Question

find the zero of the polynomial in each of the following cases 1. f(×)=×+2, 2. f(×)=×-2, 3. f(×)=2×+3, 4. f(×)=2×-3,5. f(×)=×^2. friend plz tell this solution​

Answer 1

Answer:

this is ff 2555655656591926+5+2-2*/2-*5//5-565'@5-5*5@5

Answer 2

Answer:

1. -2

2. 2

3. -3/2

4. 3/2

5. 1/1^-1

Step-by-step explanation:

1. x+2=0

x=-2, so -2 is a zero of f(x)=x+2

2. x-2=0

x=2, so 2 is a zero of f(x)=x-2

3. 2x+3=0

x= -3/2, so -3/2 is a zero of f(x)=2x+3

4. 2x-3=0

x= 3/2, so 3/2 is a zero of f(x)= 2x-3

5. x= 1/1^-1

so, 1/1^-1 is a solution of f(x)=x^2