How do you factor x^3-x^2-4x+5x3−x2−4x+5?
1 Answer
Factoring by grouping does not work with this polynomial, but read on...
Explanation:
Factoring by grouping does not immediately help with this cubic polynomial. If it were
x^3-x^2-4x+4x3−x2−4x+4
= (x^3-x^2)-(4x-4)=(x3−x2)−(4x−4)
= x^2(x-1)-4(x-1)=x2(x−1)−4(x−1)
= (x^2-4)(x-1)=(x2−4)(x−1)
= (x-2)(x+2)(x-1)=(x−2)(x+2)(x−1)
Was there a typo in the question?
Can we factor
The only possible rational zeros are
This cubic turns out to have one Real zero and two Complex zeros.
It is possible, if a little lengthy, to find them using Cardano's method, which will give you:
Real zero:
x_1 = 1/3(1-root(3)((97-3sqrt(69))/2)-root(3)((97+3sqrt(69))/2))x1=13⎛⎝1−3√97−3√692−3√97+3√692⎞⎠
Complex zeros:
x_2 = 1/3(1-omega root(3)((97-3sqrt(69))/2)-omega^2 root(3)((97+3sqrt(69))/2))x2=13⎛⎝1−ω3√97−3√692−ω23√97+3√692⎞⎠
x_3 = 1/3(1-omega^2 root(3)((97-3sqrt(69))/2)-omega root(3)((97+3sqrt(69))/2))x3=13⎛⎝1−ω23√97−3√692−ω3√97+3√692⎞⎠
where
Then:
x^3-x^2-4x+5=(x-x_1)(x-x_2)(x-x_3)x3−x2−4x+5=(x−x1)(x−x2)(x−x3)