Question

Factorise x^3-3x^2-9x-5​

Answer 1

X³ - 3x² - 9x - 5

= x³ + x² - 4x² - 4x - 5x - 5

= x²( x + 1) - 4x ( x + 1) - 5(x + 1)

= (x + 1)(x² - 4x - 5)

= (x + 1)(x² -5x + x - 5)

= (x + 1)(x - 5) (x + 1)

hence, (x +1), (x -5) and (x +1) are the factors of given polynomial .

OR

Let p(x) = x3 - 3x2 - 9x - 5

By trial, we find that

p(- 1) = (- 1)3 - 3(- 1)2- 9(- 1) -5

= - 1 - 3 + 9 - 5 = 0

∴ By Factor Theorem, x - (- 1), i.e., (x + 1) is a factor of p(x).

Now,

x3 - 3x2 - 9x - 5

= x2(x + 1) - 4x(x + 1) - 5(x + 1)

= (x + 1)(x2- 4x - 5)

= (x + 1)(x2 - 5x + x - 5)

= (x+ 1){x(x - 5) + 1 (x - 5)}

= (x + 1)(x - 5)(x + 1).

hope this helps you :)