Write the function x³-x²-7x-5 as a product of its linear factors ?

Let f(x) = x³ - x² - 7x - 5

f(-1) = (-1)³ - (-1)² - 7(-1) - 5 = 0
Thus (x + 1) is a factor of f(x).

x³ - x² - 7x - 5
= (x³ + x²) + (-2x² - 2x) + (-5x - 5)
= x²(x + 1) - 2x(x + 1) - 5(x + 1)
= (x + 1)(x² - 2x - 5)

7 - (1 + 1 + 5) = 0
-7 + (1 + 1 + 5) = 0

So we can see that x = -1 or x = 1 is a root

(-1)^3 - (-1)^2 - 7 * (-1) - 5 = -1 - 1 + 7 - 5 = 7 - 7 = 0. x = -1
(1)^3 - 1^2 - 7 * 1 - 5 = 1 - 1 - 7 - 5 = -12. x = 1 is not a root

(x + 1) * (ax^2 + bx + c) = x^3 - x^2 - 7x - 5
ax^3 + bx^2 + cx + ax^2 + bx + c = x^3 - x^2 - 7x - 5

ax^3 = x^3
a = 1
bx^2 + ax^2 = -x^2
b + a = -1
b + 1 = -1
b = -2
cx + bx = -7x
c + b = -7
c - 2 = -7
c = -5

(x + 1) * (x^2 - 2x - 5)

x = (2 +/- sqrt(4 + 25)) / 2 = (2 +/- sqrt(29)) / 2.

(x + 1) * (x^2 - 2x - 5) is the most factored form.

x³ - x² - 7x - 5
= (x + 1)(x - 2)(x - 5)