Question

Factories x^3-3x^2-9x-5

Answer 1

Hey There,

Here we are given a polynomial:

$x^3-3x^2-9x-5$

Here, Sum of coefficients of terms with odd powers of x = 1-9 = -8
And Sum of coefficients of terms with even powers of x = -3-5 = -8

Both are same. This means that (x+1) is a factor.


So, we can factorise as follows:

$x^3-3x^2-9x-5 \\ \\ =x^3+x^2-4x^2-4x-5x-5 \\ \\ = x^2(x+1) -4x(x+1) -5(x+1) \\ \\ = (x+1)(x^2-4x-5) \\ \\ = (x+1)(x^2+x-5x-5) \\ \\ =(x+1)(x(x+1)-5(x+1)) \\ \\ =(x+1)(x+1)(x-5) \\ \\ = (x+1)^2(x-5)$


Hope it helps
Purva
Brainly Community

Answer 2

Step-by-step explanation:

Given Equation is x^3 - 3x^2 - 9x - 5

= > x^3 - 4x^2 + x^2 - 5x - 4x - 5

= > x^3 - 4x^2 - 5x + x^2 - 4x - 5

= > x(x - 4x - 5) + 1(x^2 - 4x - 5)

= > (x + 1)(x^2 - 4x - 5)

= > (x + 1)(x^2 + x - 5x - 5)

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

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

= > (x + 1)^2(x - 5).

Hope this helps!