# How do you write  f(x)=x^5-3x^4-4x^3+28x^2-37x+15 in factored form?

Sep 18, 2015

Use numerical methods as shown below.

#### Explanation:

Since there is no direct formula (that i am aware of) to find the roots of a 5th degree polynomial, I would suggest using numerical methods to solve this problem, of which there are many.

A good simple formula in numerical analysis which converges rather quickly, is Newton's Method of root finding.
It involves selecting an arbitrary starting value ${x}_{0}$ and then applying the algorithmic iterations which will converge to the nearest root value.

The algorithm to use is : x_n=x_(n-1)-f(x_(n-1))/(f'(x_(n-1)) ; $n \ge 1$

Then select another arbitrary value (a distance away from the first root found) and repeat Newton's algorithm and converge to a 2nd root.

Continuing in this way, you will eventually find all 5 roots (if they exist) and would have the full factors of the quintic polynomial.

I have performed the iterations with initial value 0 and after 2 iterations found convergence to root value approximately 0,657 and hence (x-0,657) is a factor.

You may continue in this fashion to find the other factors.

Sep 18, 2015

$f \left(x\right) = {x}^{5} - 3 {x}^{4} - 4 {x}^{3} + 28 {x}^{2} - 37 x + 15$

$= \left(x - 1\right) \left(x - 1\right) \left(x + 3\right) \left({x}^{2} - 4 x + 5\right)$

$= \left(x - 1\right) \left(x - 1\right) \left(x + 3\right) \left(x - 2 - i\right) \left(x - 2 + i\right)$

#### Explanation:

$f \left(x\right) = {x}^{5} - 3 {x}^{4} - 4 {x}^{3} + 28 {x}^{2} - 37 x + 15$

Trevor is correct that quintic polynomials are not solvable by simple formulae - though there are some spectacularly complex methods to find solutions in terms of a special kind of radical called a Bring radical. However, this is not a general quintic and this particular one is solvable.

By the rational root theorem, any rational roots of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $q > 0$, $p$ a divisor of the constant term $15$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots of $f \left(x\right) = 0$ are:

$\pm 1$, $\pm 3$, $\pm 5$, $\pm 15$

Let's try some:

$f \left(1\right) = 1 - 3 - 4 + 28 - 37 + 15 = 0$
$f \left(- 1\right) = - 1 - 3 + 4 + 28 + 37 + 15 = 80$
$f \left(3\right) = 243 - 243 - 108 + 252 - 111 + 15 = 148$
$f \left(- 3\right) = - 243 - 243 + 108 + 252 - 111 + 15 = 0$

So $\left(x - 1\right)$ and $\left(x + 3\right)$ are factors

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

Divide $f \left(x\right)$ by $\left({x}^{2} + 2 x - 3\right)$ to find:

${x}^{5} - 3 {x}^{4} - 4 {x}^{3} + 28 {x}^{2} - 37 x + 15$

$= \left({x}^{2} + 2 x - 3\right) \left({x}^{3} - 5 {x}^{2} + 9 x - 5\right)$

The remaining cubic factor is also zero when $x = 1$, since $1 - 5 + 9 - 5 = 0$.

So there's another factor of $\left(x - 1\right)$.

$\left({x}^{3} - 5 {x}^{2} + 9 x - 5\right) = \left(x - 1\right) \left({x}^{2} - 4 x + 5\right)$

The discriminant of ${x}^{2} - 4 x + 5$ is ${\left(- 4\right)}^{2} - \left(4 \times 1 \times 5\right) = 16 - 20 = - 4 < 0$.

So this quadratic factor has no simpler factors with Real coefficients.

We can write:

${x}^{2} - 4 x + 5 = {\left(x - 2\right)}^{2} + 1$, which has Complex zeros $x = 2 \pm i$

So

${x}^{2} - 4 x + 5 = \left(x - 2 - i\right) \left(x - 2 + i\right)$