How do you factor #x^3-9x^2+25x-21#, given that #x=3# is a zero ?

1 Answer
Nov 2, 2016

Answer:

#x^3-9x^2+25x-21 = (x-3)(x-3-sqrt(2))(x-3+sqrt(2))#

Explanation:

#h(x) = x^3-9x^2+25x-21#

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this later with #a = (x-3)# and #b=sqrt(2)#

We are told that #3# is a zero, so #(x-3)# must be a factor:

#x^3-9x^2+25x-21 = (x-3)(x^2-6x+7)#

Then, completing the square we find:

#x^2-6x+7 = x^2-6x+9-2#

#color(white)(x^2-6x+7) = (x-3)^2-(sqrt(2))^2#

#color(white)(x^2-6x+7) = ((x-3)-sqrt(2))((x-3)+sqrt(2))#

#color(white)(x^2-6x+7) = (x-3-sqrt(2))(x-3+sqrt(2))#

Putting it all together:

#x^3-9x^2+25x-21 = (x-3)(x-3-sqrt(2))(x-3+sqrt(2))#