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

Nov 18, 2017

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

#### 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.