How do you factor x^3-x^2-4x+5x3x24x+5?

1 Answer
May 20, 2016

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+4x3x24x+4 then it would be much easier:

x^3-x^2-4x+4x3x24x+4

= (x^3-x^2)-(4x-4)=(x3x2)(4x4)

= x^2(x-1)-4(x-1)=x2(x1)4(x1)

= (x^2-4)(x-1)=(x24)(x1)

= (x-2)(x+2)(x-1)=(x2)(x+2)(x1)

Was there a typo in the question?

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Can we factor f(x) = x^3-x^2-4x+5f(x)=x3x24x+5 ?

The only possible rational zeros are +-1±1, +-5±5, neither of which work.

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=1313973692397+3692

Complex zeros:

x_2 = 1/3(1-omega root(3)((97-3sqrt(69))/2)-omega^2 root(3)((97+3sqrt(69))/2))x2=131ω3973692ω2397+3692

x_3 = 1/3(1-omega^2 root(3)((97-3sqrt(69))/2)-omega root(3)((97+3sqrt(69))/2))x3=131ω23973692ω397+3692

where omega = -1/2+sqrt(3)/2iω=12+32i is the primitive Complex cube root of 11.

Then:

x^3-x^2-4x+5=(x-x_1)(x-x_2)(x-x_3)x3x24x+5=(xx1)(xx2)(xx3)