Question

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

Answer 1

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>=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.

Please ask me if you still need further assistance.

Answer 2

`f(x) = x^5 - 3x^4-4x^3+28x^2-37x+15`

`= (x-1)(x-1)(x+3)(x^2-4x+5)`

`= (x-1)(x-1)(x+3)(x-2-i)(x-2+i)`

Explanation:

`f(x) = x^5 - 3x^4-4x^3+28x^2-37x+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(x)` must be expressible in the form `p/q` in lowest terms, where `p, q in ZZ`, `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(x) = 0` are:

`+-1`, `+-3`, `+-5`, `+-15`

Let's try some:

`f(1) = 1 - 3 - 4 + 28 - 37 + 15 = 0`
`f(-1) = -1-3+4 + 28+37+15 = 80`
`f(3) = 243 - 243 - 108+252-111+15 = 148`
`f(-3) = -243-243+108+252-111+15 = 0`

So `(x-1)` and `(x+3)` are factors

`(x-1)(x+3) = x^2+2x-3`

Divide `f(x)` by `(x^2+2x-3)` to find:

`x^5 - 3x^4-4x^3+28x^2-37x+15`

`= (x^2+2x-3)(x^3-5x^2+9x-5)`

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

So there's another factor of `(x-1)`.

`(x^3-5x^2+9x-5) = (x-1)(x^2-4x+5)`

The discriminant of `x^2-4x+5` is `(-4)^2 - (4xx1xx5) = 16-20 = -4 < 0`.

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

We can write:

`x^2-4x+5 = (x-2)^2+1`, which has Complex zeros `x = 2+-i`

So

`x^2-4x+5 = (x-2-i)(x-2+i)`