How do you solve the polynomial #x^4-4x^3+5x^2-4x+4=0#?

1 Answer
Dec 6, 2015

Answer:

Use the rational root theorem to help find one of the roots. Separate out the corresponding factor, then factor by grouping to find roots #2#, #2#, #i# and #-i#.

Explanation:

Let #f(x) = x^4-4x^3+5x^2-4x+4#

By the rational root theorem, any rational roots of #f(x) = 0# must be expressible in lowest terms as #p/q# for some integers #p#, #q#, where #p# is a divisor of the constant term #4# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational roots are:

#+-1#, #+-2#, #+-4#

Also since the signs of #f(x)# alternate for odd and even powers of #x#, #f(x) = 0# has no negative roots. So that only leaves the following possible rational roots:

#1#, #2#, #4#

Let's try each in turn:

#f(1) = 1-4+5-4+4 = 2#
#f(2) = 16-32+20-8+4 = 0#

So #x=2# is a root and #(x-2)# a factor.

#x^4-4x^3+5x^2-4x+4 = (x-2)(x^3-2x^2+x-2)#

The remaining cubic can be factored by grouping:

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

Putting it all together:

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

#= (x-2)^2(x-i)(x+i)#

So the #4# roots are #2#, #2#, #i# and #-i#

graph{x^4-4x^3+5x^2-4x+4 [-9.625, 10.375, -1.36, 8.64]}