Question

factorise x³-3x²-9x-5​

Answer 1

Given :

• A polynomial x³-3x²-9x-5

To Find :

• Three factors of polynomial x³-3x²-9x-5

Solution :

Constant Term = -5

Factors of Constant Term = 1,5

Putting x = 1 :

$\longmapsto\tt{{(1)}^{3}-3{(1)}^{2}-9(1)-5}$

$\longmapsto\tt{1-3-9-5}$

$\longmapsto\tt{1-17}$

$\longmapsto\tt\bold{-16}$

Putting x = -1 :

$\longmapsto\tt{{(-1)}^{3}-3{(-1)}^{2}-9(-1)-5}$

$\longmapsto\tt{-1-3+9-5}$

$\longmapsto\tt{-4+9-5}$

$\longmapsto\tt{5-5}$

$\longmapsto\tt\bold{0}$

• x = -1 is the root .
• x + 1 is the factor.

(Refer to the attachment)

Now :

$\longmapsto\tt\bold{{x}^{2}-4x-5}$

$\longmapsto\tt{{x}^{2}-(5x-1x)-5}$

$\longmapsto\tt{{x}^{2}-5x+1x-5}$

$\longmapsto\tt{x(x-5)+1(x-5)}$

$\longmapsto\tt\bold{(x+1)(x-5)}$

So , The three factors are (x+1) , (x-5) , (x+1)...