Question

is 1/3 a zero of p(x)=2x^3+3x^2-6x+7 explain

Answer 1

Answer:

No

Step-by-step explanation:

P(x)= 2x^3+3x^2-6x+7

If 1/3 is a zero of P(x), then, P(1/3)=0

P(1/3)= 2(1/3)^3+3(1/3)^2-6(1/3)+7

P(1/3)= 2/27+1/3-2+7

P(1/3)= 2/27+9/27-54/27+189/27

P(1/3)= 146/27

Therefore it is found out that P(1/3) is not equal to zero. So, 1/3 is not a zero of P(x).