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## How do you divide #(x^4+5x^3+6x^2-x-2)-:(x+2)#?

Jim G.
Featured 5 months ago

${x}^{3} + 3 {x}^{2} - 1$

#### Explanation:

$\text{one way is to use the divisor as a factor in the numerator}$

$\text{consider the numerator}$

$\textcolor{red}{{x}^{3}} \left(x + 2\right) \textcolor{m a \ge n t a}{- 2 {x}^{3}} + 5 {x}^{3} + 6 {x}^{2} - x - 2$

$= \textcolor{red}{{x}^{3}} \left(x + 2\right) \textcolor{red}{+ 3 {x}^{2}} \left(x + 2\right) \cancel{\textcolor{m a \ge n t a}{- 6 {x}^{2}}} \cancel{+ 6 {x}^{2}} - x - 2$

$= \textcolor{red}{{x}^{3}} \left(x + 2\right) \textcolor{red}{+ 3 {x}^{2}} \left(x + 2\right) \textcolor{red}{- 1} \left(x + 2\right) \cancel{\textcolor{m a \ge n t a}{+ 2}} \cancel{- 2}$

$\text{quotient "=color(red)(x^3+3x^2-1)," remainder } = 0$

$\Rightarrow \frac{{x}^{4} + 5 {x}^{3} + 6 {x}^{2} - x - 2}{x + 2}$

$= \frac{\cancel{\left(x + 2\right)} \left({x}^{3} + 3 {x}^{2} - 1\right)}{\cancel{\left(x + 2\right)}} = {x}^{3} + 3 {x}^{2} - 1$

## How do I find the center, vertices, foci, and eccentricity of the ellipse? #x^2 + 8y^2 âˆ’ 8x âˆ’ 16y âˆ’ 40 = 0#

Douglas K.
Featured 4 months ago

One should complete the squares so that the equation may be written in one of the two following forms:

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [1]}$

${\left(y - k\right)}^{2} / {a}^{2} + {\left(x - h\right)}^{2} / {b}^{2} = 1 \text{ [2]}$

where $a > b$

#### Explanation:

Given:

#x^2 + 8y^2 âˆ’ 8x âˆ’ 16y âˆ’ 40 = 0#

#x^2 + 8y^2 âˆ’ 8x âˆ’ 16y = 40#

Group the x terms and the y terms together:

#(x^2 âˆ’ 8x) + (8y^2 âˆ’ 16y) = 40#

We cannot complete the square unless the leading coefficient is 1, therefore, we remover a factor of 8 from the y terms:

#(x^2 âˆ’ 8x) + 8(y^2 âˆ’ 2y) = 40#

Because ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ we want to insert an ${h}^{2}$ into the x term's group but we must, also add ${h}^{2}$ to the right side so that equality is maintained:

#(x^2 âˆ’ 8x+h^2) + 8(y^2 âˆ’ 2y) = 40+h^2#

Matching the x terms with the general pattern, ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, we observe that the equation

$- 2 h x = - 8 x$

will allow us to solve for the value of h:

$h = 4$

This means that ${h}^{2}$ on the right side becomes 16 and the group of x terms become ${\left(x - 4\right)}^{2}$

#(x âˆ’ 4)^2 + 8(y^2 âˆ’ 2y) = 40+16#

Combine like terms:

#(x âˆ’ 4)^2 + 8(y^2 âˆ’ 2y) = 56#

We want insert ${k}^{2}$ into the y terms but to maintain equality we must add $8 {k}^{2}$ to the right side:

#(x âˆ’ 4)^2 + 8(y^2 âˆ’ 2y+ k^2) = 56+8k^2#

Matching the y terms with the general pattern, ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$, we observe that the equation

$- 2 k y = - 2 y$

will allow us to solve for the value of k:

$k = 1$

This means that $8 {k}^{2}$ on the right side becomes 8 and the group of y terms become ${\left(y - 1\right)}^{2}$

#(x âˆ’ 4)^2 + 8(y âˆ’ 1)^2 = 56+8#

Combine like terms:

#(x âˆ’ 4)^2 + 8(y âˆ’ 1)^2 = 64#

Divide both sides by 64:

#(x âˆ’ 4)^2/64 + (y âˆ’ 1)^2/8 = 1#

Write the denominators as squares:

#(x âˆ’ 4)^2/8^2 + (y âˆ’ 1)^2/(2sqrt2)^2 = 1#

This is the same form as equation [1],

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [1]}$

The center, $\left(h , k\right) = \left(4 , 1\right)$
The vertices are, $\left(h - a , k\right) = \left(- 4 , 1\right)$ and $\left(h + a , k\right) = \left(12 , 1\right)$
The foci are, $\left(h - \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(4 - \sqrt{56} , 1\right)$ and $\left(h + \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(4 + \sqrt{56} , 1\right)$
The eccentricity is $\frac{\sqrt{{a}^{2} - {b}^{2}}}{a} = \frac{\sqrt{56}}{8}$

## How do you find the maximum or minimum of #f(x)=-1/2x^2-2x+3#?

Jim G.
Featured 4 months ago

$\left(- 2 , 5\right) \text{ is the maximum}$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

#â€¢ " if "a>0" then f(x) is a minimum"#

#â€¢ " if "a<0" then f(x) is a maximum"#

$\left(h , k\right) \text{ will be the max/min turning point of f(x)}$

$\text{to express "f(x)=-1/2x^2-2x+3" in vertex form}$

$\text{use the method of "color(blue)"completing the square}$

#â€¢ " the coefficient of the "x^2" term must be 1"#

$\Rightarrow f \left(x\right) = - \frac{1}{2} \left({x}^{2} + 4 x\right) + 3$

#â€¢ " add/subtract "(1/2"coefficient of x-term")^2" to"#
${x}^{2} + 4 x$

$f \left(x\right) = - \frac{1}{2} \left({x}^{2} + 2 \left(2\right) x \textcolor{red}{+ 4} \textcolor{red}{- 4}\right) + 3$

$\textcolor{w h i t e}{f \left(x\right)} = - \frac{1}{2} {\left(x + 2\right)}^{2} + 2 + 3$

$\textcolor{w h i t e}{f \left(x\right)} = - \frac{1}{2} {\left(x + 2\right)}^{2} + 5 \leftarrow \textcolor{b l u e}{\text{in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex "=(-2,5)" and }} a < 0$

$\Rightarrow \text{maximum turning point } = \left(- 2 , 5\right)$
graph{(y+1/2x^2+2x-3)((x+2)^2+(y-5)^2-0.04)=0 [-10, 10, -5, 5]}

## What is the domain and range of the function #f(x) = x/(x^2-4)# ?

Featured 3 months ago

The domain is $x \in \left(- \infty , - 2\right) \cup \left(- 2 , 2\right) \cup \left(2 , + \infty\right)$.
The range is $y \in \mathbb{R}$

#### Explanation:

The function is $f \left(x\right) = \frac{x}{{x}^{2} - 4} = \frac{x}{\left(x + 2\right) \left(x - 2\right)}$

As the denominator must be $\ne 0$, so

$x \ne 2$ and $x \ne - 2$

The domain is $x \in \left(- \infty , - 2\right) \cup \left(- 2 , 2\right) \cup \left(2 , + \infty\right)$

To find the range, proceed as follows :

$y = \frac{x}{{x}^{2} - 4}$

$y \left({x}^{2} - 4\right) = x$

$y {x}^{2} - 4 y = x$

$y {x}^{2} - x - 4 y = 0$

In order for this quadratic equation in $x$ to have solutions, the discriminant $\Delta \ge 0$

${b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \left(y\right) \left(- 4 y\right) \ge 0$

$1 + 16 {y}^{2} \ge 0$

Therefore,

$\forall y \in \mathbb{R} ,$, $\Delta \ge 0$

The range is $y \in \mathbb{R}$

graph{x/(x^2-4) [-10, 10, -5, 5]}

## How do you write an equation for the hyperbola with asymptote #y=+-(4x)/3#, contains #(3sqrt(2,)4)#?

Ratnaker Mehta
Featured 2 months ago

${x}^{2} / 9 - {y}^{2} / 16 = 1$.

#### Explanation:

Recall that, for the Hyperbola $S : {x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$, the

asymptotes are given by, #y=+-b/a*x;" where, "a,b gt 0#.

In our Case, the asymptotes are, $y = \pm \frac{4}{3} \cdot x$.

$\therefore \frac{b}{a} = \frac{4}{3.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$.

Next, $\left(3 \sqrt{2} , 4\right) \in S \Rightarrow {\left(3 \sqrt{2}\right)}^{2} / {a}^{2} - {4}^{2} / {b}^{2} = 1$.

$\therefore \frac{18}{a} ^ 2 - \frac{16}{b} ^ 2 = 1$.

Multiplying by ${b}^{2}$, we get, $18 \cdot {b}^{2} / {a}^{2} - 16 = {b}^{2}$.

By $\left(1\right)$, then, $18 \cdot {\left(\frac{4}{3}\right)}^{2} - 16 = {b}^{2} , i . e . , {b}^{2} = 16$.

$\therefore b = 4 \left(\because , b > 0\right) , \mathmr{and} , \left(1\right) \Rightarrow a = 3 , \mathmr{and} , {a}^{2} = 9$.

With ${a}^{2} = 9 , \mathmr{and} {b}^{2} = 16$, the reqd. eqn. of the Hyperbola is,

${x}^{2} / 9 - {y}^{2} / 16 = 1$.

## How do you solve the system #r+s+t=15, r+t=12, s+t=10# using matrices?

Cem Sentin
Featured 2 months ago

$r = 5$, $s = 3$ and $t = 7$

#### Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

$A = \left(\begin{matrix}1 & 1 & 1 & | & 15 \\ 1 & 0 & 1 & | & 12 \\ 0 & 1 & 1 & | & 10\end{matrix}\right)$

I have written the equations not in the sequence as in the question in order to get $1$ as pivot.

Perform the folowing operations on the rows of the matrix

$R 2 \leftarrow R 2 - R 1$

$A = \left(\begin{matrix}1 & 1 & 1 & | & 15 \\ 0 & - 1 & - 0 & | & - 3 \\ 0 & 1 & 1 & | & 10\end{matrix}\right)$

$R 1 \leftarrow R 1 + R 2$; $R 3 \leftarrow R 3 + R 2$

$A = \left(\begin{matrix}1 & 0 & 1 & | & 12 \\ 0 & - 1 & - 0 & | & - 3 \\ 0 & 0 & 1 & | & 7\end{matrix}\right)$

$R 1 \leftarrow R 1 - R 3$

$A = \left(\begin{matrix}1 & 0 & 0 & | & 5 \\ 0 & - 1 & - 0 & | & - 3 \\ 0 & 0 & 1 & | & 7\end{matrix}\right)$

$R 2 \leftarrow \left(R 2\right) \cdot \left(- 1\right)$

$A = \left(\begin{matrix}1 & 0 & 0 & | & 5 \\ 0 & 1 & 0 & | & 3 \\ 0 & 0 & 1 & | & 7\end{matrix}\right)$

Thus $r = 5$, $s = 3$ and $t = 7$

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