Question

find integral zeroes of polynomial x³+3x²-x-3​

Answer 1

Answer:

The integral zeroes are : -1 , 1 , 3

Step-by-step explanation:

Given :

To find the integral zeroes of the polynomial :

$x^3 + 3x^2 - x - 3$

Solution :

$x^3 + 3x^2 - x - 3$

⇒ $x^2(x + 3) - 1(x + 3)$

⇒ $(x^2-1)(x+3)$

⇒ $(x^2+x - x-1)(x+3)$

⇒ $(x(x+1) - (x+1))(x+3)$

⇒$(x+1)(x-1)(x+3)$

If their product equals 0,.

Atleast one of the factors must be 0,

i.e.,

$(x+1)(x-1)(x+3)=0$

Either $(x+1)=0$ (or) $(x-1)=0$ (or) $(x+3)=0$

⇒ $x = -1 , 1 , 3$

∴ The integral zeroes are : -1 , 1 , 3

Answer 2

To find the integral zeroes of the polynomial :

x^3 + 3x^2 - x - 3x3+3x2−x−3

Solution :

x^3 + 3x^2 - x - 3x3+3x2−x−3

⇒ x^2(x + 3) - 1(x + 3)x2(x+3)−1(x+3)

⇒ (x^2-1)(x+3)(x2−1)(x+3)

⇒ (x^2+x - x-1)(x+3)(x2+x−x−1)(x+3)

⇒ (x(x+1) - (x+1))(x+3)(x(x+1)−(x+1))(x+3)

⇒(x+1)(x-1)(x+3)(x+1)(x−1)(x+3)

If their product equals 0,.

Atleast one of the factors must be 0,

i.e.,

(x+1)(x-1)(x+3)=0(x+1)(x−1)(x+3)=0

Either (x+1)=0(x+1)=0 (or) (x-1)=0(x−1)=0 (or) (x+3)=0(x+3)=0

⇒ x = -1 , 1 , 3x=−1,1,3

∴ The integral zeroes are : -1 , 1 , 3