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## Factorize #16x^3-24x^2-15x-2=0#?

Shwetank Mauria
Featured 6 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 solve #1/x+1>=0# graphically?

Somebody N.
Featured 5 months ago

See below.

#### Explanation:

First graph the line $y = \frac{1}{x} + 1$. This is a $\le$ inequality so use a solid line, because the line is an included region.

With graph plotted, there will be four possible regions: A, B , C , D

We have to test coordinates in each region to identify which are included regions and which are excluded regions:

Region A

$\left(- 1 , 4\right)$

$4 \le \frac{1}{- 1} + 1$

$4 \le 0 \textcolor{w h i t e}{888}$ FALSE

Region A is an excluded region.

Region B

$\left(1 , 6\right)$

$6 \le \frac{1}{1} + 1$

$6 \le 2 \textcolor{w h i t e}{888}$ FALSE

Region B is an excluded region.

Region C

$\left(- 1 , - 6\right)$

$- 6 \le \frac{1}{- 1} + 1$

$- 6 \le 0 \textcolor{w h i t e}{888}$ TRUE

Region C is an included region.

Region D

$\left(1 , - 6\right)$

$- 6 \le \frac{1}{1} + 1$

$- 6 \le 2 \textcolor{w h i t e}{888}$ TRUE

Region D is an included region.

Shade included regions C and D

## 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 5 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.

## What is the relationship between the roots and coefficients of a polynomial?

George C.
Featured 2 months ago

A few more...

#### Explanation:

Elementary symmetric polynomials

The coefficients of a monic polynomial are (modulo alternating signs), the elementary symmetric polynomials in the zeros.

For example:

$\left(x - \alpha\right) \left(x - \beta\right) = {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta$

$\left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right) = {x}^{3} - \left(\alpha + \beta + \gamma\right) {x}^{2} + \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) x - \alpha \beta \gamma$

etc.

In particular, given a polynomial:

${a}_{n} {x}_{n} + {a}_{n - 1} {x}_{n - 1} + \ldots + {a}_{1} x + {a}_{0}$

the sum of its zeros is $- {a}_{n - 1} / {a}_{n}$ and the product of its zeros is ${\left(- 1\right)}^{n} {a}_{0} / {a}_{n}$.

Since any symmetric polynomial can be constructed from the elementary symmetric polynomials, we can find a polynomial that has zeros that are (say) the squares of the zeros of another polynomial - without finding what the zeros actually are.

For a substantial and important application of this, see https://socratic.org/s/aPGxwybx

Coefficient sum shortcuts

If the sum of the coefficients of a polynomial is $0$, then you can infer that $1$ is a zero.

If inverting the signs on terms of odd degree results in coefficients that sum to $0$, then you can infer that $- 1$ is a zero.

Reversing the order and reciprocals

Given a polynomial:

${a}_{n} {x}_{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}$

with ${a}_{n} \ne 0$ and ${a}_{0} \ne 0$

Then the polynomial:

${a}_{0} {x}_{n} + {a}_{1} {x}_{n - 1} + \ldots + {a}_{n - 1} x + {a}_{n}$

has zeros which are reciprocals of the original polynomial.

To see why that is so, note that:

$\frac{1}{x} _ n \left({a}_{n} {x}_{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}\right)$

$= {a}_{n} + {a}_{n - 1} \frac{1}{x} + \ldots + {a}_{1} \frac{1}{x} ^ \left(n - 1\right) + {a}_{0} \frac{1}{x} ^ n$

Symmetry and reciprocals

From the preceding property, we can deduce that:

If the coefficients of a polynomial are symmetrical then you can infer that the reciprocal of any zero is also a zero.

For example:

$6 {x}^{4} + 5 {x}^{3} - 38 {x}^{2} + 5 x + 6$

has zeros $2 , \frac{1}{2} , - 3 , - \frac{1}{3}$

## A corridor of width #a# meets a corridor of width #b# at right angles. Workmen wish to push a heavy beam on dollies around the corner, but they want to be sure it will be able to make the turn before starting. How long a beam will go around the corner ?

George C.
Featured 2 months ago

${\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{\frac{3}{2}}$

#### Explanation:

Let us locate the outer corner of the corridor at $\left(0 , 0\right)$ and the inner corner at $\left(a , b\right)$.

Consider lines of negative slope passing through $\left(a , b\right)$ and the distance between their $x$ and $y$ intercepts.

The minimum distance then models the maximum feasible length of beam.

graph{((x-14)^100+(y-13)^100-10^100)((x-20)^100+(y-20)^100-20^100)(y+x/2-5)((x-4)^2+(y-3)^2-0.02)(x^2+(y-5)^2-0.02)((x-10)^2+y^2-0.02) = 0 [-5.625, 14.375, -3, 7]}

The equation of such a line can be written in point slope form as:

$y - b = m \left(x - a\right)$

The $x$ intercept is then given by putting $y = 0$ to find:

$x = a - \frac{b}{m}$

and the $y$ intercept by putting $x = 0$ to find:

$y = b - a m$

The square of the distance between these intercepts is:

${\left(a - \frac{b}{m}\right)}^{2} + {\left(b - a m\right)}^{2} = {a}^{2} - 2 a b \frac{1}{m} + {b}^{2} \frac{1}{m} ^ 2 + {b}^{2} - 2 a b m + {a}^{2} {m}^{2}$

The minimum will occur when the derivative of this with respect to $m$ is zero. That is:

$0 = 2 a b \frac{1}{m} ^ 2 - 2 {b}^{2} \frac{1}{m} ^ 3 - 2 a b + 2 {a}^{2} m$

Multiplying by ${m}^{3} / 2$ this becomes:

$0 = a b m - {b}^{2} - a b {m}^{3} + {a}^{2} {m}^{4}$

$\textcolor{w h i t e}{0} = {a}^{2} {m}^{4} - a b {m}^{3} + a b m - {b}^{2}$

$\textcolor{w h i t e}{0} = a {m}^{3} \left(a m - b\right) + b \left(a m - b\right)$

$\textcolor{w h i t e}{0} = \left(a {m}^{3} + b\right) \left(a m - b\right)$

$\textcolor{w h i t e}{0} = \left({a}^{\frac{1}{3}} m + {b}^{\frac{1}{3}}\right) \left({a}^{\frac{2}{3}} {m}^{2} - {a}^{\frac{1}{3}} {b}^{\frac{1}{3}} m + {b}^{\frac{2}{3}}\right) \left(a m - b\right)$

Note here that the last linear factor gives $m = \frac{b}{a} > 0$, so is not suitable.

Also the quadratic factor has only non-real solutions.

So we require $m = - {b}^{\frac{1}{3}} {a}^{- \frac{1}{3}}$

With this value of $m$, the square of the distance between the intercepts is:

${\left(a - \frac{b}{m}\right)}^{2} + {\left(b - a m\right)}^{2} = {\left(a + {a}^{\frac{1}{3}} {b}^{\frac{2}{3}}\right)}^{2} + {\left(b + {a}^{\frac{2}{3}} {b}^{\frac{1}{3}}\right)}^{2}$

$\textcolor{w h i t e}{{\left(a - \frac{b}{m}\right)}^{2} + {\left(b - a m\right)}^{2}} = {a}^{\frac{2}{3}} {\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{2} + {b}^{\frac{2}{3}} {\left({b}^{\frac{2}{3}} + {a}^{\frac{2}{3}}\right)}^{2}$

$\textcolor{w h i t e}{{\left(a - \frac{b}{m}\right)}^{2} + {\left(b - a m\right)}^{2}} = {\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{3}$

So the distance is:

$\sqrt{{\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{3}} = {\left({a}^{\frac{2}{3}} + {b}^{\frac{2}{3}}\right)}^{\frac{3}{2}}$

## How do you graph #y=-2^x#?

Sridhar V.
Featured 3 weeks ago

$\text{ }$
Create a data table to graph the exponential function : #color(red)(y=-2^x#

#### Explanation:

$\text{ }$
#color(green)("Step 1 :"#

Create a data table:

Closely examine the values for $x$ and $y$

No value of $x$ makes $y$ zero.

So, the graph gets closer and closer to the x-axis but never touches the x-axis.

Hence, #color(brown)(y=0# is the horizontal asymptote.

#color(green)("Step 2 :"#

Using the data table, graph the function $y = f \left(x\right) = - {2}^{x}$

For easy comprehension, graph the exponential function $y = f \left(x\right) = {2}^{x}$, it's parent function.

Compare both the graphs to understand the behavior of the given exponential function $y = f \left(x\right) = - {2}^{x}$

The graph of $y = f \left(x\right) = - {2}^{x}$ is reflected about the x-axis when the sign of $f \left(x\right)$ is negative.

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