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

##### 1 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

#### Explanation:

Let

By the rational root theorem, any rational roots of

So the only possible rational roots are:

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

Also since the signs of

#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^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

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