac { 4 } { x - \sqrt { 3 } }$

Accepted Answer

The answer of Factor the denominator of the given rational...

Question 23

Factor the denominator of the given rational expression, and then find the partial fraction decomposition. $\frac { 2 x ^ { 3 } + 11 x - 3 } { x ^ { 4 } + 6 x ^ { 2 } + 9 }$&#10;A) $\frac { - 5 x + 3 } { \left( x ^ { 2 } + 3 \right) ^ { 2 } } + \frac { - 2 x } { x ^ { 2 } + 3 }$&#10;B) $\frac { 5 x - 3 } { \left( x ^ { 2 } + 3 \right) ^ { 2 } } + \frac { x } { x ^ { 2 } + 3 }$&#10;C) $\frac { 5 x - 3 } { \left( x ^ { 2 } + 3 \right) ^ { 2 } } + \frac { - 2 x } { x ^ { 2 } + 3 }$&#10;D) $\frac { - 5 x + 3 } { \left( x ^ { 2 } + 3 \right) ^ { 2 } } + \frac { 2 x } { x ^ { 2 } + 3 }$&#10;E) $\frac { 5 x - 3 } { \left( x ^ { 2 } + 3 \right) ^ { 2 } } + \frac { 2 x } { x ^ { 2 } + 3 }$

Accepted Answer

The answer of Factor the denominator of the given rational...

Question 24

Determine which of the following polynomials is not reducible.&#10;A) $x ^ { 2 } + 18$&#10;B) $x ^ { 2 } - 4$&#10;C) $x ^ { 2 } - 1$&#10;D) $x ^ { 2 } - 21$&#10;E) $x ^ { 2 } - 18$

Accepted Answer

The answer of Determine which of the following polynomials is...

Question 25

Factor the denominator of the given rational expression, and then find the partial fraction decomposition. $\frac { 1 } { x ^ { 3 } + x ^ { 2 } - 44 x + 96 }$&#10;A) $\frac { \frac { 1 } { 11 } } { x - 3 } + \frac { - \frac { 1 } { 12 } } { x - 4 } + \frac { \frac { 1 } { 132 } } { x + 8 }$&#10;B) $\frac { - \frac { 1 } { 11 } } { x - 3 } + \frac { - \frac { 1 } { 12 } } { x - 4 } + \frac { \frac { 1 } { 132 } } { x + 8 }$&#10;C) $\frac { \frac { 1 } { 11 } } { x - 3 } + \frac { - \frac { 1 } { 12 } } { x - 4 } + \frac { - \frac { 1 } { 132 } } { x + 8 }$&#10;D) $\frac { - \frac { 1 } { 11 } } { x - 3 } + \frac { \frac { 1 } { 12 } } { x - 4 } + \frac { \frac { 1 } { 132 } } { x + 8 }$&#10;E) $\frac { - \frac { 1 } { 11 } } { x - 3 } + \frac { \frac { 1 } { 12 } } { x - 4 } + \frac { - \frac { 1 } { 132 } } { x + 8 }$

Accepted Answer

The answer of Factor the denominator of the given rational...

Deck 13: Roots of Polynomial Equations