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

2 Answers
Sep 18, 2015

Answer:

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.

Sep 18, 2015

Answer:

#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)#