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## Question #45dbd

Shwetank Mauria
Featured 3 months ago

$16 {x}^{3} - 24 {x}^{2} - 15 x - 2 = {\left(4 x + 1\right)}^{2} \left(x - 2\right) = 0$

#### Explanation:

According to Factor theorem, for a polynomial $p \left(x\right)$ of degree greater than or equal to one, $x - \alpha$ is a factor of $p \left(x\right)$, if $p \left(\alpha\right) = 0$, then $\left(x - \alpha\right)$ is a factor of $p \left(x\right)$, where $\alpha$ is a real number. In fact $\alpha$ is also a factor of the constant term in the polynomial $p \left(x\right)$.

Hence, if we have a polynomial say $p \left(x\right) = {x}^{3} + l {x}^{2} + m x + n$ and $p \left(\alpha\right) = 0$, then $\alpha$ is a factor of $n$. For example if the constant term is say $6$, then our $\alpha$’s could be $\pm 1 , \pm 2 , \pm 3 , \pm 6$. Hence, we seek a number among these for which $p \left(\alpha\right) = 0$ and then we have $\left(x - \alpha\right)$ a factor of $p \left(x\right)$.

Observe that coefficient of highest power of $x$ in $p \left(x\right)$ is $1$. In case, the coefficient of highest power of $x$ in $p \left(x\right)$ is more than $1$, say $a$ like in polynomial $a {x}^{3} + b {x}^{2} + c x + d$, then our $\alpha$ is a factor of $\frac{d}{a}$.

In practice, Factor Theorem is used for factorising polynomials completely. So our steps in factorising a polynomial $p \left(x\right)$ would be
1 Check the constant term in $p \left(x\right)$
2. Find its all possible factors.
3. Take one of the factors, say $\alpha$ and replace $x$ by this factor in the given polynomial $p \left(x\right)$.
4. Try more factors whose number should be equal to the degree of polynomial.

And you have got all the factors.

Coming to your question, we have $2 \ln \left(4 x - 5\right) + \ln \left(x + 1\right) = 3 \ln 3$, which can be written as $\ln \left\{{\left(4 x - 5\right)}^{2} \left(x + 1\right)\right\} = \ln 27$

or #(16x^2-40x+25)(x+1)}=ln27#

or $16 {x}^{3} - 24 {x}^{2} - 15 x - 2 = 0$

$16 {x}^{3} - 24 {x}^{2} - 15 x - 2 = 0$

This is the same as in $\left(b\right)$ so we use factor theorem as follows.

We should have zeros among $\pm 2 , \pm 1 , \pm \frac{1}{2} , \pm \frac{1}{4} , \pm \frac{1}{8}$.Observe that for $x = 2 , - \frac{1}{4}$, it is $0$ and hence $x - \left(- \frac{1}{4}\right)$ or $x + \frac{1}{4}$ i.e. $\left(4 x + 1\right)$ and $\left(x - 2\right)$ are two factors. We do not get any other number for which it is $0$, but dividing $16 {x}^{3} - 24 {x}^{2} - 15 x - 2$ by $\left(4 x + 1\right)$ and $\left(x - 2\right)$, we get $\left(4 x + 1\right)$, hence factors are

${\left(4 x + 1\right)}^{2} \left(x - 2\right) = 0$

Note that it is easier to check whether $\pm 1$ are zeros or not, as for $+ 1$, we should have sum of all coefficients as $z e r o$ and for $- 1$, we change sign of alternate coefficient and find if sum is zero or not. As it is not so in given polynomial one can avoid using them.

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

Douglas K.
Featured 3 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 − 8x − 16y − 40 = 0#

#x^2 + 8y^2 − 8x − 16y = 40#

Group the x terms and the y terms together:

#(x^2 − 8x) + (8y^2 − 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 − 8x) + 8(y^2 − 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 − 8x+h^2) + 8(y^2 − 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 − 4)^2 + 8(y^2 − 2y) = 40+16#

Combine like terms:

#(x − 4)^2 + 8(y^2 − 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 − 4)^2 + 8(y^2 − 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 − 4)^2 + 8(y − 1)^2 = 56+8#

Combine like terms:

#(x − 4)^2 + 8(y − 1)^2 = 64#

Divide both sides by 64:

#(x − 4)^2/64 + (y − 1)^2/8 = 1#

Write the denominators as squares:

#(x − 4)^2/8^2 + (y − 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 zeros and use a sign chart to sketch the polynomial #F(x)=x^3(x+2)^2#?

Alan P.
Featured 2 months ago

Determination of the zeros and a sign chart of $f \left(x\right)$ will only provide some limited information and not enough to sketch this equation. (see below)

#### Explanation:

Finding the zeros of $f \left(x\right) = {x}^{3} {\left(x + 2\right)}^{2}$
Remember that a function has the value zero at (and only at) points where some factor of the function has the value zero.

The factors of $f \left(x\right) = {x}^{2} {\left(x + 2\right)}^{2}$
are $x \times x \times x \times \left(x + 2\right) \times \left(x + 2\right)$
and the only unique factors are $x$ and $\left(x + 2\right)$

Therefore $f \left(x\right)$ only has zeros where
#{:(x=0,color(white)("xxx")andcolor(white)("xxx"),(x+2)=0), (,,rarr x=-2):}#

For non-zero values of $f \left(x\right)$ these points divide the domain into 3 intervals:
$x < - 2 \textcolor{w h i t e}{\text{xxx") x in (-2,0)color(white)("xxx}} x > 0$

We can pick arbitrary values within each interval to determine if $f \left(x\right)$ is positive or negative in that interval.

#{: (x=-3,color(white)("xxx"),x=-1,color(white)("xxx"),x=1), (f(-3)=-27,,f(-1)=-1,,f(1)=+9), (f(x)" negative",,f(x)" negative",,f(x)" positive") :}#

Unfortunately this does not** tell us any detailed information about the behavior of $f \left(x\right)$ except whether it is positive or negative within these ranges.

Just for reference, here is what the graph should look like, but you would need to use something beyond the zeros and a sign chart to sketch this.

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

Ratnaker Mehta
Featured 1 month 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 #x^2-4x>=5#?

Nimo N.
Featured 1 month ago

Solution set
# color(red) (-oo <= x <= -1"," or 5 <= x <= oo #
Alternate notation:
# color(red) ( x in (-oo, -1] uu [5, oo) #

#### Explanation:

Solve:
# x^2 − 4x ≥ 5 #

1) Examine the related equation to find the boundaries.
# x^2 − 4x ≥ 5 #
# x^2 − 4x - 5 = 0 #
$\left(x - 5\right) \left(x + 1\right) = 0$
$\left(x - 5\right) = 0$
$x = 5$
or
$\left(x + 1\right) = 0$
$x = - 1$

The solutions to the related equation define boundary points that divide the number line into 5 parts, 3 regions and the 2 boundary points.

• Since the inequality is inclusive ("non-strict") the boundary points are in the solution set.

A graph of the number line, with solid dots on (-1), and (5) will help in thinking about the problem.

To check for the solution, use test points, one from each region, in the inequality to see if they qualify:

Try (-2), (0), and (6). It is best not to pick numbers that make the arithmetic difficult.

In region, $- \infty < x < - 1$
# ( -2 )^2 − 4( -2) ≥ 5 " ?"#
# 4 + 8 ≥ 5 " ?"#
# 12 ≥ 5 " YES"#

In region, $- 1 < x < 5$
# ( 0 )^2 − 4( 0 ) ≥ 5 " ?" #
# 0 ≥ 5 " NO" #

In region, $5 < x < \infty$
# ( 6 )^2 − 4( 6 ) ≥ 5 " ?" #
# 36 − 24 ≥ 5 " ?" #
# 12 ≥ 5 " YES" #

We must be careful in stating the solution inequalities. x can not be in both regions at once, since they are disconnected. There is a full region between the two regions in the solution, so we can't use "and", we must use "or".

Solution set
# color(red) (-oo <= x <= -1"," or 5 <= x <= oo #
Alternate notation:
# color(red) ( x in (-oo, -1] uu [5, oo) #

2-D Graph of the inequality:

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

Cem Sentin
Featured 1 month 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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