How do you factor x3−4x2−11x+2=0?

1 Answer
George C.
Aug 18, 2016

x3−4x2−11x+2=(x+2)(x−3−2√2)(x−3+2√2)

Explanation:

f(x)=x3−4x2−11x+2

By the rational roots theorem, any rational zeros of f(x) must be expressible in the form pq for integers p,q with p a divisor of the constant term 2 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational zeros are:

±1,±2

Trying each in turn, we find:

f(−2)=−8−4(4)+11(2)+2=−8−16+22+2=0

So x=−2 is a zero and (x+2) a factor:

x3−4x2−11x+2=(x+2)(x2−6x+1)

Factor the remaining quadratic by completing the square and using the difference of squares identity:

a2−b2=(a−b)(a+b)

as follows:

x2−6x+1

=x2−6x+9−8

=(x−3)2−(2√2)2

=((x−3)−2√2)((x−3)+2√2)

=(x−3−2√2)(x−3+2√2)

Putting it all together:

x3−4x2−11x+2=(x+2)(x−3−2√2)(x−3+2√2)

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