Math, asked by riddschauhan06, 5 months ago

The number of zeroes that polynomial f(x) = (x – 2)^2 + 4 can have is:

(i)1

(ii) 2

(iii) 0

(iv) 3

answer is zero.
plz tell why​

Answers

Answered by Anonymous
3

Step-by-step explanation:

ii} 2 Will be the answer

2 - 4

= 2

Answered by bhowmik2018piu
2

Step-by-step explanation:

f(x) = (x – 2)^2 + 4

f(x)= x^2-4x+4+4

f(x)= x^2-4x+8

f(0)= 0^2-4*0+8

f(0)= 8

f(1)= 1^2-4*1+8

f(1)= 1-4+8

f(1)= 5

f(2)= 2^2-4*2+8

f(2)= 4-4+8

f(2)= 8

f(3)= 3^2-4*3+8

f(3)= 9-4+8

f(3)= 13

The number of zeroes that polynomial f(x) = (x – 2)^2 + 4 can have is:

iii) 0

because we put the value of x= 0,1,2,3,......

but the polynomial f(x) = (x – 2)^2 + 4 is not 0

so the zeros of that polynomial f(x) = (x – 2)^2 + 4 is 0 as we don't get the value zero for the polynomial at any value of x.

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