Describe x^3 - 2 x^2 -14 x +40 as product of linear factors.

Hey there!
There are multiple methods to factor polynomials and get there linear factors. One method that will give you the real and rational roots is to use the rational root theorem. This states that all the possible rational roots of a polynomial can be found by taking the factors of constant term and dividing it by the factors of the leading coefficient.
In your example the constant is 40 and the leading coefficient is 1 , so all the possible rational roots are all the factors of 40/1 or just 40 .
+-1,+-2,+-4,+-5,+-8,+-10,+-20,+-40
we can test these roots by trial and error using synthetic division and find that x=-4
is a root. Now we can divide out x-(-4) or x+4.
Our polynomial now becomes:
(x+4)(x^2-6x+10).


We can now use the quadratic formula on the remaining polynomial, which will yield the two remaining roots of 3+-i.
giving us our completely factored polynomial: (x+4)(x-3-i)(x-3+i).

Hello!
To factor a polynomial it is useful to find its roots. If a polynomial P(x) has a root a, then it may be expressed as P(x) = (x - a)Q(x), where Q(x) is a polynomial of the degree one less than the degree of P(x).
Let's try to find an integer root of our P(x) = x^3 - 2x^2 - 14x + 40. It must be a divisor of the constant term, 40. Trial and error gives us a root x_1 = -4:
(-4)^3 - 2*(-4)^2 - 14*(-4) + 40 = -64 - 32 + 56 + 40 = 0.
Therefore we can divide P(x) by (x+4):
x^3 - 2x^2 - 14x + 40 =x^2(x+4) - 4x^2 - 2x^2 - 14x + 40 =x^2(x+4) - 6x^2 - 14x + 40 =
= x^2(x+4) - 6x(x+4) + 24x - 14x + 40 =x^2(x+4) - 6x(x+4) + 10x + 40 =
= x^2(x+4) - 6x(x+4) + 10(x + 4) = (x^2 - 6x + 10)(x + 4).
The polynomial Q(x) =x^2 - 6x + 10 has no real roots because its discriminant is negative: (-6)^2-4*10=-4lt0. Its complex roots are 3+-i. So we cannot make it a product of linear factors under real numbers,
x^3 - 2x^2 - 14x + 40 =(x^2 - 6x + 10)(x + 4).
Under complex numbers we have the answer,
x^3 - 2x^2 - 14x + 40 =(x-3-i)(x-3+i)(x + 4).