Question

The number of zeroes of the given polynomial, (x+1)(x+2)(x-3)² is
a) 2
b) 3
c) 4
d) 1


please give me A correct and step by step solution...​

Answer 1

Answer:

The number of zeros of the polynomial $(x+1)(x+2)(x-3)^2$ is 3.

Step-by-step explanation:

The zeros of a polynomial $p(x)$ are those values of $x$ for which $p(x)$ is zero.

Given the polynomial,

$p(x)=(x+1)(x+2)(x-3)^2$

Then$P(x)=0\implies (x+1)(x+2)(x-3)^2=0$

The product is zero if either of the factors is zero. That is

$(x+1)=0$ or $x+2=0$ or $(x-3)^2=0$

$\implies x=-1$ or $x=-2$ or $x=3$

Here $x=3$ is a repeated root. So the number of zeros of the given polynomial is 3.

Answer 2

Given : polynomial, (x+1)(x+2)(x-3)²

To Find : The number of zeroes of the given polynomial

Solution:

(x+1)(x+2)(x-3)²

Zeroes of a polynomial where graph of the polynomial intersects the x axis or touches the x axis

x + 1 = 0

=> x = - 1   is one of the zero

x + 2 = 0

=> x = - 2  is another zero

(x-3)² = 0

=> | x -  3|  = 0

=> x = 3    is one more zero   ( here graph of polynomial touches x axis and does not cross as multiplicity of zero is 2 ( even ) )

Total zeroes are    3

( - 1 , - 2 ,  3)

Correct option is  b) 3

The number of zeroes of the given polynomial (x+1)(x+2)(x-3)²  are 3

Learn More:

if 2 + root 3 and 2 minus root 3 are the two zeros of the polynomial 2 ...

brainly.in/question/8973942

If one of the zero is 7times the other find the other zeros 0 - Brainly.in

brainly.in/question/8158084