Define $\displaystyle g(x)= f(x) - 8 =5x- \cos 2x - 8$. Then, if $\displaystyle g(c)= f(c) - 8 = 0$, it follows that $\displaystyle f(c) = 8$.

By the Intermediate Value Theorem (IVT), if $\displaystyle g(x)$ is a continuous function, and $\displaystyle g(a)$ and $\displaystyle g(b)$ are of unlike sign, then $\displaystyle g(c) = 0$ for some $\displaystyle c \in (a, b)$. $\displaystyle 5x$ and $\displaystyle \cos 2x$ are both continuous everywhere, so $\displaystyle g(x)$ is a continuous function, so the IVT applies here.

Evaluate $\displaystyle g(x)$ for each of the following values: $\displaystyle \left \{ 1.0, 1.1, 1.2, 1.3, 1.4, 1.5\right \}$

$\displaystyle g(1) =5 (1)- \cos 2 (1)- 8$

$\displaystyle =5 - \cos 2 - 8$

$\displaystyle \approx 5 - (-0.42) - 8$

$\displaystyle \approx -2.58$

$\displaystyle g(1.1) =5 (1.1)- \cos 2 (1.1)- 8$

$\displaystyle =5.5 - \cos 2.2 - 8$

$\displaystyle \approx 5.5 - (-0.59) - 8$

$\displaystyle \approx -1.91$

$\displaystyle g(1.2) =5 (1.2)- \cos 2 (1.2)- 8$

$\displaystyle =6 - \cos 2.4 - 8$

$\displaystyle \approx 6 - (-0.74) - 8$

$\displaystyle \approx -1.26$

$\displaystyle g(1.3) =5 (1.3)- \cos 2 (1.3)- 8$

$\displaystyle =6.5 - \cos 2.6 - 8$

$\displaystyle \approx 6.5 -(-0.86) - 8$

$\displaystyle \approx -0.64$

$\displaystyle g(1.4) =5 (1.4)- \cos 2 (1.4)- 8$

$\displaystyle =7 - \cos 2.8 - 8$

$\displaystyle \approx 7 - (-0.94) - 8$

$\displaystyle \approx -0.06$

$\displaystyle g(1.5) =5 (1.5)- \cos 2 (1.5)- 8$

$\displaystyle =7.5 - \cos 3 - 8$

$\displaystyle \approx 7.5 -(-0.99)- 8$

$\displaystyle \approx 0.49$

Only in the case of $\displaystyle (1.4, 1.5)$ does it hold that $\displaystyle g (x)$ assumes a different sign at both endpoints - $\displaystyle g(1.4) < 0 < g(1.5)$. By the IVT, $\displaystyle g(c) = 0$, and $\displaystyle f(c) = 8$, for some $\displaystyle c\in (1.4, 1.5)$.

The absolute value equation $\displaystyle |x + 7| = 13$ can be rewritten as the compound equation

$\displaystyle x+ 7 =- 13$  $\displaystyle \or$ $\displaystyle x+ 7 = 13$

Each simple equation can be solved by subtracting 7 from both sides:

$\displaystyle x+ 7 =- 13$

$\displaystyle x+ 7 - 7 =- 13 - 7$

$\displaystyle x = -20$

$\displaystyle x+ 7 = 13$

$\displaystyle x+ 7 - 7 = 13 - 7$

$\displaystyle x = 6$

The absolute value equation has solution set $\displaystyle \left \{ - 20 , 6 \right \}$.

Suppose $\displaystyle P(x)$ is a polynomial. Then, by the Factor Theorem, $\displaystyle x - c$ is a factor of $\displaystyle P(x)$ if and only if  $\displaystyle P(c) = 0$ - that is, if $\displaystyle c$ is a solution of $\displaystyle P(x) = 0$. We are seeking a quadratic equation with solution set the same as that of the absolute value equation, $\displaystyle \left \{ - 20 , 6 \right \}$, we want polynomial $\displaystyle P(x)$ to have as its linear binomial factors $\displaystyle x- (-20)$, or $\displaystyle x+ 20$, and $\displaystyle x - 6$. 

The correct polynomial will be the product of these two:

$\displaystyle P(x) = (x+20)(x-6)$

Using the FOIL technique to rewrite this:

$\displaystyle P(x) = x^{2}-6x + 20x - 120$

$\displaystyle P(x) = x^{2} + 14x - 120$

The correct choice is the equation $\displaystyle x^{2} + 14x - 120 = 0$.

Set $\displaystyle y = x^{\frac{1}{3}}$. Then $\displaystyle x^{\frac{2}{3}} = \left (x^{\frac{1}{3}} \right )^{2} = y ^{2}$. 

$\displaystyle 8 x^{\frac{2}{3}} -2 x ^{\frac{1}{3}} - 3 = 0$ can be rewritten as

$\displaystyle 8 \left (x^{\frac{1}{3}} \right )^{2} -2 x ^{\frac{1}{3}} - 3 = 0$

Substituting $\displaystyle y^{2}$ for $\displaystyle x$ and $\displaystyle y$ for $\displaystyle x^{\frac{1}{3}}$, the equation becomes 

$\displaystyle 8y^{2} -2 y - 3 = 0$,

a quadratic equation in $\displaystyle y$.

This can be solved using the $\displaystyle ac$ method. We are looking for two integers whose sum is $\displaystyle -2$ and whose product is $\displaystyle 8 (-3) = -24$. Through some trial and error, the integers are found to be $\displaystyle -6$ and 4, so the above equation can be rewritten, and solved using grouping, as

$\displaystyle 8y^{2}-6y + 4 y - 3 = 0$

$\displaystyle 2y ( 4 y - 3)+1 ( 4 y - 3) = 0$

$\displaystyle (2y +1) ( 4 y - 3) = 0$

By the Zero Product Principle, one of these factors is equal to zero:

Either:

$\displaystyle 2y + 1 = 0$

$\displaystyle 2y + 1 - 1 = 0 - 1$

$\displaystyle 2y = - 1$

$\displaystyle \frac{2y}{2} = \frac{- 1}{2}$

$\displaystyle y = - \frac{1}{2}$

Substituting back:

$\displaystyle x^{\frac{1}{3}} = -\frac{1}{2}$, or $\displaystyle \sqrt[3]{x} = - \frac{1}{2}$

Cubing both sides:

$\displaystyle (\sqrt[3]{x} )^{3} = \left ( - \frac{1}{2} \right )^{3}$

$\displaystyle x = -\frac{1}{8}$

Or:

$\displaystyle 4y - 3 = 0$

$\displaystyle 4y - 3+ 3 = 0+ 3$

$\displaystyle 4 y = 3$

$\displaystyle \frac{4 y}{4} = \frac{3}{4}$

$\displaystyle y= \frac{3}{4}$

Substituting back:

$\displaystyle x^{\frac{1}{3}} = \frac{3}{4}$, or $\displaystyle \sqrt[3]{x} = \frac{3}{4}$

Cubing both sides:

$\displaystyle (\sqrt[3]{x} )^{3} = \left (\frac{3}{4} \right )^{3}$

$\displaystyle x = \frac{27}{64}$.

The solution set is $\displaystyle \left \{ -\frac{1}{8} , \frac{27}{64} \right \}$.

Set $\displaystyle y = x^{2}$. Then $\displaystyle x^{4 } = \left (x^{2} \right )^{2} = y ^{2}$. 

$\displaystyle 8x^{4} - 2x^{2} - 3 = 0$ can be rewritten as

$\displaystyle 8 \left (x^{2} \right )^{2} -2 x ^{2} - 3 = 0$

Substituting $\displaystyle y^{2}$ for $\displaystyle x^{4 }$ and $\displaystyle y$ for $\displaystyle x^{2}$, the equation becomes 

$\displaystyle 8y^{2} -2 y - 3 = 0$,

a quadratic equation in $\displaystyle y$.

This can be solved using the $\displaystyle ac$ method. We are looking for two integers whose sum is $\displaystyle -2$ and whose product is $\displaystyle 8 (-3) = -24$. Through some trial and error, the integers are found to be $\displaystyle -6$ and 4, so the above equation can be rewritten, and solved using grouping, as

$\displaystyle 8y^{2}-6y + 4 y - 3 = 0$

$\displaystyle 2y ( 4 y - 3)+1 ( 4 y - 3) = 0$

$\displaystyle (2y +1) ( 4 y - 3) = 0$

By the Zero Product Principle, one of these factors is equal to zero:

Either:

$\displaystyle 2y + 1 = 0$

$\displaystyle 2y + 1 - 1 = 0 - 1$

$\displaystyle 2y = - 1$

$\displaystyle \frac{2y}{2} = \frac{- 1}{2}$

$\displaystyle y = - \frac{1}{2}$

Substituting back:

$\displaystyle x ^{2}= - \frac{1}{2}$

However, there are no real numbers whose squares are negative. Since we are looking only for real solutions, none are yielded here.

Or:

$\displaystyle 4y - 3 = 0$

$\displaystyle 4y - 3+ 3 = 0+ 3$

$\displaystyle 4 y = 3$

$\displaystyle \frac{4 y}{4} = \frac{3}{4}$

$\displaystyle y= \frac{3}{4}$

Substituting back:

$\displaystyle x^{2} = \frac{3}{4}$

Taking both square roots of both sides and simplifying using the Quotient of Radicals Rule:

$\displaystyle x = \pm \sqrt{ \frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}$

The solution set is $\displaystyle \left \{ -\frac{\sqrt{3}}{2} , \frac{\sqrt{3}}{2} \right \}$.

Define $\displaystyle g(x) = f(x) + 22 = x^{3}+3x+ 22$. Then, if $\displaystyle g(c)= f(c)+22 = 0$, it follows that $\displaystyle f(c) = -22$.

By the Intermediate Value Theorem (IVT), if $\displaystyle g(x)$ is a continuous function, and $\displaystyle g(a)$ and $\displaystyle g(b)$ are of unlike sign, then $\displaystyle g(c) = 0$ for some $\displaystyle c \in (a, b)$. As a polynomial, $\displaystyle g(x)$ is a continuous function, so the IVT applies here. 

Evaluate $\displaystyle g(x)$ for each of the following values: $\displaystyle \left \{ -2.5, -2.4, -2.3, -2.2, -2.1, -2.0 \right \}$:

$\displaystyle g(-2.5) = (-2.5)^{3}+3(-2.5)+ 22 =-1.125$

$\displaystyle g(-2.4) = (-2.4)^{3}+3(-2.4)+ 22 =0.976$

$\displaystyle g(-2.3) = (-2.3)^{3}+3(-2.3)+ 22 =2.933$

$\displaystyle g(-2.2) = (-2.2)^{3}+3(-2.2)+ 22 =4.752$

$\displaystyle g(-2.1) = (-2.1)^{3}+3(-2.1)+ 22 =6.439$

$\displaystyle g(-2.0) = (-2.0)^{3}+3(-2.0)+ 22 =8$

Only in the case of $\displaystyle (-2.5, -2.4)$ does it hold that $\displaystyle g (x)$ assumes different signs at the endpoints - $\displaystyle g(-2.5) < 0 < g(2.4)$. By the IVT, $\displaystyle g(c) = 0$, and $\displaystyle f(c) = -22$, for some $\displaystyle c \in (-2.5, -2.4)$.

Subtract 18 on both sides.

$\displaystyle 3x+18 -18= -3-18$

$\displaystyle 3x=-21$

Divide by three on both sides.

$\displaystyle \frac{3x}{3}=\frac{-21}{3}$

The answer is:  $\displaystyle -7$

Subtract seven from both sides.

$\displaystyle -9x+7-7 = -16-7$

$\displaystyle -9x = -23$

Divide both sides by negative nine.

$\displaystyle \frac{-9x}{-9} =\frac{ -23}{-9}$

The answer is:  $\displaystyle \frac{23}{9}$

Add $\displaystyle 2x$ on both sides of the equation.

$\displaystyle -2x+4 +2x= 18-x+2x$

$\displaystyle 4=18+x$

Subtract 18 on both sides of the equation.

$\displaystyle 4-18=18+x-18$

$\displaystyle x=-14$

The answer is:  $\displaystyle -14$

Multiply both sides by the least common denominator $\displaystyle 6x$.  This will eliminate the fractions on both sides of the equation.

$\displaystyle 6x(\frac{2}{3x}+4 )= \frac{5}{6x}\cdot 6x$

$\displaystyle 4+24x = 5$

Subtract 4 on both sides.

$\displaystyle 4+24x -4= 5-4$

$\displaystyle 24x=1$

Divide by 24 on both sides.

$\displaystyle \frac{24x}{24}=\frac{1}{24}$

The answer is:  $\displaystyle \frac{1}{24}$

A cubic polynomial has three zeroes, if a zero of multiplicity $\displaystyle n$ is counted $\displaystyle n$ times. Since its lead term is $\displaystyle x^{3}$, we know that

$\displaystyle p(x) = (x-b_{1}) (x-b_{2}) (x-b_{3})$, 

where $\displaystyle b_{1}$, $\displaystyle b_{2}$, and $\displaystyle b_{3}$ are its zeroes.

 $\displaystyle p(3) = p(5) = 0$, so 3 and 5 are zeroes of $\displaystyle p(x)$, and, by the Factor Theorem, $\displaystyle x-3$ and $\displaystyle x-5$ are among the linear factors - that is,

$\displaystyle p(x) = (x- 3) (x-5) (x-b_{3})$

for some $\displaystyle b_{3}$.

However, we are not given the multiplicity of either 3 or 5, so we do not know which, if either, of the two are equal to $\displaystyle b_{3}$. Without this information, we cannot determine $\displaystyle p(x)$ for certain.