Question

how to 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 .

Answer 2

Factorise = x³ - 3x² - 9x - 5

Let, P(x) = x³ - 3x² - 9x - 5

By taking (x + 1) as a factor of the polynomial P(x) = x³ - 3x² - 9x - 5
=> (x + 1) = 0
=> x = (-1)

P(x) = x³ - 3x² - 9x - 5
P(-1) = (-1)³ - 3×(-1)² - 9×(-1) - 5
P(-1) = -1 - 3 + 9 - 5
P(-1) = 9 - 9
P(-1) = 0

So, P(-1) = 0

Now,

We shall now look for all factors of (-5). Some of these are : (1) , (-1) , (5) , (-5)

By taking (x - 1) as a factor of the polynomial P(x) = x³ - 3x² - 9x - 5
=> (x - 1) = 0
=> x = (1)

P(x) = x³ - 3x² - 9x - 5
P(1) = (1)³ - 3×(1)² - 9×(1) - 5
P(1) = 1 - 3 - 9 - 5
P(1) = -16

By trial, we found that P(1) = -16

Now,
divide (x³ - 3x² - 9x - 5) by (x+1) Long division method.

...... [ see in the attached image ].....

(x³ - 3x² - 9x - 5) ÷ (x+1) = (x² - 4x - 5)

Now, (x² - 4x - 5) is also a another factor of polynomial P(x) = (x³ - 3x² - 9x - 5)

Now,
factorising (x² - 4x - 5)
= x² + (1 - 5)x - 5
= x² + 1x - 5x - 5
= x(x+1) -5(x +1)
= (x+1)(x-5)

Hence,
(x³ - 3x² - 9x - 5) = (x+1)(x² - 4x - 5)

OR

(x³ - 3x² - 9x - 5) = (x+1)(x+1)(x-5)