# How do you factor completely P(k)=k^5+3k^4-5k^3-15k^2+4k+12?

Jan 9, 2017

$\textcolor{g r e e n}{\left(x - 2\right) \left(x - 1\right) \left(x + 1\right) \left(x + 2\right) \left(x + 3\right)}$

#### Explanation:

Using the Rational Factor Theorem we know that any solutions to
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{1} {k}^{5} + 3 {k}^{4} - 5 {k}^{3} - 15 {k}^{2} + 4 k + \textcolor{b l u e}{12}$
will be factors of $\frac{\textcolor{b l u e}{12}}{\textcolor{red}{1}}$

That is rational zeros of the given function will be
$\textcolor{w h i t e}{\text{XXX}} k \in \left\{\pm 12 , \pm 6 , \pm 4 , \pm 3 , \pm 2 , \pm 1\right\}$

Evaluating $P \left(k\right)$ for these values (I used a spreadsheet):

gives us 5 distinct zeros: $k \in \left\{2 , 1 , - 1 , - 2 , - 3\right\}$

Since $P \left(k\right)$ is a polynomial of degree $5$, it can have at most $5$ distinct zeros; so this method has given us all solutions.

The factors are therefore:
$\textcolor{w h i t e}{\text{XXX}} \left(x - 2\right) \left(x - 1\right) \left(x - \left(- 1\right)\right) \left(x - \left(- 2\right)\right) \left(x - \left(- 3\right)\right)$

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

Jan 9, 2017

$P \left(k\right) = \left(k + 3\right) \left(k + 1\right) \left(k - 1\right) \left(k + 2\right) \left(k - 2\right)$.

#### Explanation:

We find that, the sum of the co-effs. is $0 , \text{ so } \left(k - 1\right)$ is a factor.

Again, the sum of the co-effs. of the even-powered terms

(i.e., $3 - 15 + 12 = 0$) & that of odd-powered (i.e., $1 - 5 + 4 = 0$)

are equal.

$\therefore \left(k + 1\right) \text{ is a factor}$.

Thus, $\left({k}^{2} - 1\right) | P \left(k\right) .$

Now, $P \left(k\right) = \underline{{k}^{5} + 3 {k}^{4}} - \underline{5 {k}^{3} - 15 {k}^{2}} + \underline{4 k + 12}$

$= {k}^{4} \left(k + 3\right) - 5 {k}^{2} \left(k + 3\right) + 4 \left(k + 3\right)$

$= \left(k + 3\right) \left({k}^{4} - 5 {k}^{2} + 4\right)$

Now, knowing that $\left({k}^{2} - 1\right)$ is a factor, we have,

$P \left(k\right) = \left(k + 3\right) \left({k}^{4} - {k}^{2} - 4 {k}^{2} + 4\right)$

$= \left(k + 3\right) \left\{{k}^{2} \left({k}^{2} - 1\right) - 4 \left({k}^{2} - 1\right)\right\}$

$= \left(k + 3\right) \left({k}^{2} - 1\right) \left({k}^{2} - 4\right)$

$\therefore P \left(k\right) = \left(k + 3\right) \left(k + 1\right) \left(k - 1\right) \left(k + 2\right) \left(k - 2\right)$.

Enjoy Maths.!