How do you factor x^3 - 4x^2 - 11x + 2 = 0x3−4x2−11x+2=0?
1 Answer
Explanation:
f(x) = x^3-4x^2-11x+2f(x)=x3−4x2−11x+2
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
+-1, +-2±1,±2
Trying each in turn, we find:
f(-2) = -8-4(4)+11(2)+2 = -8-16+22+2 = 0f(−2)=−8−4(4)+11(2)+2=−8−16+22+2=0
So
x^3-4x^2-11x+2 = (x+2)(x^2-6x+1)x3−4x2−11x+2=(x+2)(x2−6x+1)
Factor the remaining quadratic by completing the square and using the difference of squares identity:
a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b)
as follows:
x^2-6x+1x2−6x+1
=x^2-6x+9-8=x2−6x+9−8
=(x-3)^2-(2sqrt(2))^2=(x−3)2−(2√2)2
=((x-3)-2sqrt(2))((x-3)+2sqrt(2))=((x−3)−2√2)((x−3)+2√2)
=(x-3-2sqrt(2))(x-3+2sqrt(2))=(x−3−2√2)(x−3+2√2)
Putting it all together:
x^3-4x^2-11x+2 = (x+2)(x-3-2sqrt(2))(x-3+2sqrt(2))x3−4x2−11x+2=(x+2)(x−3−2√2)(x−3+2√2)
