Question

Explain how to find the zeros of the given polynomial: x3+3x2–x−3

What are the zeros?​

Answer 1

Answer:

The zeroes of the function are x = -3, x = -1, and x = 1.30-Jan-2021

i hope this will be helpfull for youu

Answer 2

The given polynomial is

P(x)=x^3-3x^2-x+3P(x)=x

3

−3x

2

−x+3

We need to find the zeroes of the given polynomial.

Equate the polynomial equal to 0.

P(x)=0P(x)=0

x^3-3x^2-x+3=0x

3

−3x

2

−x+3=0

(x^3-3x^2)+(-x+3)=0(x

3

−3x

2

)+(−x+3)=0

Taking common factors from each parenthesis.

x^2(x-3)-(x-3)=0x

2

(x−3)−(x−3)=0

(x^2-1)(x-3)=0(x

2

−1)(x−3)=0

(x-1)(x+1)(x-3)=0(x−1)(x+1)(x−3)=0 \because a^2-b^2=(a-b)(a+b)∵a

2

−b

2

=(a−b)(a+b)

Using zero product property we get

x-1=0\Rightarrow x=1x−1=0⇒x=1

x+1=0\Righ tarrow x=-1x+1=0⇒x=−1

x-3=0\Right arrow x=3x−3=0⇒x=3

It is given that one of its zero is 1.

Therefore -

the remaining zeroes are -1 and 3.